Chapter 2 Exercise SolutionsSeveral exercises in this chapter differ from those in the 4th edition. An “*” following the exercise number indicates that the description has changed (e.g., new values). A second exercise number in parentheses indicates that the exercise number has changed. For example, “2-16* (2-9)” means that exercise 2-16 was 2-9 in the 4th edition, and that the description also differs from the 4th edition (in this case, asking for a time series plot instead of a digidot plot). New exercises are denoted with an “☺”. 2-1*. (a) x = ∑ xi n = (16.05 + 16.03 + n i =1 (b) n ( ) n ∑ x − ∑ xi i =1 s= 2 i + 16.07 ) 12 = 16.029 oz 2 n i =1 n −1 = (16.052 + + 16.07 2 ) − (16.05 + 12 − 1 + 16.07) 2 12 = 0.0202 oz MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex2-1 Variable Ex2-1 Variable Ex2-1 N N* Mean 12 0 16.029 Maximum 16.070 SE Mean 0.00583 StDev 0.0202 Minimum 16.000 Q1 16.013 Median 16.025 Q3 16.048 2-2. (a) x = ∑ xi n = ( 50.001 + 49.998 + n i =1 (b) n s= ( ) n 2 ∑ xi − ∑ xi i =1 i =1 n −1 + 50.004 ) 8 = 50.002 mm 2 n = (50.0012 + + 50.0042 ) − (50.001 + 8 −1 + 50.004) 2 8 = 0.003 mm MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex2-2 Variable Ex2-2 Variable Ex2-2 N N* Mean 8 0 50.002 Maximum 50.006 SE Mean 0.00122 StDev 0.00344 Minimum 49.996 Q1 49.999 Median 50.003 Q3 50.005 2-1 Chapter 2 Exercise Solutions 2-3. (a) x = ∑ xi n = ( 953 + 955 + n i =1 (b) n s= ( ) n ∑ x − ∑ xi i =1 2 i i =1 n −1 + 959 ) 9 = 952.9 °F 2 n = (9532 + + 9592 ) − (953 + 9 −1 + 959) 2 9 = 3.7 °F MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex2-3 Variable Ex2-3 Variable Ex2-3 N N* Mean 9 0 952.89 Maximum 959.00 SE Mean 1.24 StDev 3.72 Minimum 948.00 Q1 949.50 Median 953.00 Q3 956.00 2-4. (a) In ranked order, the data are {948, 949, 950, 951, 953, 954, 955, 957, 959}. The sample median is the middle value. (b) Since the median is the value dividing the ranked sample observations in half, it remains the same regardless of the size of the largest measurement. 2-5. MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex2-5 Variable Ex2-5 Variable Ex2-5 N N* Mean 8 0 121.25 Maximum 156.00 SE Mean 8.00 StDev 22.63 Minimum 96.00 Q1 102.50 Median 117.00 Q3 144.50 2-2 Chapter 2 Exercise Solutions 2-6. (a), (d) MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex2-6 Variable Ex2-6 Variable Ex2-6 N N* Mean 40 0 129.98 Maximum 160.00 SE Mean 1.41 StDev 8.91 Minimum 118.00 Q1 124.00 Median 128.00 Q3 135.25 (b) Use √n = √40 ≅ 7 bins MTB > Graph > Histogram > Simple Histogram of Time to Failure (Ex2-6) 20 Frequency 15 10 5 0 112 120 128 136 Hours 144 152 160 (c) MTB > Graph > Stem-and-Leaf Stem-and-Leaf Display: Ex2-6 Stem-and-leaf of Ex2-6 Leaf Unit = 1.0 2 11 89 5 12 011 8 12 233 17 12 444455555 19 12 67 (5) 12 88999 16 13 0111 12 13 33 10 13 10 13 677 7 13 7 14 001 4 14 22 HI 151, 160 N = 40 2-3 Chapter 2 Exercise Solutions 2-7. Use n = 90 ≅ 9 bins MTB > Graph > Histogram > Simple Histogram of Process Yield (Ex2-7) 18 16 14 Frequency 12 10 8 6 4 2 0 84 88 92 96 Yield 2-4 Chapter 2 Exercise Solutions 2-8. (a) Stem-and-Leaf Plot 2 12o|68 6 13*|3134 12 13o|776978 28 14*|3133101332423404 (15) 14o|585669589889695 37 15*|3324223422112232 21 15o|568987666 12 16*|144011 6 16o|85996 1 17*|0 Stem Freq|Leaf (b) Use n = 80 ≅ 9 bins MTB > Graph > Histogram > Simple Histogram of Viscosity Data (Ex 2-8) 20 Frequency 15 10 5 0 13 14 15 Viscosity 16 17 Note that the histogram has 10 bins. The number of bins can be changed by editing the X scale. However, if 9 bins are specified, MINITAB generates an 8-bin histogram. Constructing a 9-bin histogram requires manual specification of the bin cut points. Recall that this formula is an approximation, and therefore either 8 or 10 bins should suffice for assessing the distribution of the data. 2-5 Chapter 2 Exercise Solutions 2-8(c) continued MTB > %hbins 12.5 17 .5 c7 Row 1 2 3 4 5 6 7 8 9 10 11 Intervals 12.25 to 12.75 12.75 to 13.25 13.25 to 13.75 13.75 to 14.25 14.25 to 14.75 14.75 to 15.25 15.25 to 15.75 15.75 to 16.25 16.25 to 16.75 16.75 to 17.25 Totals Frequencies 1 2 7 9 16 18 12 7 4 4 80 Percents 1.25 2.50 8.75 11.25 20.00 22.50 15.00 8.75 5.00 5.00 100.00 (d) MTB > Graph > Stem-and-Leaf Stem-and-Leaf Display: Ex2-8 Stem-and-leaf of Ex2-8 N = 80 Leaf Unit = 0.10 2 12 68 6 13 1334 12 13 677789 28 14 0011122333333444 (15) 14 555566688889999 37 15 1122222222333344 21 15 566667889 12 16 011144 6 16 56899 1 17 0 median observation rank is (0.5)(80) + 0.5 = 40.5 x0.50 = (14.9 + 14.9)/2 = 14.9 Q1 observation rank is (0.25)(80) + 0.5 = 20.5 Q1 = (14.3 + 14.3)/2 = 14.3 Q3 observation rank is (0.75)(80) + 0.5 = 60.5 Q3 = (15.6 + 15.5)/2 = 15.55 (d) 10th percentile observation rank = (0.10)(80) + 0.5 = 8.5 x0.10 = (13.7 + 13.7)/2 = 13.7 90th percentile observation rank is (0.90)(80) + 0.5 = 72.5 x0.90 = (16.4 + 16.1)/2 = 16.25 2-6 Chapter 2 Exercise Solutions 2-9 ☺. MTB > Graph > Probability Plot > Single Probability Plot of Liquid Detergent (Ex2-1) Normal 99 Mean StDev N AD P-Value 95 90 16.03 0.02021 12 0.297 0.532 Percent 80 70 60 50 40 30 20 10 5 1 15.98 16.00 16.02 16.04 Fluid Ounces 16.06 16.08 When plotted on a normal probability plot, the data points tend to fall along a straight line, indicating that a normal distribution adequately describes the volume of detergent. 2-10 ☺. MTB > Graph > Probability Plot > Single Probability Plot of Furnace Temperatures (Ex2-3) Normal 99 Mean StDev N AD P-Value 95 90 952.9 3.723 9 0.166 0.908 Percent 80 70 60 50 40 30 20 10 5 1 945.0 947.5 950.0 952.5 955.0 957.5 Temperature (deg F) 960.0 962.5 When plotted on a normal probability plot, the data points tend to fall along a straight line, indicating that a normal distribution adequately describes the furnace temperatures. 2-7 Chapter 2 Exercise Solutions 2-11 ☺. MTB > Graph > Probability Plot > Single Probability Plot of Failure Times (Ex2-6) Normal 99 Mean StDev N AD P-Value 95 90 130.0 8.914 40 1.259 <0.005 Percent 80 70 60 50 40 30 20 10 5 1 110 120 130 140 150 160 Hours When plotted on a normal probability plot, the data points do not fall along a straight line, indicating that the normal distribution does not reasonably describe the failure times. 2-12 ☺. MTB > Graph > Probability Plot > Single Probability Plot of Process Yield Data (Ex2-7) Normal 99.9 Mean StDev N AD P-Value 99 95 Percent 90 89.48 4.158 90 0.956 0.015 80 70 60 50 40 30 20 10 5 1 0.1 80 85 90 Yield 95 100 105 When plotted on a normal probability plot, the data points do not fall along a straight line, indicating that the normal distribution does not reasonably describe process yield. 2-8 Chapter 2 Exercise Solutions 2-13 ☺. MTB > Graph > Probability Plot > Single (In the dialog box, select Distribution to choose the distributions) Probability Plot of Viscosity Data (Ex2-8) Normal 99.9 Mean StDev N AD P-Value 99 Percent 95 90 14.90 0.9804 80 0.249 0.740 80 70 60 50 40 30 20 10 5 1 0.1 12 13 14 15 Viscosity 16 17 18 Probability Plot of Viscosity Data (Ex2-8) Lognormal 99.9 Loc Scale N AD P-Value 99 Percent 95 90 2.699 0.06595 80 0.216 0.841 80 70 60 50 40 30 20 10 5 1 0.1 12 13 14 15 Viscosity 16 17 18 19 2-9 10 15.1 10 11 12 13 14 Viscosity 15 16 17 18 Both the normal and lognormal distributions appear to be reasonable models for the data. with no bends or curves. 2-10 . the plot points on the Weibull probability plot are not straight—particularly in the tails— indicating it is not a reasonable model.36 80 1.032 <0.010 10 5 3 2 1 0.Chapter 2 Exercise Solutions 2-13 continued Probability Plot of Viscosity Data (Ex2-8) Weibull Percent 99.9 99 Shape Scale N AD P-Value 90 80 70 60 50 40 30 20 16. the plot points tend to fall along a straight line. However. Chapter 2 Exercise Solutions 2-14 ☺. MTB > Graph > Probability Plot > Single (In the dialog box.137 Percent 80 70 60 50 40 30 20 10 5 1 -5000 0 5000 10000 15000 Cycles to Failure 20000 25000 Probability Plot of Cycles to Failure (Ex2-14) Lognormal 99 Loc Scale N AD P-Value 95 90 8.8537 20 0.521 0.549 0.776 0. select Distribution to choose the distributions) Probability Plot of Cycles to Failure (Ex2-14) Normal 99 Mean StDev N AD P-Value 95 90 8700 6157 20 0.163 Percent 80 70 60 50 40 30 20 10 5 1 1000 10000 Cycles to Failure 100000 2-11 . Chapter 2 Exercise Solutions 2-14 continued Probability Plot of Cycles to Failure (Ex2-14) Weibull 99 Shape Scale N AD P-Value Percent 90 80 70 60 50 40 1.336 >0. though the Weibull distribution appears to best fit the data in the tails. 2-12 .250 30 20 10 5 3 2 1 1000 10000 Cycles to Failure Plotted points do not tend to fall on a straight line on any of the probability plots.464 9624 20 0. 201 0.651 40 0.56 40 8. ppm 100 150 Probability Plot of Concentration (Ex2-15) Lognormal 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 0.1 1.470 22.Chapter 2 Exercise Solutions 2-15 ☺. ppm 100.005 Loc Scale N AD P-Value 0. MTB > Graph > Probability Plot > Single (In the dialog box.0 Concentration.426 <0.873 Percent 80 70 60 50 40 30 20 10 5 1 -50 0 50 Concentration.0 2-13 . select Distribution to choose the distributions) Probability Plot of Concentration (Ex2-15) Normal 99 95 90 Mean StDev N AD P-Value 9.9347 1.0 10. ppm 100.001 0.6132 5.Chapter 2 Exercise Solutions 2-15 continued Probability Plot of Concentration (Ex2-15) Weibull 99 Shape Scale N AD P-Value Percent 90 80 70 60 50 40 0.000 Concentration. Plotted points on the normal and Weibull probability plots tend to fall off a straight line.100 1.000 The lognormal distribution appears to be a reasonable model for the concentration data.637 0.000 10.010 0. 2-14 .782 40 0.091 30 20 10 5 3 2 1 0. there are no trends. MTB > Graph > Time Series Plot > Single (or Stat > Time Series > Time Series Plot) Time Series Plot of Yield Data (Ex2-7) 100 Ex2-7 95 90 85 1 9 18 27 36 45 54 63 Time Order of Collection 72 81 90 Time may be an important source of variability. shifts or obvious patterns in the data. 2-15 .Chapter 2 Exercise Solutions 2-16* (2-9). MTB > Graph > Time Series Plot > Single (or Stat > Time Series > Time Series Plot) Time Series Plot of Viscosity Data (Ex2-8) 17 Ex2-8 16 15 14 13 12 1 8 16 24 32 40 48 56 Time Order of Collection 64 72 80 From visual examination. as evidenced by potentially cyclic behavior. 2-17* (2-10). indicating that time is not an important source of variability. 438 StDev 4. The order in which the data were collected may be an important source of variability. 2-19 (2-11).250 Q3 93.476 Maximum 98. there are two unusually large readings (9. MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex2-7 Variable Ex2-7 Variable Ex2-7 N N* Mean 90 0 89.125 2-16 .000 SE Mean 0.600 Q1 86.158 Minimum 82. MTB > Graph > Time Series Plot > Single (or Stat > Time Series > Time Series Plot) Time Series Plot of Concentration Data (Ex2-15) 140 120 100 Ex2-15 80 60 40 20 0 4 8 12 16 20 24 28 Time Order of Collection 32 36 40 Although most of the readings are between 0 and 20. as well as occasional “spikes” around 20.100 Median 89. 35).Chapter 2 Exercise Solutions 2-18 ☺. for example. 6.07 16.02 16.Chapter 2 Exercise Solutions 2-20 (2-12). 4.04 16. those ending in 1. 2-21 (2-13). The stem-and-leaf plot suggests that certain values may occur more frequently than others.01 16. The data appear to be fairly well scattered.06 Fluid Ounces 16.05 16. MTB > Graph > Boxplot > Simple Boxplot of Detergent Data (Ex2-1) 16.10 2 82 69 6 83 0167 14 84 01112569 20 85 011144 30 86 1114444667 38 87 33335667 43 88 22368 (6) 89 114667 41 90 0011345666 31 91 1247 27 92 144 24 93 11227 19 94 11133467 11 95 1236 7 96 1348 3 97 38 1 98 0 N = 90 Neither the stem-and-leaf plot nor the frequency histogram reveals much about an underlying distribution or a central tendency in the data.00 2-17 . and 7. MTB > Graph > Stem-and-Leaf Stem-and-Leaf Display: Ex2-7 Stem-and-leaf of Ex2-7 Leaf Unit = 0.03 16. x = 9 3 / 36.0000 49.Chapter 2 Exercise Solutions 2-22 (2-14). x = ∑ xi p( xi ) = 2 (1 36 ) + 3 ( 2 36 ) + 11 i =1 n ∑ xi p( xi ) − ⎡⎢ ∑ xi p ( xi ) ⎤⎥ ⎣ i =1 ⎦ S = i =1 n −1 n + 12 (1 36 ) = 7 2 n = 5. x = 3 3 / 36. 2} + Pr{2. 10. x = 8 4 / 36. 12} Pr{x = 2} = Pr{1. 8.1} = 1 × 1 + 1 × 1 + 1 × 1 = 3 6 6 6 6 6 6 36 . 7. MTB > Graph > Boxplot > Simple Boxplot of Bearing Bore Diameters (Ex2-2) 50.1} = 1 × 1 = 1 6 6 36 Pr{x = 3} = Pr{1.3} + Pr{2. x = 6 6 / 36. x = 11 1/ 36. 3. x = 4 4 / 36. x = 12 0. x = 10 2 / 36.0025 50. 4. 2} + Pr{3. x = 2 2 / 36.9975 49. x: {the sum of two up dice faces} sample space: {2. ( ) ( ( ) ) ( ) ( ) ⎧1/ 36.0075 50.1} = 1 × 1 + 1 × 1 = 2 6 6 6 6 36 Pr{x = 4} = Pr{1. 11.92 − 7 2 11 = 0.. 6. 9. 5.38 10 2-18 . x = 7 p ( x) = ⎨ ⎩5 / 36.9950 2-23 (2-15). otherwise 2-24 (2-16).. x = 5 5 / 36.0050 mm 50. µ = 1/λ = 1 (Eqn.01 (0.5 + 5k ) / 3. This is a Poisson distribution with parameter λ = 0. 2-26 (2-18). ∫ f ( x)dx must equal unity. x = 3 0. x = 2 p ( x) = ⎨ ⎩(0. 2-33) 2-27 (2-19). x ~ POI(0.05 2-19 .02 (0.02).0196 1! (b) Pr{x ≥ 1} = 1 − Pr{x = 0} = 1 − p(0) = 1 − e −0. from 0. ⎧(1 + 3k ) / 3.01)0 = 1 − 0.02)1 = 0. x ~ POI(0. 2-32) σ2 = 1/λ2 = 1 (Eqn.02. (a) Pr{x = 1} = p(1) = e −0.0198 to 0. e −0.0198 0! (c) This is a Poisson distribution with parameter λ = 0.02 (0.5 + 5k ) =1 3 10k = 0.01).Chapter 2 Exercise Solutions 2-25 (2-17).5 k = 0.02)0 = 1 − 0.0100 Pr{x ≥ 1} = 1 − Pr{x = 0} = 1 − p(0) = 1 − 0! Cutting the rate at which defects occur reduces the probability of one or more defects by approximately one-half.0100. otherwise (a) ∞ To solve for k. use F ( x) = ∑ p( xi ) = 1 i =1 (1 + 3k ) + (1 + 2k ) + (0.9900 = 0. x = 1 (1 + 2k ) / 3. −∞ ∞ −x −x ∞ ∫ ke dx = [−ke ]0 = − k[0 − 1] = k ⇒ 1 0 This is an exponential distribution with parameter λ=1. +∞ For f(x) to be a probability distribution.01.9802 = 0. … ∞ F ( x) = ∑ kr x = 1 by definition i =0 k ⎡⎣1 (1 − r ) ⎤⎦ = 1 k = 1− r 2-29 (2-21). x = 3 ⎪ 3 ⎩ 2-28 (2-20).750.118 Approximately 11.75 = 1. The effect of warranty replacements is to decrease profit by $5.05) ⎤ ⎡ 0. (b) Mfg.5 + 5(0.1. x = 1 ⎪ 3 ⎪ 1.Chapter 2 Exercise Solutions 2-27 continued (b) ⎡1 + 3(0.05) ⎤ ⎡1 + 2(0. x = 0.15 + 1.8% will fail during the first year. 2.367) + 32 (0.383) + 22 (0.15 + 1.1 + 0.125(1) = 0.615 i =1 (c) 1.15 ⎧ = 0.10/calculator.000. p ( x) = kr x . 2-20 .867 3 3 3 ⎣ ⎦ ⎣ ⎦ ⎣ 3 µ = ∑ xi p( xi ) = 1× ⎢ i =1 3 σ 2 = ∑ xi2 p( xi ) − µ 2 = 12 (0. (a) This is an exponential distribution with parameter λ = 0.90/calculator.250) − 1.1 ⎪ F ( x) = ⎨ = 0.867 2 = 0. x = 2 3 ⎪ ⎪1.05) ⎤ + 2× ⎢ + 3× ⎢ ⎥ ⎥ ⎥⎦ = 1. 0 < r < 1.383.125: Pr{x ≤ 1} = F (1) = 1 − e −0.118) + 75]/calculator = $19. cost = $50/calculator Sale profit = $25/calculator Net profit = $[-50(1 + 0. 875 − 11.02)0 (1 − 0. 4 ⎛ 50 ⎞ Pr{ pˆ ≤ 0.75 = 0.02) x (1 − 0. 12 Pr{x < 12} = F (12) = ∫ −∞ 12 f ( x) dx = ∫ 4( x − 11.02 and n = 50.921 ⎝4⎠ 2-21 .02) 4 (1 − 0. n. x ~ BIN(25. and the process will be stopped to look for a cause.75) dx = 11.01 and n = 25.75 4 x2 2 12 − 47 x 11.64 ⎝0⎠ 2-33* (2-25).75 2-31* (2-23).04) Stop process if x ≥ 1.75 = 11.36 = 0. This is a binomial distribution with parameter p = 0.01)0 (1 − 0.02)(50− x ) x =0 ⎝ x ⎠ ⎛ 50 ⎞ ⎛ 50 ⎞ = ⎜ ⎟ (0. The process is stopped if x ≥ 1.04} = Pr{x ≤ 2} = ∑ ⎜ ⎟(0.777821 2-32* (2-24).Chapter 2 Exercise Solutions 2-30 (2-22).22 ⎝0⎠ This decision rule means that 22% of the samples will have one or more nonconforming units.78 = 0. ⎛ 25 ⎞ Pr{x ≥ 1} = 1 − Pr{x < 1} = 1 − Pr{x = 0} = 1 − ⎜ ⎟ (0. This is a binomial distribution with parameter p = 0. This is a somewhat difficult operating situation.01 x P( X <= x ) 0 0. TRUE) (2) MTB > Calc > Probability Distributions > Binomial Cumulative Distribution Function Binomial with n = 25 and p = 0.01) 25 = 1 − 0. 0.02) 49 + ⎝0⎠ ⎝1⎠ ⎛ 50 ⎞ + ⎜ ⎟ (0. p. This exercise may also be solved using Excel or MINITAB: (1) Excel Function BINOMDIST(x.02)50 + ⎜ ⎟ (0. ⎛ 25 ⎞ Pr{x ≥ 1} = 1 − Pr{x < 1} = 1 − Pr{x = 0} = 1 − ⎜ ⎟ (0.02)1 (1 − 0.04)0 (1 − 0.125 12 11.02) 46 = 0.04) 25 = 1 − 0. 992] = 0.0100 Pr{ pˆ > kσ + p} = 1 − Pr{ pˆ ≤ kσ + p} = 1 − Pr{x ≤ n(kσ + p)} k=1 1 − Pr{x ≤ n(kσ + p )} = 1 − Pr{x ≤ 100(1(0.003 2-22 .01)} = 1 − Pr{x ≤ 2} 2 ⎛ 100 ⎞ 100 − x x = 1− ∑ ⎜ ⎟(0.982 + ⎜ (0.921 + ⎜ (0.01) (0.99)97 ⎥ = 1− ∑ ⎜ ⎟ ⎟ x =0 ⎝ x ⎠ ⎝ 3 ⎠ ⎣ ⎦ = 1 − [0.01) (0.01) x (0.99)100− x = 1 − ⎢0.01(1 − 0.Chapter 2 Exercise Solutions 2-34* (2-26).0100) + 0.99) ⎥ ⎝ 1 ⎠ ⎝ 2 ⎠ ⎣⎝ 0 ⎠ ⎦ = 1 − [0.01 and n = 100. This is a binomial distribution with parameter p = 0.018 k=3 1 − Pr{x ≤ n(kσ + p )} = 1 − Pr{x ≤ 100(3(0. σ = 0.01) x =0 ⎝ x ⎠ ⎡⎛100 ⎞ ⎤ ⎛ 100 ⎞ ⎛100 ⎞ 0 100 1 99 2 98 = 1 − ⎢⎜ ⎟ (0.99) + ⎜ ⎟ (0.99) + ⎜ ⎟ (0.01) x (0.0100) + 0.079 k=2 1 − Pr{x ≤ n(kσ + p )} = 1 − Pr{x ≤ 100(2(0.99)100− x = 1 − ⎢0.01) (1 − 0.01) (0.01) 4 (0.01)} = 1 − Pr{x ≤ 4} 4 ⎛ 100 ⎞ ⎡ ⎤ ⎛100 ⎞ (0.921] = 0.01) 100 = 0.0100) + 0.99)96 ⎥ = 1− ∑ ⎜ ⎟ ⎟ x =0 ⎝ x ⎠ ⎝ 4 ⎠ ⎣ ⎦ = 1 − [0.01)3 (0.982] = 0.01)} = 1 − Pr{x ≤ 3} 3 ⎛ 100 ⎞ ⎡ ⎤ ⎛100 ⎞ (0. 659 ⎝0⎠ This approximation.033 ≤ 0. so the binomial approximation would be a satisfactory approximation the hypergeometric in this case. D.Chapter 2 Exercise Solutions 2-35* (2-27). ⎛5⎞ Pr{acceptance} = p (0) = ⎜ ⎟ (0. The binomial approximation is not satisfactory in this case. M = 2. p = D/N = 2/25 = 0.08)0 (1 − 0.1. This is a hypergeometric distribution with N = 25 and n = 5. n.1. n/N = 5/150 = 0.20 be less than the suggested 0.08 and n = 5. and n = 5 x P( X <= x ) 0 0. violates the rule-ofthumb that n/N = 5/25 = 0. N) (2) MTB > Calc > Probability Distributions > Hypergeometric Cumulative Distribution Function Hypergeometric with N = 25. 649) Pr{Acceptance} = p (0) = ⎝ ⎠ ⎝ = = 0.633333 (b) For the binomial approximation to the hypergeometric.130) ⎛ 25 ⎞ ⎜ ⎟ ⎝5⎠ This exercise may also be solved using Excel or MINITAB: (1) Excel Function HYPGEOMDIST(x. (c) For N = 150. without replacement. (a) Given D = 2 and x = 0: ⎛ 2 ⎞ ⎛ 25 − 2 ⎞ ⎜ ⎟⎜ ⎟ 0 5 − 0 ⎠ (1)(33. though close to the exact solution for x = 0.08)5 = 0.633 (53. 2-23 . 057 p(0) = ⎝ ⎠ ⎝ ⎠ = (3. 457.95. 760) ⎛ 25 ⎞ ⎜ ⎟ ⎝ 10 ⎠ try n = 11 ⎛ 5 ⎞ ⎛ 20 ⎞ ⎜ ⎟⎜ ⎟ 0 11 (1)(167. ⎛ 3 ⎞⎛ 30 − 3 ⎞ ⎜ ⎟⎜ ⎟ 1 ⎠⎝ 5 − 1 ⎠ (3)(17.038 p(0) = ⎝ ⎠ ⎝ ⎠ = (4.550) ⎝ Pr{x = 1} = p (1) = = = 0. ⎛ 5 ⎞ ⎛ 25 − 5 ⎞ ⎛ 5 ⎞ ⎛ 20 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ 0⎠⎝ n − 0 ⎠ ⎝ 0⎠⎝ n ⎠ ⎝ = p(0) = ⎛ 25 ⎞ ⎛ 25 ⎞ ⎜ ⎟ ⎜ ⎟ ⎝n⎠ ⎝n⎠ try n = 10 ⎛ 5 ⎞ ⎛ 20 ⎞ ⎜ ⎟⎜ ⎟ 0 10 (1)(184.369 (142. 268. or equivalently Pr{x = 0 | D = 5} < 0. This is a hypergeometric distribution with N = 30. 756) = 0.506) ⎛ 30 ⎞ ⎜ ⎟ ⎝5⎠ ⎛ 3 ⎞ ⎛ 27 ⎞ ⎜ ⎟⎜ ⎟ 0 5 Pr{x ≥ 1} = 1 − Pr{x = 0} = 1 − p(0) = 1 − ⎝ ⎠ ⎝ ⎠ = 1 − 0.567 = 0.433 ⎛ 30 ⎞ ⎜ ⎟ ⎝5⎠ 2-24 . 400) ⎛ 25 ⎞ ⎜ ⎟ ⎝ 11 ⎠ Let sample size n = 11.Chapter 2 Exercise Solutions 2-35 continued (d) Find n to satisfy Pr{x ≥ 1 | D ≥ 5} ≥ 0. and D = 3.960) = 0. 2-36 (2-28). n = 5.05. 00001)0 = 1 − 0. e −0. This is a Poisson distribution with λ = 0.01)1 = 0. This is a hypergeometric distribution with N = 500 pages. TRUE) (2) MTB > Calc > Probability Distributions > Poisson Cumulative Distribution Function Poisson with mean = 0. This is a Poisson distribution with λ = 0.364 − 0.0099 Pr{x = 1} = p (1) = 1 2-25 .00001 stones/bottle.364) = 0.1 defects/unit.1)0 = 1 − 0.1.904837 2-39 (2-31). x.020)0 (1 − 0.264 ⎝1⎠ 2-38 (2-30). ⎛ 50 ⎞ Pr{x = 0} = p(0) = ⎜ ⎟ (0.020)50−0 = (1)(1)(0. the binomial distribution can be used to approximate the hypergeometric.01 (0. Checking n/N = 50/500 = 0.020.00001 (0.905 = 0.99999 = 0.1 (0. and D = 10 errors.1 x P( X <= x ) 0 0. This is a Poisson distribution with λ = 0.372 = 0. with p = D/N = 10/500 = 0.020)50−1 = 1 − 0. Pr{x ≥ 1} = 1 − Pr{x = 0} = 1 − e −0. e −0.095 Pr{x ≥ 1} = 1 − Pr{x = 0} = 1 − p(0) = 1 − 0! This exercise may also be solved using Excel or MINITAB: (1) Excel Function POISSON(λ.020)1 (1 − 0.00001 0! 2-40 (2-32).Chapter 2 Exercise Solutions 2-37 (2-29).364 − ⎜ ⎟ (0.364 ⎝0⎠ Pr{x ≥ 2} = 1 − Pr{x ≤ 1} = 1 − [Pr{x = 0} + Pr{x = 1}] = 1 − p(0) − p(1) ⎛ 50 ⎞ = 1 − 0.1 ≤ 0.01 errors/bill. n = 50 pages. 000001)(0.010 0.… ∞ d ∞ 1 µ = ∑ t ⎡⎣ p(1 − p)t −1 ⎤⎦ = p ⎡ ∑ q t ⎤ = ⎢ ⎥ t =1 dq ⎣ t =1 ⎦ p 2-42 (2-34).158655 How many exceed 48 lb.0110. LSL = 35 lb. and x = 1 + (5000/100) = 51.055 So. x ~ N (40.000) × (0.6) = 1 − 0.6} 5 ⎭ ⎩ = 1 − Φ (1. n = 50. σ.000) × (0.9949 ⎜ ⎟ 0. 2. is (50.000 How many fail the minimum specification. ⎛ 51 − 1⎞ 3 51−3 Pr{x = 51} = p(51) = ⎜ = (1225)(0.159) = 7950.01. t = 1. µ.945 = 0.0008 ⎟ (0.3.? 48 − 40 ⎫ ⎧ Pr{x > 48} = 1 − Pr{x ≤ 48} = 1 − Pr ⎨ z ≤ ⎬ = 1 − Pr{z ≤ 1.9862 ⎝0⎠ ⎝1⎠ ⎝2⎠ 2-43* (2-35). Pr(t ) = p(1 − p)t −1 .159 5 ⎭ ⎩ So. r = 3 welds.01) 3 1 − ⎝ ⎠ Pr{x > 51} = Pr{r = 0} + Pr{r = 1} + Pr{r = 2} ⎛ 50 ⎞ ⎛ 50 ⎞ ⎛ 50 ⎞ = ⎜ ⎟ 0. 52). the number that fail the minimum specification are (50.012 0.9950 + ⎜ ⎟ 0.Chapter 2 Exercise Solutions 2-41 (2-33). This is a Pascal distribution with Pr{defective weld} = p = 0. the number that exceed 48 lb.617290) = 0. 2-26 .01) (1 − 0.055) = 2750. TRUE) (2) MTB > Calc > Probability Distributions > Normal Cumulative Distribution Function Normal with mean = 40 and standard deviation = 5 x P( X <= x ) 35 0.9948 = 0. This exercise may also be solved using Excel or MINITAB: (1) Excel Function NORMDIST(X.? 35 − 40 ⎫ ⎧ Pr{x ≤ 35} = Pr ⎨ z ≤ ⎬ = Pr{z ≤ −1} = Φ (−1) = 0. 0021 2-27 . x ~ N(5.0005 z = Φ −1 (0.29 z= LSL − µ σ . The process. Find µ such that Pr{x < 32} = 0.95 V. x ~ N ( µ .0 2-47 (2-39).0228) = −1. USL = 5.001. Assume that the process remains centered between the specification limits at 5 V. 0.9991) + 32 = 40. with mean 5 V.95 − 5 ⎞ = Φ⎜ ⎟−Φ⎜ ⎟ = Φ (2.02 ⎠ ⎝ 0. 2-46 (2-38).0152 to have at least 999 of 1000 conform to specification. is currently centered between the specification limits (target = 5 V).0005) = −3. 352) Pr{x > 1000} = 1 − Pr{x ≤ 1000} 1000 − 900 ⎫ ⎧ = 1 − Pr ⎨ x ≤ ⎬ 35 ⎩ ⎭ = 1 − Φ (2. x ~ N(900. so σ = LSL − µ 4.Chapter 2 Exercise Solutions 2-44* (2-36). Φ −1 (0.05 − 5 ⎞ ⎛ 4.00621 = 0. Need Pr{x ≤ LSL} = 0.0228. Desire Pr{Conformance} = 1 / 1000 = 0.015 z −3.022).9991 4 µ = −4(−1. 42 ).8571) = 1 − 0. Φ ( z ) = 0.001 / 2 = 0.98758 ⎝ 0. LSL = 4.5) = 0.9979 = 0.5) − Φ (−2.29 Process variance must be reduced to 0.99379 − 0.05 V Pr{Conformance} = Pr{LSL ≤ x ≤ USL} = Pr{x ≤ USL} − Pr{x ≤ LSL} ⎛ 5.95 − 5 = = 0.9991 32 − µ = −1.02 ⎠ 2-45* (2-37).0005. Shifting the process mean in either direction would increase the number of nonconformities produced. 9380 – c1 For Process 2 proportion defective = p2 = 1 − Pr{LSL ≤ x2 ≤ USL} = 1 − Pr{x2 ≤ USL} + Pr{x2 ≤ LSL} 10.0124 profit for process 1 = 10 (1 – 0. σ12 = 10002).0000) + 5 (0. x ~ N(5000.0124) + 5 (0. x2 ~ N(7500.9938 + 0.5) + Φ (−2.0000 = 0. 000 1. 000 ⎩ ⎭ ⎩ ⎭ = 1 − Φ (2.005) = −2. σ22 = 5002).0124) – c1 = 9.000 h.000 h sales = $10/unit.0000) – c2 = 10 – c2 If c2 > c1 + 0. 000 − 7.0000 profit for process 2 = 10 (1 – 0. defect = $5/unit.005 Φ −1 (0.Chapter 2 Exercise Solutions 2-48 (2-40). 502). USL = 10. x1 ~ N(7500.5) = 1 − 0. profit = $10 × Pr{good} + $5 × Pr{bad} – c For Process 1 proportion defective = p1 = 1 − Pr{LSL ≤ x1 ≤ USL} = 1 − Pr{x1 ≤ USL} + Pr{x1 ≤ LSL} 10.0062 = 0.500 ⎫ ⎧ ⎧ = 1 − Pr ⎨ z1 ≤ ⎬ + Pr ⎨ z1 ≤ ⎬ 1.500 ⎫ 5. 000 − 7.500 ⎫ 5. LSL = 5.5758 LSL − 5000 = −2. Find LSL such that Pr{x < LSL} = 0.5758) + 5000 = 4871 2-49 (2-41).0620. 000 − 7. 000 − 7.500 ⎫ ⎧ ⎧ = 1 − Pr ⎨ z2 ≤ ⎬ + Pr ⎨ z2 ≤ ⎬ 500 500 ⎩ ⎭ ⎩ ⎭ = 1 − Φ (5) + Φ (−5) = 1 − 1. then choose process 1 2-28 .5758 50 LSL = 50(−2.0000 + 0. Chapter 2 Exercise Solutions 2-50 (2-42). 2 ⎣ ⎝ C0 + C2 ⎠ ⎦ 2-29 . Proportion less than lower specification: 6−µ⎫ ⎧ pl = Pr{x < 6} = Pr ⎨ z ≤ ⎬ = Φ (6 − µ ) 1 ⎭ ⎩ Proportion greater than upper specification: 8−µ ⎫ ⎧ pu = Pr{x > 8} = 1 − Pr{x ≤ 8} = 1 − Pr ⎨ z ≤ ⎬ = 1 − Φ (8 − µ ) 1 ⎭ ⎩ Profit = +C0 pwithin − C1 pl − C2 pu = C0 [Φ (8 − µ ) − Φ (6 − µ )] − C1[Φ (6 − µ )] − C2 [1 − Φ (8 − µ )] = (C0 + C2 )[Φ (8 − µ )] − (C0 + C1 )[Φ (6 − µ )] − C2 d d ⎡ 8− µ 1 ⎤ [Φ (8 − µ )] = exp(−t 2 / 2)dt ⎥ ∫ ⎢ dµ d µ ⎣ −∞ 2π ⎦ Set s = 8 – µ and use chain rule d d ⎡s 1 1 ⎤ ds =− [Φ (8 − µ )] = ⎢ ∫ exp(−t 2 / 2)dt ⎥ exp −1/ 2 × (8 − µ ) 2 dµ ds ⎣ −∞ 2π 2π ⎦ dµ ( ) d (Profit) ⎡ 1 ⎤ ⎡ 1 ⎤ = −(C0 + C2 ) ⎢ exp −1/ 2 × (8 − µ ) 2 ⎥ + (C0 + C1 ) ⎢ exp −1/ 2 × (6 − µ ) 2 ⎥ dµ ⎣ 2π ⎦ ⎣ 2π ⎦ ( ) ( ) Setting equal to zero 2 C0 + C1 exp −1/ 2 × (8 − µ ) = = exp(2µ − 14) C0 + C2 exp −1/ 2 × (8 − µ ) 2 ( ( So µ = ) ) ⎤ 1 ⎡ ⎛ C0 + C1 ⎞ ⎢ln ⎜ ⎟ + 14 ⎥ maximizes the expected profit. Chapter 2 Exercise Solutions 2-51 (2-43).… x! ⎞ − λ ∞ λ ( x −1) =e− λ λ eλ = λ ⎟=e λ∑ x = 0 ( x − 1)! ⎠ ( ) σ 2 = E[( x − µ ) 2 ] = E ( x 2 ) − [ E ( x)]2 ⎛ e− λ λ x ⎞ 2 E ( x ) = ∑ x p( xi ) = ∑ x ⎜ ⎟=λ +λ i =1 x =0 ⎝ x! ⎠ 2 ∞ ∞ 2 i 2 σ 2 = (λ 2 + λ ) − [ λ ] = λ 2 2-30 . p( x) = ∞ ∞ ⎛ e− λ λ x ⎝ x! µ = E[ x] = ∑ xi p( xi ) = ∑ x ⎜ i =1 x =0 e− λ λ x .1. x = 0. x = 0. For the Poisson distribution. ⎛n⎞ For the binomial distribution... p ( x) = ⎜ ⎟ p x (1 − p) n − x ... n ⎝ x⎠ ∞ n ⎡⎛ n ⎞ ⎤ n −1 µ = E ( x) = ∑ xi p ( xi ) = ∑ x ⎢⎜ ⎟ p x (1 − p ) n − x ⎥ = n ⎡⎣ p + (1 − p ) ⎤⎦ p = np i =1 x =0 ⎣⎝ x ⎠ ⎦ σ 2 = E[( x − µ ) 2 ] = E ( x 2 ) − [ E ( x)]2 ∞ n ⎡⎛ n ⎞ n− x ⎤ E ( x 2 ) = ∑ xi2 p ( xi ) = ∑ x 2 ⎢⎜ ⎟ p x (1 − p ) ⎥ =np + (np ) 2 − np 2 i =1 x =0 ⎣⎝ x ⎠ ⎦ σ 2 = ⎡⎣ np + (np) 2 − np 2 ⎤⎦ − [ np ] = np(1 − p) 2 2-52 (2-44).1. setting u = x 2 and dv = λ exp(−λ x) uv − ∫ vdu = ⎡⎣ x exp(−λ x) ⎤⎦ 2 +∞ 0 +∞ + 2 ∫ x exp(−λ x)dx = (0 − 0) + 0 σ2 = 2 λ 2 − 1 λ 2 = 2 λ2 1 λ2 2-31 . setting u = x and dv = λ exp(−λ x) +∞ +∞ 1 1 uv − ∫ vdu = ⎡⎣ − x exp ( −λ x ) ⎤⎦ 0 + ∫ exp ( −λ x ) dx = 0 + = λ 0 λ For the variance: ⎛1⎞ σ = E[( x − µ ) ] = E ( x ) − [ E ( x) ] = E ( x ) − ⎜ ⎟ ⎝λ⎠ 2 2 2 2 +∞ +∞ −∞ 0 2 2 E ( x 2 ) = ∫ x 2 f ( x)dx = ∫ x 2 λ exp(−λ x)dx Integrate by parts. f ( x ) = λ e − λ x .Chapter 2 Exercise Solutions 2-53 (2-45). x ≥ 0 For the mean: +∞ +∞ 0 0 µ = ∫ xf ( x)dx = ∫ x ( λ e − λ x )dx Integrate by parts. For the exponential distribution. 25.25.78 σ n 0. H1: µ ≠ 8. Reject H0 if Z0 > Zα.96 0.00000] = 0 (c) ⎞ ≤ µ ≤ x + Z ⎛σ ⎞ x − Zα / 2 ⎛⎜ σ ⎟ ⎟ α /2 ⎜ n⎠ n⎠ ⎝ ⎝ ( 8. H1: µ > 125.002 ) ( 15 ≤ µ ≤ 8. x = 8.Chapter 3 Exercise Solutions 3-1. and conclude that the mean tensile strength exceeds 125 psi.25 + 1. σ = 0.251 MTB > Stat > Basic Statistics > 1-Sample Z > Summarized data One-Sample Z Test of mu = 8.645 Reject H0: µ = 125. x = 127 psi. x − µ0 8.78 P 0.25.000 3-2. 8.25101) Z -6.002 15 ) 8. Reject H0 if |Z0| > Zα/2. (b) P-value = 2[1 − Φ(Z0)] = 2[1 − Φ(6. x − µ0 127 − 125 Z0 = = = 2. σ = 2 psi (a) µ0 = 125. α = 0.96 0.05 Test H0: µ = 8.002 N Mean SE Mean 95% CI 15 8.25000 0.05 = 1.05 Test H0: µ = 125 vs.96 Reject H0: µ = 8.2535 The assumed standard deviation = 0.249 ≤ µ ≤ 8.25 Z0 = = = 6. n = 8. and conclude that the mean bearing ID is not equal to 8.025 = 1.002 cm (a) µ0 = 8.2535 − 8. n = 15.78)] = 2[1 − 1.25 cm.00052 (8.25 vs.24899.05/2 = Z0.2535 cm.2535 vs not = 8.002 15 Zα/2 = Z0.828 σ n 2 8 Zα = Z0. α = 0.25 − 1. 3-1 . s = 1. A one-sided hypothesis test lets us do this.042 3-2 . x ~ N(µ.0. 10−1 = 1. x − µ0 26. n = 10 (a) x = 26.62.837 2. Reject H0 if t0 > tα.05.5138 95% Lower Bound 25.00234 (c) In strength tests. and conclude that the mean life exceeds 25 h. we usually are interested in whether some minimum requirement is met.833 Reject H0: µ = 25.952 S n 1.0 − 25 t0 = = = 1.Chapter 3 Exercise Solutions 3-2 continued (b) P-value = 1 − Φ(Z0) = 1 − Φ(2.6248 SE Mean 0. µ0 = 25.05 Test H0: µ = 25 vs.83 P 0. MTB > Stat > Basic Statistics > 1-Sample t > Samples in columns One-Sample T: Ex3-3 Test of mu = 25 vs > 25 Variable Ex3-3 N 10 Mean 26. not simply that the mean does not equal the hypothesized value.828) = 1 − 0.707 125.0581 T 1.62 10 tα. (d) ( n) ≤ µ 127 − 1.002 3-3.000 0. σ). n−1 = t0.0000 StDev 1. H1: µ > 25. α = 0.95 P 0.99766 = 0.645 ( 2 8 ) ≤ µ x − Zα σ 125.8 ≤ µ MTB > Stat > Basic Statistics > 1-Sample Z > Summarized data One-Sample Z Test of mu = 125 vs > 125 The assumed standard deviation = 2 95% Lower N Mean SE Mean Bound Z 8 127. 0000 1.62 n ≤ µ ≤ x + tα / 2.833 1. Hours 30 32 The plotted points fall approximately along a straight line.62 10 ) 25.114 0.06 ≤ µ ≤ 26.0 + 1.Chapter 3 Exercise Solutions 3-3 continued (b) α = 0. 26. 3-3 . so the assumption that battery life is normally distributed is appropriate.94 MTB > Stat > Basic Statistics > 1-Sample t > Samples in columns One-Sample T: Ex3-3 Test of mu = 25 vs not = 25 Variable N Mean StDev Ex3-3 10 26.95% CI 99 Mean StDev N AD P-Value 95 90 26 1.625 10 0.9419) T 1.95 P 0.833 1.6248 SE Mean 0.083 (c) MTB > Graph > Probability Plot > Single Probability Plot of Battery Service Life (Ex3-3) Normal .n −1 S ) n ( 10 ≤ µ ≤ 26.5138 90% CI (25.986 Percent 80 70 60 50 40 30 20 10 5 1 20 22 24 26 28 Lifetime.0 − 1.0581.10 x − tα / 2.n −1 S ( 26. 9 = 2. Reject H0 if |t0| > tα/2.39216 ≤ µ ≤ 13.00391 95% CI (13.n −1 S 10 ) 13.05.4 Variable N Mean StDev SE Mean Ex3-5 10 13. x = 26.3990) T -3.39618 + 3.4 vs not = 13.0 − 1.4. α = 0.62 25.013 3-4 .2498 ( 0.05.06 ≤ µ The manufacturer might be interested in a lower confidence interval on mean battery life when establishing a warranty policy.025. 13.4 Variable N Mean StDev SE Mean Ex3-5 10 13. α = 0.089 S n 0.n −1 S ( 13. n−1 = t0.09 P 0.4 vs. σ).833 1. n = 10.0 h.3934.0039 0. and conclude that the mean thickness differs from 13. 13.4 vs not = 13.4002) T -3.3922.9 = 1. x ~ N(µ. n = 10. MTB > Stat > Basic Statistics > 1-Sample t > Samples in columns One-Sample T: Ex3-5 Test of mu = 13.0012 99% CI (13. H1: µ ≠ 13.3962 0.Chapter 3 Exercise Solutions 3-4. x − µ0 13.4.00391 10 tα/2.4 × 1000 Å.013 n ≤ µ ≤ x + tα / 2. tα.4 t0 = = = −3.0039 0.62 h.05 Test H0: µ = 13.39618 − 3. σ).262 Reject H0: µ = 13.09 ) ( n) 10 ) ≤ µ ≤ 13.00391 µ0 = 13.0012 (b) α = 0.39618 × 1000 Å. (a) x ~ N(µ.3962 0.39618 − 13.40020 MTB > Stat > Basic Statistics > 1-Sample t > Samples in columns One-Sample T: Ex3-5 Test of mu = 13.01 ( x − tα / 2. s = 1. 3-5.2498 0.00391 P 0.n −1 S ( 26.4 × 1000 Å. x = 13. s = 0. n−1 = t0.833 ( ) 10 ) ≤ µ n ≤µ x − tα . MTB > Stat > Basic Statistics > 1-Sample t > Samples in columns One-Sample T: Ex3-6 Test of mu = 12 vs > 12 Variable Ex3-6 N 10 Mean 12.0096 99% Lower Bound 11.250 Do not reject H0: µ = 12.57 P 0.380 13.Chapter 3 Exercise Solutions 3-5 continued (c) MTB > Graph > Probability Plot > Single Probability Plot of Photoresist Thickness (Ex3-5) Normal . 9 = 3.5655 S n 0. (a) x ~ N(µ.030 Test H0: µ = 12 vs.015 − 12 t0 = = = 1. so the assumption that battery life is normally distributed is not appropriate.390 13.9880 T 1.410 The plotted points form a reverse-“S” shape. H1: µ > 12. s = 0.40 0.015. α = 0. instead of a straight line.405 Thickness.003909 10 0.237 0. n−1 = t0. x1000 Angstroms 13.01 n = 10. σ).0150 StDev 0.0303 SE Mean 0.005.400 13. x = 12.711 Percent 80 70 60 50 40 30 20 10 5 1 13.95% CI 99 Mean StDev N AD P-Value 95 90 13.385 13.395 13.0303 10 tα/2. µ0 = 12. Reject H0 if t0 > tα.076 3-5 . and conclude that there is not enough evidence that the mean fill volume exceeds 12 oz. 3-6. x − µ0 12. 95% CI 99 Mean StDev N AD P-Value 95 90 12.262 ( S 10 ) ≤ µ ≤ 12.0096 95% CI (11. 12.025 = 1.274 0. 3-7.95 12.015 + 2. 9 = 2.262 x − tα / 2.582 Percent 80 70 60 50 40 30 20 10 5 1 11.9933.10 12.0303 SE Mean 0.025. find n ( ( 2 ⎡ Zα / 2 σ ⎣ 2 ⎡1. α = 0. σ = 4 lb. ounces 12.02 0.9600.993 ≤ µ ≤ 12.Chapter 3 Exercise Solutions 3-6 continued (b) α = 0.57 P 0.n −1 S n ≤ µ ≤ x + tα / 2.00 12.015 − 2.037 MTB > Stat > Basic Statistics > 1-Sample t > Samples in columns One-Sample T: Ex3-6 Test of mu = 12 vs not = 12 Variable N Mean StDev Ex3-6 10 12.05. total confidence interval width = 1 lb. so the assumption that fill volume is normally distributed is appropriate.0367) T 1.n −1 S ( ) ( n) 12.0150 0.05 Fill Volume.90 11. Zα/2 = Z0.9600 4 ⎣ ) ) n ⎤ = total width ⎦ n ⎤ =1 ⎦ n = 246 3-6 .15 The plotted points fall approximately along a straight line.05 tα/2.03028 10 0. n−1 = t0.152 (c) MTB > Graph > Probability Plot > Single Probability Plot of Fill Volume (Ex3-6) Normal .62 ( S 10 ) 11. σ = 0.5046 in.5025. x = 0.025 = 1.05 n = 25. x = 10.259 V. 15 = 2.000020 (0.960 ( 0.05/2 = Z0.5025.0001 n ≤ µ ≤ x + Zα / 2 σ 25 ) 0. α = 0.0001 ) ( n) 25 ) ≤ µ ≤ 0.7917) T -6. 0. x ~ N(µ. σ).0001 25 Zα/2 = Z0.960 0.5046 + 1.5046 − 0.000 3-7 .5025 vs not = 0. and conclude that the mean rod diameter differs from 0.7270.0001 in Test H0: µ = 0.131 Reject H0: µ = 12. n−1 = t0.504639) Z 105. MTB > Stat > Basic Statistics > 1-Sample t > Samples in columns One-Sample T: Ex3-9 Test of mu = 12 vs not = 12 Variable N Mean StDev Ex3-9 16 10.971 S n 0.0001 N Mean SE Mean 95% CI 25 0.9990 SE Mean 0. α = 0. 10. Reject H0 if |Z0| > Zα/2.50456 ≤ µ ≤ 0.5046 − 1.5025 The assumed standard deviation = 0. (a) x ~ N(µ.50464 3-9.97 P 0.2594 0. µ0 = 0. H1: µ ≠ 0.96 Reject H0: µ = 0.2498 95% CI (9. σ).00 P 0. MTB > Stat > Basic Statistics > 1-Sample Z > Summarized data One-Sample Z Test of mu = 0. x −µ 0 = 0.259 − 12 t0 = = = −6. H1: µ ≠ 12. and conclude that the mean output voltage differs from 12V. s = 0.5025 = 105 Z0 = σ n 0.999 16 tα/2.Chapter 3 Exercise Solutions 3-8.999 V (a) µ0 = 12.5025. Reject H0 if |t0| > tα/2.504600 0.5025 vs.000 (b) P-value = 2[1 − Φ(Z0)] = 2[1 − Φ(105)] = 2[1 − 1] = 0 (c) ( x − Zα / 2 σ ( 0.05 Test H0: µ = 12 vs. n = 16. x − µ0 10.5025.504561.025. 488 χ21−α/2.391 2 0.n −1 S 16 ) 9. the null hypothesis.975.262 Do not reject H0: σ 2 = 1.025. 3-8 . and conclude that there is insufficient evidence that the variance differs from 1. σ 02 = 1.999 n ≤ µ ≤ x + tα / 2.546 Since the 95% confidence interval on σ contains the hypothesized value.792 (c) σ 02 = 1.259 + 2. n−1 = χ20. H1: σ 2 ≠ 1.131( 0.259 − 2.n −1 ≤σ2 ≤ (n − 1) S 2 χ2 1−α / 2.488 6. (16 − 1)0.05 Test H0: σ 2 = 1 vs. n −1 (16 − 1)0.999 ) ( n) 16 ) ≤ µ ≤ 10.9992 = 14. n−1 = χ20.738 ≤ σ ≤ 1.n −1 S ( 10.16−1 = 27. α = 0.262 2 0.727 ≤ µ ≤ 10.545 ≤ σ ≤ 2.999 (16 − 1)0. χ 02 = (n − 1) S 2 = (d) (n − 1) S 2 χα2 / 2. H0: σ 2 = 1.970 1 σ 02 χ2α/2.16−1 = 6. cannot be rejected. n-1.9992 2 ≤σ ≤ 27. n-1 or χ20 < χ 21-α/2. Reject H0 if χ 20 > χ 2α/2.Chapter 3 Exercise Solutions 3-9 continued (b) ( x − tα / 2.131 0. 15 σ2 ≤ (n − 1) S 2 χ12−α .727 10.792 95% C onfidence Interv al for M edian 9.23 0.116487 -0.430 10.75 11.50 9.259 0.546 Mean Median 9.492793 16 M inimum 1st Q uartile M edian 3rd Q uartile M aximum 12 8.n −1 = χ 0.999 0.000 95% C onfidence Interv al for M ean 9.00 10.767 M ean S tD ev V ariance S kew ness Kurtosis N 10.9992 σ ≤ 7.140 11.2609 α = 0.062 σ ≤ 1.75 10.2609 2 σ ≤ 2.05.50 10.436 2 3-9 .Chapter 3 Exercise Solutions 3-9 (d) continued MTB > Stat > Basic Statistics > Graphical Summary Summary for Output Voltage (Ex3-9) A nderson-D arling N ormality Test 8 9 10 11 A -S quared P -V alue 0.533 10.150 12.95.25 10.738 1.945 95% C onfidence Interv al for S tD ev 95% Confidence Intervals 0.370 9.00 (e) 2 = 7.n −1 (16 − 1)0.998 0. χ12−α . 05.00000] = 0 3-10 .025 = 1.015 l.26 0. x1 = 2. ( x1 − x2 ) − ∆ 0 (2.04 − 2. ∆ 0 = 0 Test H0: µ1 – µ2 = 0 versus H0: µ1 – µ2 ≠ 0.0102 25 + 0.07) − 0 Z0 = = = −7. σ2 = 0. (b) P-value = 2[1 − Φ(Z0)] = 2[1 − Φ(−7. n2 = 20.9990 16 0.010 l. (a) α = 0. σ1 = 0.682)] = 2[1 − 1.682 2 2 0. 3-10. x2 = 2.230 0. Reject H0 if Z0 > Zα/2 or Z0 < –Zα/2.04 l.767 Percent 80 70 60 50 40 30 20 10 5 1 7 8 9 10 11 Output Voltage 12 13 14 From visual examination of the plot.96 Reject H0: µ1 – µ2 = 0.0152 20 σ 1 n1 + σ 2 n2 Zα/2 = Z0. n1 = 25.05/2 = Z0.07 l. the assumption of a normal distribution for output voltage seems appropriate.95% CI 99 Mean StDev N AD P-Value 95 90 10. and conclude that there is a difference in mean net contents between machine 1 and machine 2.96 −Zα/2 = −1.Chapter 3 Exercise Solutions 3-9 continued (f) MTB > Graph > Probability Plot > Single Probability Plot of Output Voltage (Ex3-9) Normal . We can conclude that there is no difference in measurements obtained by the two technicians.1204 P-Value = 0.010 σ 12 25 n1 + σ 22 + 0.n + n 1 2 −2 S p 1 n1 + 1 n2 (1.05.07) + 1. Ex3-11T2 Two-sample T for Ex3-11T1 vs Ex3-11T2 N Mean StDev SE Mean Ex3-11T1 7 1.07) − 1. and an investigation should be undertaken to understand why.383 − 1. 3-11 .11 Both use Pooled StDev = 0.mu (Ex3-11T2) Estimate for difference: 0.376.9600 0.141 The confidence interval for the difference contains zero.1604(0.376) + 2. and conclude that there is not sufficient evidence of a difference between measurements obtained by the two technicians. (a) MTB > Stat > Basic Statistics > 2-Sample t > Samples in different columns Two-Sample T-Test and CI: Ex3-11T1.038 ≤ ( µ1 − µ 2 ) ≤ −0.9600 0.04 − 2. (c) n1 = 7.115 0. x2 = 1. x1 = 1. S1 = 0.Chapter 3 Exercise Solutions 3-10 continued (c) ( x1 − x2 ) − Zα / 2 2 (2.120 n1 + n2 − 2 7+8−2 ( x1 − x2 ) − tα / 2.n + n 1 2 −2 S p 1 n1 + 1 n2 ≤ ( µ1 − µ 2 ) ≤ ( x1 − x2 ) + tα / 2.127969. tα/2. S2 = 0.376 0.1604(0.1152 + (8 − 1)0.383.141183) T-Test of difference = 0 (vs not =): T-Value = 0. 0.1252 = = 0.043 Ex3-11T2 8 1. We can conclude that the machines do not fill to the same volume.010 20 −0.015 2 20 The confidence interval for the difference does not contain zero.917 DF = 13 Do not reject H0: µ1 – µ2 = 0.376) − 2.383 − 1. 13 = 2. 3-11.04 − 2.015 2 n2 σ 12 ≤ ( µ1 − µ 2 ) ≤ ( x1 − x2 ) + Zα / 2 n1 + σ 22 n2 2 ≤ ( µ1 − µ 2 ) ≤ (2. the readings will be the same. we would have been concerned that the technicians obtained different measurements. n1+n2−2 = t0. (b) The practical implication of this test is that it does not matter which technician measures parts.125 0. If the null hypothesis had been rejected.125 α = 0.120) 1 7 + 1 8 −0.1604 Sp = (n1 − 1) S12 + (n2 − 1) S22 (7 − 1)0.006607 95% CI for difference: (-0.120) 1 7 + 1 8 ≤ ( µ1 − µ 2 ) ≤ (1.383 0. n2 = 8.022 25 + 0.115.127 ≤ ( µ1 − µ 2 ) ≤ 0.044 Difference = mu (Ex3-11T1) .025. n −1.n −1.854 Lev ene's Test Test Statistic P-Value Ex3-11T2 0.5 1.7 −1.176 MTB > Stat > Basic Statistics > 2 Variances > Summarized data Test for Equal Variances for Ex3-11T1. 3-12 .920 0.n −1 = F0.85 0. we would have been concerned about the difference in measurement variability between the technicians.n 1 2 −1 = F1−0. Ex3-11T2 F-Test Test Statistic P-Value Ex3-11T1 0.7 = 0.Chapter 3 Exercise Solutions 3-11 continued (d) α = 0.n −1.25 95% Bonferroni Confidence Intervals for StDevs 0.15 0.8464 Fα / 2.10 0.n −1.01 0.8−1 = F0. and an investigation should be undertaken to understand why.05/ 2.2 1.30 Ex3-11T1 Ex3-11T2 1.115 0.n −1. If the null hypothesis is rejected. and conclude that there is no difference in variability of measurements obtained by the two technicians.05/ 2.6.4 Data 1.n 1 2 −1 2 or F0 < F1−α / 2. 1 2 F0 = S S = 0.05 Test H 0 : σ 12 = σ 22 versus H1 : σ 12 ≠ σ 22 .20 0.025.119 2 1 1 2 2 2 2 F1−α / 2.6 Do not reject H0.975.8−1 = F0.7 = 5.125 = 0.7 −1.3 1. Reject H 0 if F0 > Fα / 2.6. 065 (g) MTB > Graph > Probability Plot > Multiple Probability Plot of Surface Finish by Technician (Ex3-11T1.2 1. Fα / 2.943 0. n1 −1 2 1 S2 S σ σ 12 0.235 0.125 2 2 = 16.0 1.n −1 = χ 0.6955 S12 σ S F Fα / 2.165 ≤ σ 12 ≤ 4.376 Percent 80 StDev N AD P 0.4 Data 1.1148 7 0.0128 1.n2 −1 = χ 0. χα2 / 2.975.1252 (8 − 1)0.05.6 = 5.1249 8 0. x2 = 1.142 0. S2 = 0.n 2 −1.0128. 3-13 .693 70 60 50 40 30 20 10 5 1 1.007 ≤ σ ≤ 0.n −1 ≤ ≤ 2 1−α / 2.1252 0.n −1 (8 − 1)0.6 = 0.7 = 1.025.7 2 (n − 1) S 2 χα2 / 2. n1 −1 2 2 1 1 2 2 2 2 = F0.6899 2 0.7.Chapter 3 Exercise Solutions 3-11 continued (e) α = 0.8 The normality assumption seems reasonable for these readings.1252 σ 22 0. n1 −1 = F0.1954. χ12−α / 2.975.6899 α = 0.383 1.6955) ≤ ≤ 0.n −1 ≤σ2 ≤ (n − 1) S 2 χ12−α / 2.1152 0.6 1.05 F1−α / 2.376.n 2 −1.n −1.1152 (0. Ex3-11T2) Normal .025.1954) (5. n2 −1.821 σ 22 (f) n2 = 8.1252 2 ≤σ ≤ 16.7.95% CI 99 Variable Ex3-11T1 Ex3-11T2 95 90 Mean 1. 22) 1 10 + 1 10 −6.4) − 2.451 DF = 18 Do not reject H0.05. x1 = 147.4) + 2.97 1.40 5. the test statistic is 2 2 2 + S n S n x − x 1 1 2 2 1 2 with degrees of freedom ν = −2 t0* = 2 2 2 2 2 S1 n1 + S 22 n2 S1 n1 S 2 n2 + ( n1 + 1) ( n2 + 1) ( ( ) ) ( ) A 100(1-α)% confidence interval on the difference in means would be: ( x1 − x2 ) − tα / 2.4.60 4. 3. x2 = 149.n + n 1 2 −2 S p 1 n1 + 1 n2 ≤ ( µ1 − µ 2 ) ≤ ( x1 − x2 ) + tα / 2.22) 1 10 + 1 10 ≤ ( µ1 − µ 2 ) ≤ (147. (b) α = 0.22 10 + 10 − 2 n1 + n2 − 2 ( x1 − x2 ) − tα / 2. tα/2.025. From Eqn.6.6 − 149.ν S12 n1 + S 22 n2 ≤ ( µ1 − µ2 ) ≤ ( x1 − x2 ) + tα / 2.7 Difference = mu (Ex3-13SQ) .1009 Sp = (n1 − 1) S12 + (n2 − 1) S 22 (10 − 1)4. Ex3-13OQ Two-sample T for Ex3-13SQ vs Ex3-13OQ N Mean StDev SE Mean Ex3-13SQ 10 147. 18 = 2. for σ 12 ≠ σ 22 and both unknown.n + n 1 2 −2 S p 1 n1 + 1 n2 (147.mu (Ex3-13OQ) Estimate for difference: -1. and conclude that there is no difference between the quenching processes. Saltwater quench: n1 = 10.6 − 149.1009(5.2217 P-Value = 0.10615) T-Test of difference = 0 (vs not =): T-Value = -0.97 Oil quench: n2 = 10.6 Ex3-13OQ 10 149. n1+n2−2 = t0.46 1.Chapter 3 Exercise Solutions 3-12.80000 95% CI for difference: (-6.7 ≤ ( µ1 − µ 2 ) ≤ 3.462 = = 5.77 Both use Pooled StDev = 5.1 3-14 .ν S12 n1 + S 22 n2 3-13.97 2 + (10 − 1)5. 3-54 and 3-55.70615. S1 = 4.1009(5. S2 = 5.46 (a) Assume σ 12 = σ 22 MTB > Stat > Basic Statistics > 2-Sample t > Samples in different columns Two-Sample T-Test and CI: Ex3-13SQ. 2484) (4.9 2 −1.Chapter 3 Exercise Solutions 3-13 continued (c) α = 0.9.971 10 0. 2 1.169 0.9.34 σ 22 Since the confidence interval includes the ratio of 1.05 F1−α / 2.906 90 Percent 80 70 60 50 40 30 20 10 5 1 130 140 150 Hardness 160 170 The normal distribution assumptions for both the saltwater and oil quench methods seem reasonable. 1 1 σ 22 S S22 σ 12 4.779 149. n2 −1.21 ≤ σ 12 ≤ 3. Ex3-13OQ) Normal .95% CI 99 Variable Ex3-13SQ Ex3-13OQ 95 AD P Mean StDev N 147.0260 S12 σ 12 S ≤ ≤ F F − α n − n − 1 / 2.97 2 4. Fα / 2.462 0. 3-15 .97 2 ≤ ≤ (0.025. n1 −1 = F0.n =F = 0. (d) MTB > Graph > Probability Plot > Multiple Probability Plot of Quench Hardness (Ex3-13SQ. the assumption of equal variances seems reasonable.461 10 0.218 0.2484.n 0.0260) σ 22 5.6 4.4 5. n1 −1 2 1 2 α / 2. n1 −1 2 2 −1.975.462 5.9 = 4. not an approximation.Chapter 3 Exercise Solutions 3-14.05 = 1.10.5) − np0 (18 + 0. Test H0: p = 0. (b) α = 0.057 ≤ p ≤ 0.6382] = 0.050338. pˆ = x/n = 18/200 = 0.10/2 = Z0. Z0 = ( x + 0.090000 (0.09) 200 0. 0.10.10) Zα/2 = Z0.09 (a) p0 = 0. Reject H0 if |Z0| > Zα/2.09 − 1. x = 18.96 Do not reject H0. Zα/2 = Z0.1 vs p not = 0.645 0.645 0.10) = 20 Since (x = 18) < (np0 = 20).09 + 1.5) − 20 = = −0.10 versus H1: p ≠ 0. α = 0. and conclude that the sample process fraction nonconforming does not differ from 0. P-value = 2[1 − Φ|Z0|] = 2[1 − Φ|−0.7236 MTB > Stat > Basic Statistics > 1 Proportion > Summarized data Test and CI for One Proportion Test of p = 0.645 pˆ − Zα / 2 pˆ (1 − pˆ ) n ≤ p ≤ pˆ + Zα / 2 pˆ (1 − pˆ ) n 0.129662) Z-Value -0.10.637 Note that MINITAB uses an exact method.123 3-16 .09) 200 ≤ p ≤ 0.025 = 1.05/2 = Z0. np0 = 200(0.05.10. use the normal approximation to the binomial for x < np0.09(1 − 0.1 Sample X N Sample p 95% CI 1 18 200 0.09(1 − 0.3536|] = 2[1 − 0.47 P-Value 0.3536 np0 (1 − p0 ) 20(1 − 0. n = 200. 99997] = 0.13(1 − 0.0387 np0 (1 − p0 ) 40(1 − 0.05/2 = Z0. α = 0. Reject H0 if |Z0| > Zα/2. x = 65.08.08) Zα/2 = Z0.08.025 = 1. pˆ = x/n = 65/500 = 0. Zα = Z0.130 (a) p0 = 0.96 Reject H0. not an approximation.159478) Z-Value 4.Chapter 3 Exercise Solutions 3-15. MTB > Stat > Basic Statistics > 1 Proportion > Summarized data Test and CI for One Proportion Test of p = 0.645 0.00006 (c) α = 0. np0 = 500(0.05 = 1. n = 500.08 vs p not = 0. Test H0: p = 0.08) = 40 Since (x = 65) > (np0 = 40).100522.000 Note that MINITAB uses an exact method.05.12 P-Value 0. ( x − 0. and conclude the sample process fraction nonconforming differs from 0.130000 (0.05.08 versus H1: p ≠ 0.5) − np0 (65 − 0. use the normal approximation to the binomial for x > np0.08.13 + 1. (b) P-value = 2[1 − Φ|Z0|] = 2[1 − Φ|4. 0.08 Sample X N Sample p 95% CI 1 65 500 0.0387|] = 2[1 − 0.155 3-17 .645 p ≤ pˆ + Zα pˆ (1 − pˆ ) n p ≤ 0.13) 500 p ≤ 0.5) − 40 Z0 = = = 4. 05/2 = Z0.96 Zα/2 = Z0.645 0.066667 Difference = p (1) . Test H0: p1 = p2 versus H1: p1 ≠ p2.06(1 − 0.05 − 0.05(1 − 0.0246745) Test for difference = 0 (vs not = 0): Z = -0.05) 200 + 0.050 − 0.025 = 1.067) 300 pˆ 1 (1 − pˆ 1 ) + pˆ 2 (1 − pˆ 2 ) n1 n2 ≤ ( p1 − p2 ) ≤ (0. 0. x2 = 20.067) − 1.067 (b) Use α = 0.050000 2 20 300 0.018 3-18 .05 n2 = 300.06) (1 200 + 1 300 ) = −0. pˆ1 = x1/n1 = 10/200 = 0. Conclude there is no strong evidence to indicate a difference between the fraction nonconforming for the two processes.77 P-Value = 0.0580079.442 (c) ( pˆ 1 − pˆ 2 ) − Z α / 2 pˆ 1 (1 − pˆ 1 ) + pˆ 2 (1 − pˆ 2 ) n1 n2 ≤ ( p1 − p2 ) ≤ ( pˆ 1 − pˆ 2 ) + Z α / 2 (0. MTB > Stat > Basic Statistics > 2 Proportions > Summarized data Test and CI for Two Proportions Sample X N Sample p 1 10 200 0. Reject H0 if Z0 > Zα/2 or Z0 < –Zα/2 pˆ = x1 + x2 10 + 20 = = 0.p (2) Estimate for difference: -0. pˆ 2 = x2/n2 = 20/300 = 0.067 = 0.067(1 − 0.067) 300 −0.067) + 1.05.05(1 − 0.7842 −Zα/2 = −1.Chapter 3 Exercise Solutions 3-16.052 ≤ ( p1 − p2 ) ≤ 0.05 − 0. (a) n1 = 200.06 n1 + n2 200 + 300 Z0 = pˆ1 − pˆ 2 pˆ (1 − pˆ ) (1 n1 + 1 n2 ) 0.05) 200 + 0.96 Do not reject H0.645 0.0166667 95% CI for difference: (-0.067(1 − 0. x1 = 10. tα. Reject H0 if t0 > tα. x1 = 9.7 = 0.85 2.n1 −1.05.0987 F0 = 1.0987 < 4.08 = = 1.18 (a) Test H 0 : σ 12 = σ 22 versus H1 : σ 12 ≠ σ 22 .082 DF = 16 The mean impurity after installation of the new purification unit is not less than before.n1 −1.48596 5.08 2.n2 −1 Fα / 2.85. 10+8−2 = 1.9.55525 2.* before: n1 = 10.85 − 8. so do not reject H 0 MTB > Stat > Basic Statistics > 2 Variances > Summarized data Test for Equal Variances 95% Bonferroni confidence intervals for standard deviations Sample N Lower StDev Upper 1 10 1.34856 T-Test of difference = 0 (vs >): T-Value = 1.88 Difference = mu (1) .7 = 4. F1−α / 2.79 6.9. at α = 0.49 0.461 1 n1 + 1 n2 2.24710 2 8 1.n2 − 2 = F0.922 The impurity variances before and after installation are the same.18 = 2. 3-19 .554 1 10 + 1 8 MTB > Stat > Basic Statistics > 2-Sample t > Summarized data Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 10 9.5582 P-Value = 0.77000 95% lower bound for difference: -0.61 0.n2 −1 = F0.10.70449 2. (b) Test H0: µ1 = µ2 versus H1: µ1 > µ2.60576 5.975.83 2 8 8.8232.554 10 + 8 − 2 x1 − x2 9.8232 and > 0.n1+n2−2.05.n2 − 2 or F0 < F1−α / 2.18 = 1.n1 −1.46 Both use Pooled StDev = 2. α = 0. S12 = 6.n1+n2−2 = t0.n1 −1.05 Reject H 0 if F0 > Fα / 2.2383.Chapter 3 Exercise Solutions 3-17.mu (2) Estimate for difference: 1.69405 F-Test (normal distribution) Test statistic = 1.746 SP = t0 = SP ( n1 − 1) S12 + ( n2 − 1) S22 n1 + n2 − 2 = (10 − 1) 6.025. p-value = 0. x2 = 8.2383 F0 = S12 S 22 = 6.79 after: n2 = 8. S 22 = 6.08.79 + (8 − 1) 6. 75 Difference = mu (1) .3 psi.645 x x − ( 1 2 ) − ∆ 0 = (175.30 3.320 DF = 30 Note: For equal variances and sample sizes.50000 95% upper bound for difference: -3. The P-values are close due to the sample sizes. x2 = 181.0000 P-Value = 0.8 psi. n1 = 16. test H0: µ1 − µ2 = − 5 versus H1: µ1 − µ2 < − 5. x1 = 175. so H1: µ1 + 5 < µ2.80 3.8 − 181.75 2 16 181.47 Both use Pooled StDev = 3.05 = −1.4714 Z0 = 32 16 + 32 16 σ 12 n1 + σ 22 n2 (Z0 = −0. n2 = 16. σ1 = σ2 = 3.00 0.3) − (−5) = −0.645. ∆0 = −5 −Zα = −Z0.00 0. so do not reject H0.69978 T-Test of difference = -5 (vs <): T-Value = -0. Reject H0 if Z0 < − Zα .4714) > −1. So test a difference ∆0 = −5.0 psi Want to demonstrate that µ2 is greater than µ1 by at least 5 psi. the Z-value is the same as the t-value.3187 MTB > Stat > Basic Statistics > 2-Sample t > Summarized data Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 16 175. P-value = Φ(Z0) = Φ(−0.mu (2) Estimate for difference: -5. 3-20 .4714) = 0. The mean strength of Design 2 does not exceed Design 1 by 5 psi.Chapter 3 Exercise Solutions 3-18. 001250. Ex3-19VC Paired T for Ex3-19MC .22 = 2.001311 0.0013112 ( ) n = −0.000835 0.10 (|t0| = 1.151) + " + ( 0.001311 12 = −1. n1 + n2 − 2.10 P-Value = 0.001621 0.152 ) ⎤⎦ = −0.000417) T-Test of mean difference = 0 (vs not = 0): T-Value = -1.000417 12 ⎛ n ⎞ − d ⎜∑dj ⎟ ∑ j =1 ⎝ j =1 ⎠ S d2 = ( n − 1) n 2 j t0 = d (S d ) 2 n = 0. tα/2. Test H0: µd = 0 versus H1: µd ≠ 0.151 − 0.10) < 2. j − xVernier.295 3-21 .000379 95% CI for mean difference: (-0.8188. There is no strong evidence to indicate that the two calipers differ in their mean measurements.000417 0.151583 0. so do not reject H0. 0. j ) = 1 ⎡⎣( 0.Ex3-19VC N Mean StDev SE Mean Ex3-19MC 12 0.000417 0. n1 + n2 − 2 = t0.000241 Ex3-19VC 12 0.151167 0.000468 Difference 12 -0.005.8188 d=1 ( n∑ j =1 n xMicrometer. MTB > Stat > Basic Statistics > Paired t > Samples in Columns Paired T-Test and CI: Ex3-19MC.Chapter 3 Exercise Solutions 3-19. Reject H0 if |t0| > tα/2.150 − 0. The standard deviation of the fill volume is not less than 1ml. t0 = ( x − µ ) S n = (153.083 3-21. χ21-α/2.05.05 (a) Test H0: σ2 = 1 versus H1: σ2 < 1. (b) χ2α/2.n −1 ≤ σ 2 ≤ (n − 1) S 2 χ12−α / 2.80 1.025. n-1 = χ20.44 P 0. (n − 1) S 2 χα2 / 2.4389 ( ) ( ) (t0 = 1.95. MTB > Stat > Basic Statistics > 1-Sample t > Summarized data One-Sample T Test of mu = 150 vs > 150 N 20 Mean 153. so do not reject H0.7 − 150 ) 11.7.19 3-22 .700 StDev 11.1170 χ 02 = ⎡⎣(n − 1) S 2 ⎤⎦ σ 02 = ⎡⎣(20 − 1)1.52 32.7291.19 = 1.05 Test H0: µ = 150 versus H1: µ > 150. Reject H0 if χ20 < χ21-α.7291. s = 1.Chapter 3 Exercise Solutions 3-20.500 SE Mean 2.5.n −1 (20 − 1)1. n = 20.85. n −1. (a) The alternative hypothesis H1: µ > 150 is preferable to H1: µ < 150 we desire a true mean weld strength greater than 150 psi. In order to achieve this result. so do not reject H0.571 95% Lower Bound 149.85 ≤ σ 2 ≤ (20 − 1)1.19 = 8.19 = 32. n −1 = t0. α = 0. x = 153.1170.52 ⎤⎦ 1 = 42. n-1.91 1.5.14 ≤ σ ≤ 2. Reject H0 if t0 > tα. α = 0.75 χ20 = 42. µ > 150. s = 11.975. H0 must be rejected in favor of the alternative H1. χ21-α.30 ≤ σ 2 ≤ 4.19 = 10. There is insufficient evidence to indicate that the mean strength is greater than 150 psi. n-1 = χ20.91.52 8.5 20 = 1.75 > 10.254 T 1. (b) n = 20. n-1 = χ20. x = 752.4389) < 1.6 ml. tα. ml (Ex3-21) A nderson-D arling N ormality Test 749 750 751 752 753 754 755 A -S quared P -V alue 0.0 752.00 756.191843 20 M inimum 1st Q uartile M edian 3rd Q uartile M aximum 756 750.511 0.17 2.95% CI 99 Mean StDev N AD P-Value 95 90 752.538 20 0.54 2.00 95% C onfidence Interv al for S tD ev 95% Confidence Intervals 1.4 752.51 0.Chapter 3 Exercise Solutions 3-21 (b) continued MTB > Stat > Basic Statistics > Graphical Summary Summary for Pinot Gris Fill Volume.281321 0. 3-23 .25 753. ml 756 758 The plotted points do not fall approximately along a straight line.8 753.55 1.2 (c) MTB > Graph > Probability Plot > SIngle Probability Plot of Pinot Gris Fill Volume (Ex3-21) Normal .37 0.27 95% C onfidence Interv al for M edian 752.172 M ean S tD ev V ariance S kew ness Kurtosis N 752.6 1.00 753.00 751.83 753.172 Percent 80 70 60 50 40 30 20 10 5 1 748 750 752 754 Fill Volume. so the assumption that battery life is normally distributed is not appropriate.00 753.00 95% C onfidence Interv al for M ean 751.25 Mean Median 752. 96 − 5 4 3 ( ) ( ) ( ) ( ) = Φ (−1. So. Check: β = Φ Zα / 2 − δ n σ − Φ − Zα / 2 − δ n σ = Φ 1. 3-46. β = Φ Zα / 2 − δ n σ − Φ − Zα / 2 − δ n σ ( ) ( ) ( ) If δ > 0.Chapter 3 Exercise Solutions 3-22.05. From Eqn. µ1 = 20. n = 4. µ0 = 15.0848 MTB > Stat > Power and Sample Size > 1-Sample Z Power and Sample Size 1-Sample Z Test Testing mean = null (versus not = null) Calculating power for mean = null + difference Alpha = 0.05.0848 − 0.2933) = 0. is less than or equal to 0.0000 = 0.9 0. σ2 = 9. Test H0: µ = 15 versus H1: µ ≠ 15.3733) − Φ (−5.915181 3-23. Let µ1 = µ0 + δ.05 Assumed standard deviation = 3 Sample Target Difference Size Power Actual Power 5 4 0. ( β ≈ Φ Zα / 2 − δ n σ ( ) Φ ( β ) ≈ Φ −1 Zα / 2 − δ n σ ) − Z β ≈ Zα / 2 − δ n σ n ≈ ⎡⎣( Zα / 2 + Z β )σ δ ⎤⎦ 2 3-24 .96 − 5 4 3 − Φ −1.10? δ = µ1 − µ2 = 20 − 15 = 5 d = δ σ = 5 9 = 1. α = 0. the operating characteristic curve for two-sided at α = 0. What n is needed such that the Type II error.6667 From Figure 3-7. then Φ − Zα / 2 − δ n σ is likely to be small compared with β.0. β. E (Q) = E ( x1 − 2 x2 ) = µ1 − 2µ2 = 0 var(Q) = var( x1 − 2 x2 ) = var( x1 ) + var(2 x2 ) = var( x1 ) + 22 var( x2 ) = Z0 = var( x1 ) var( x2 ) +4 n1 n2 Q−0 x1 − 2 x2 = SD(Q) σ 12 n1 + 4 σ 22 n2 And.Chapter 3 Exercise Solutions 3-24. x1 independent of x2 . Assume µ1 = 2µ2 and let Q = ( x1 − x2 ) . reject H 0 if Z 0 > Zα / 2 3-25 . n2 . Given x ~ N . x1 . x1 − x2 Maximize: Z 0 = Subject to: n1 + n2 = N . an equivalent statement is Minimize: L = σ 12 n1 + σ 22 n2 = σ 12 n1 + σ 22 N − n1 σ 22 ⎞ dL ⎡ −1 2 dL ⎛ σ 12 −1 + n1 σ 1 + ( N − n1 ) σ 22 ⎤ ⎜ ⎟= ⎦ dn1 ⎝ n1 N − n1 ⎠ dn1 ⎣ = −1n1−2σ 12 + (−1)(−1) ( N − n1 ) σ 22 = 0 −2 =− σ 12 n12 + σ 22 ( N − n1 ) 2 =0 n1 σ 1 = n2 σ 2 Allocate N between n1 and n2 according to the ratio of the standard deviations. n1 . x2 . 3-25. σ 1 n1 + σ 2 n2 2 2 Since ( x1 − x2 ) is fixed. n = 100. …. so do not reject H0. x ~ Poi(λ). λ/n).15 ) 0. at α = 0.5758. Reject H0 if |Z0| > Zα. x = 688. Reject H0: λ = λ0 if |Z0| > Zα/2 (b) x ~ Poi(λ). n = 1000.6 Test H0: λ = 0.3162) < 1. x ~ Poi(λ). Reject H0 if |Z0| > Zα/2. n ∑x i ~ POI( nλ ) .688 Test H0: λ = 1 versus H1: λ ≠ 1.15 100 = −1.05 = 1.0328) < 2.645. Z 0 = ( x − λ ) λ0 / n .Chapter 3 Exercise Solutions 3-26. at α = 0. Zα = Z0. x = 11. Select random sample of n observations x1.05.645 Z 0 = ( x − λ0 ) λ0 n = ( 0.15.5 ) 0. x = x/n = ~ N(λ. 3-26 . (a) Wish to test H0: λ = λ0 versus H1: λ ≠ λ0.6 − 0. x = x/N = 3/5 = 0.15 versus H1: λ ≠ 0.5 5 = 0. x = x/N = 11/100 = 0. Reject H0 if Z0 > Zα.005 = 2. so do not reject H0. so reject H0.0328 (|Z0| = 1. Zα/2 = Z0. i =1 Using the normal approximation to the Poisson. n = 5.8663) > 1.96 Z 0 = ( x − λ0 ) λ0 n = ( 0. 3-27. x2. if n is large.96.110 Test H0: λ = 0.5. at α = 0. 3-28.688 − 1) 1 1000 = −9.5 versus H1: λ > 0. x = 3. Each xi ~ POI(λ).01. xn. x = x/N = 688/1000 = 0.5758 Z 0 = ( x − λ0 ) λ0 n = ( 0. Zα/2 = Z0.05.8663 (|Z0| = 9.025 = 1.3162 (Z0 = 0.110 − 0. 5231 (----------*---------) 200 6 3.Chapter 3 Exercise Solutions 3-29.59 0.053 15 7.5 4.5 125 160 C2F6 Flow (SCCM) 200 Gas flow rate of 125 SCCM gives smallest mean percentage uniformity.0 2. With Groups Boxplot of Etch Uniformity by C2F6 Flow Etch Uniformity (%) 5.80 Pooled StDev = 0.34% R-Sq(adj) = 23.5 3.60 4. the P-value is just slightly greater than 0. (a) MTB > Stat > ANOVA > One-Way One-way ANOVA: Ex3-29Obs versus Ex3-29Flow Source Ex3-29Flow Error Total S = 0.59).32% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -----+---------+---------+---------+---125 6 3.0 4.7600 (---------*----------) 160 6 4. so flow rate does not affect etch uniformity at a significance level α = 0.7132 (F0.00 3.3167 0.4167 0.630 0.15 = 3. so there is some evidence that gas flow rate affects the etch uniformity. However.509 17 11. Boxplots of data MTB > Graph > Boxplot > One Y.2.6823) > (F0 = 3.648 1.8214 (----------*---------) -----+---------+---------+---------+---3. (b) MTB > Stat > ANOVA > One-Way > Graphs.824 3.05.05.20 4.0 3.05.278 R-Sq = 32. 3-27 .7132 DF SS MS F P 2 3.9333 0. 5 0.5 Residual 1. Residuals versus fits Residuals Versus the Fitted Values (response is Etch Uniformity (Ex3-29Obs)) 1.2 3.5 -1.0 0. 3-28 .0 4. Normal plot of residuals Normal Probability Plot of the Residuals (response is Etch Uniformity (Ex3-29Obs)) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -2 -1 0 Residual 1 2 The normality assumption is reasonable.2 4.0 3.6 3.4 3.4 Residuals are satisfactory.Chapter 3 Exercise Solutions 3-29 continued (c) MTB > Stat > ANOVA > One-Way > Graphs. (d) MTB > Stat > ANOVA > One-Way > Graphs.0 -0.8 Fitted Value 4. The statistically significant difference between the mean uniformities can be seen by centering the t distribution between.9% MSE n = 0. Flow Rate 125 160 200 scale factor = Mean Etch Uniformity 3.3 3 . and noting that 160 would fall beyond the tail of the curve.4% 3.8 M e a n E t c h U n if o r m it y The graph does not indicate a large difference between the mean etch uniformity of the three different flow rates.5087 6 = 0.Chapter 3 Exercise Solutions 3-30.5 4 .9 (1 6 0 ) 4 .3 S c a le d t D is t r ib u t io n (2 0 0 ) (1 2 5 ) 3 .6 3 . say.3% 4. 3-29 . 125 and 200.0 3 .2 4 . 7 (-----------*-----------) 20 3 1606.0 52.0 (-----------*----------) 15 3 1586. (a) MTB > Stat > ANOVA > One-Way > Graphs> Boxplots of data. Normal plot of residuals One-way ANOVA: Ex3-31Str versus Ex3-31Rod Source Ex3-31Rod Error Total S = 71.0 10.7 107.214 8 40933 5117 11 69567 R-Sq = 41.9 (-----------*-----------) 25 3 1500. 3-30 .87 0.5 No difference due to rodding level at α = 0.16% R-Sq(adj) = 19.53 DF SS MS F P 3 28633 9544 1.09% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ----+---------+---------+---------+----10 3 1500.Chapter 3 Exercise Solutions 3-31.05.0 (-----------*----------) ----+---------+---------+---------+----1440 1520 1600 1680 Pooled StDev = 71. (b) Boxplot of Compressive Strength by Rodding Level 1750 Compressive Strength 1700 1650 1600 1550 1500 1450 1400 10 15 20 25 Rodding Level Level 25 exhibits considerably less variability than the other three levels.7 77. Chapter 3 Exercise Solutions 3-31 continued (c) Normal Probability Plot of the Residuals (response is Compressive Strength (Ex3-31Str)) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -150 -100 -50 0 Residual 50 100 150 The normal distribution assumption for compressive strength is reasonable. 3-31 . 3-32 . Rodding Level 10 15 20 25 scale factor = Mean Compressive Strength 1500 1587 1607 1500 MSE n = 5117 3 = 41 S c a le d t D is tr ib u tio n (10 . 25 ) (1 5 ) (2 0 ) 1418 1459 1500 1541 1582 1623 1664 M e a n C o m p re s s iv e S tr e n g th There is no difference due to rodding level.Chapter 3 Exercise Solutions 3-32. 3-33 .497 (----------*----------) --------+---------+---------+---------+41.Chapter 3 Exercise Solutions 3-33. Normal plot of residuals One-way ANOVA: Ex3-33Den versus Ex3-33T Source DF Ex3-33T 3 Error 20 Total 23 S = 0.583 0.2 0.57% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev --------+---------+---------+---------+500 6 41.2 0.25 41.6 0.8 Normality assumption is reasonable.8 -0.553 R-Sq = 17.00 Pooled StDev = 0.4 -0.700 0.89% R-Sq(adj) = 5.141 (----------*----------) 525 6 41. (a) MTB > Stat > ANOVA > One-Way > Graphs> Boxplots of data.6 -0.3238 SS MS F P 0. (b) Normal Probability Plot of the Residuals (response is Baked Density (Ex3-33Den)) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -0.45 0.324 Temperature level does not significantly affect mean baked anode density.258 2.152 1.75 42.0 Residual 0.339 (----------*----------) 575 6 41.097 0.333 0.450 0.4 0.457 0.194 (----------*----------) 550 6 41.105 2.50 41. select the temperature with the smallest variance. More uniform anodes are produced at lower temperatures.00 40.25 41. so does variability.00 Baked Desnity 41.50 -0.25 -0.50 Residual 0. MTB > Stat > ANOVA > One-Way > Graphs> Residuals versus the Variables Residuals Versus Firing Temperature (Ex3-33T) (response is Baked Density (Ex3-33Den)) 0. 3-34. which probably also incurs the smallest cost (lowest temperature).Chapter 3 Exercise Solutions 3-33 continued (c) Boxplot of Baked Density by Firing Temperature 42.75 41.75 500 510 520 530 540 550 Temperature (deg C) 560 570 580 As firing temperature increases.25 0. 500°C (see Boxplot).50 41.50 500 525 550 Firing Temperature. deg C 575 Since statistically there is no evidence to indicate that the means are different. 3-34 . Recommend 500°C for smallest variability.75 40.00 -0. 65% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ----+---------+---------+---------+----0.85 0.68 30.02 4 71.711 Orifice size does affect mean % radon release.750 2.40 1.35 23 1265.40 4 65.55% R-Sq(adj) = 86.4 orifice diameters.000 3.062 (---*---) 0.71 1.754 (---*---) ----+---------+---------+---------+----63.37 4 82.309 (---*---) 0.38 226.37 0.63 R-Sq = 89.0 84.750 2.99 Smallest % radon released at 1.99 and 1. Boxplot of Radon Released by Orifice Diameter Radon Released.826 (---*---) 1.99 4 62.51 4 77. at α = 0.0 Pooled StDev = 2.02 Orifice Diameter 1.559 (---*---) 1. (a) MTB > Stat > ANOVA > One-Way > Graphs> Boxplots of data One-way ANOVA: Ex3-35Rad versus Ex3-35Dia Source Ex3-35Dia Error Total S = 2.711 DF SS MS F P 5 1133.0 77. % (Ex3-35Rad) 85 80 75 70 65 60 0.000 2.51 0.000 18 132.05. 3-35 .71 4 75.750 3.Chapter 3 Exercise Solutions 3-35.000 1.25 7.0 70.304 (----*---) 1. 5 5.0 -2.0 60 65 70 75 Fitted Value--Radon Released 80 85 Variability in residuals does not appear to depend on the magnitude of predicted (or fitted) values.0 Residuals violate the normality distribution.99 The assumption of equal variance at each factor level appears to be violated.5 0.02 1.71 1.0 Residual 2.5 0.40.0 Residual 2.99). 1. 3-36 . Residuals Versus the Fitted Values (response is Radon Released (Ex3-35Rad)) 5. Residuals Versus Orifice Diameter (Ex3-35Dia) (response is Radon Released (Ex3-35Rad)) 5.0 Residual 2.5 -5.0 0.40 Orifice Diameter (Ex3-35Dia) 1. 1.5 0.5 -5. Residuals versus fits. with larger variances at the larger diameters (1. Residuals versus the Variables Normal Probability Plot of the Residuals (response is Radon Released ( Ex3-35Rad)) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -5.02.0 -2.Chapter 3 Exercise Solutions 3-35 continued (b) MTB > Stat > ANOVA > One-Way > Graphs> Normal plot of residuals.51 0.37 0.0 -2. 9267 0.3853 (------*------) 3 3 1.3962 n 12 (c) σˆ 2 = MSE = 0.437 S = 0. Boxplots of data One-way ANOVA: Ex3-36Un versus Ex3-36Pos Source DF SS MS F P Ex3-36Pos 3 16.3962 + 0.0484 3-37 .3067 1.407 8.8076 R-Sq = 75.3570 (------*------) --------+---------+---------+---------+1. (a) MTB > Stat > ANOVA > One-Way > Graphs.008 Error 8 5.3167 0.6522 2 σˆ uniformity = σˆτ2 + σˆ 2 = 0.4066 − 0. Boxplot of Uniformity by Wafer Position Film Thickness Uniformity (Ex3-36Un) 6 5 4 3 2 1 1 (b) σˆτ2 = 2 3 Wafer Position (Ex3-36Pos) 4 MSfactor − MSE 5.53% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev --------+---------+---------+---------+1 3 4.220 5.4366 (------*------) 4 3 1.0 Pooled StDev = 0.Chapter 3 Exercise Solutions 3-36.6522 = 1.66% R-Sq(adj) = 66.4636 (------*------) 2 3 1.652 Total 11 21.217 0.8076 There is a statistically significant difference in wafer position.7733 0.5 6.5 3.6522 = = 0. 3. 1 is different from 2.29 0.0 4. and 4. 0 -0. Residuals Versus the Fitted Values (response is Uniformity ( Ex3-36Un)) 1.5 0.5 0.5 1. Residuals versus the Variables Normal Probability Plot of the Residuals (response is Uniformity (Ex3-36Un)) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -2 -1 0 Residual 1 2 Normality assumption is probably not unreasonable.Chapter 3 Exercise Solutions 3-36 continued (d) MTB > Stat > ANOVA > One-Way > Graphs> Normal plot of residuals.0 -1.0 -1.0 3.0 Residual 0. 3-38 .5 The variability in residuals does appear to depend on the magnitude of predicted (or fitted) values.5 -1.5 3. Residuals Versus Wafer Position (Ex3-36Pos) (response is Film Thickness Uniformity (Ex3-36Un)) 1.0 2.5 Fitted Value--Film Thickness Uniformity 4.0 -0.5 -1.0 4.0 1.5 1. but there are two very unusual observations – the outliers at either end of the plot – therefore model adequacy is questionable. Residuals versus fits.5 1.0 Residual 0.5 1 2 3 Wafer Position (Ex3-36Pos) 4 Both outlier residuals are from wafer position 1.5 2. Chapter 4 Exercise Solutions Several exercises in this chapter differ from those in the 4th edition. An “*” following the exercise number indicates that the description has changed. New exercises are denoted with an “☺”. A second exercise number in parentheses indicates that the exercise number has changed. 4-1. “Chance” or “common” causes of variability represent the inherent, natural variability of a process - its background noise. Variation resulting from “assignable” or “special” causes represents generally large, unsatisfactory disturbances to the usual process performance. Assignable cause variation can usually be traced, perhaps to a change in material, equipment, or operator method. A Shewhart control chart can be used to monitor a process and to identify occurrences of assignable causes. There is a high probability that an assignable cause has occurred when a plot point is outside the chart's control limits. By promptly identifying these occurrences and acting to permanently remove their causes from the process, we can reduce process variability in the long run. 4-2. The control chart is mathematically equivalent to a series of statistical hypothesis tests. If a plot point is within control limits, say for the average x , the null hypothesis that the mean is some value is not rejected. However, if the plot point is outside the control limits, then the hypothesis that the process mean is at some level is rejected. A control chart shows, graphically, the results of many sequential hypothesis tests. NOTE TO INSTRUCTOR FROM THE AUTHOR (D.C. Montgomery): There has been some debate as to whether a control chart is really equivalent to hypothesis testing. Deming (see Out of the Crisis, MIT Center for Advanced Engineering Study, Cambridge, MA, pp. 369) writes that: “Some books teach that use of a control chart is test of hypothesis: the process is in control, or it is not. Such errors may derail self-study”. Deming also warns against using statistical theory to study control chart behavior (falsealarm probability, OC-curves, average run lengths, and normal curve probabilities. Wheeler (see “Shewhart’s Charts: Myths, Facts, and Competitors”, ASQC Quality Congress Transactions (1992), Milwaukee, WI, pp. 533–538) also shares some of these concerns: “While one may mathematically model the control chart, and while such a model may be useful in comparing different statistical procedures on a theoretical basis, these models do not justify any procedure in practice, and their exact probabilities, risks, and power curves do not actually apply in practice.” 4-1 Chapter 4 Exercise Solutions 4-2 continued On the other hand, Shewhart, the inventor of the control chart, did not share these views in total. From Shewhart (Statistical Method from the Viewpoint of Quality Control (1939), U.S. Department of Agriculture Graduate School, Washington DC, p. 40, 46): “As a background for the development of the operation of statistical control, the formal mathematical theory of testing a statistical hypothesis is of outstanding importance, but it would seem that we must continually keep in mind the fundamental difference between the formal theory of testing a statistical hypothesis and the empirical theory of testing a hypothesis employed in the operation of statistical control. In the latter, one must also test the hypothesis that the sample of data was obtained under conditions that may be considered random. … The mathematical theory of distribution characterizing the formal and mathematical concept of a state of statistical control constitutes an unlimited storehouse of helpful suggestions from which practical criteria of control must be chosen, and the general theory of testing statistical hypotheses must serve as a background to guide the choice of methods of making a running quality report that will give the maximum service as time goes on.” Thus Shewhart does not discount the role of hypothesis testing and other aspects of statistical theory. However, as we have noted in the text, the purposes of the control chart are more general than those of hypothesis tests. The real value of a control chart is monitoring stability over time. Also, from Shewhart’s 1939 book, (p. 36): “The control limits as most often used in my own work have been set so that after a state of statistical control has been reached, one will look for assignable causes when they are not present not more than approximately three times in 1000 samples, when the distribution of the statistic used in the criterion is normal.” Clearly, Shewhart understood the value of statistical theory in assessing control chart performance. My view is that the proper application of statistical theory to control charts can provide useful information about how the charts will perform. This, in turn, will guide decisions about what methods to use in practice. If you are going to apply a control chart procedure to a process with unknown characteristics, it is prudent to know how it will work in a more idealized setting. In general, before recommending a procedure for use in practice, it should be demonstrated that there is some underlying model for which it performs well. The study by Champ and Woodall (1987), cited in the text, that shows the ARL performance of various sensitizing rules for control charts is a good example. This is the basis of the recommendation against the routine use of these rules to enhance the ability of the Shewhart chart to detect small process shifts. 4-2 Chapter 4 Exercise Solutions 4-3. Relative to the control chart, the type I error represents the probability of concluding the process is out of control when it isn't, meaning a plot point is outside the control limits when in fact the process is still in control. In process operation, high frequencies of false alarms could lead could to excessive investigation costs, unnecessary process adjustment (and increased variability), and lack of credibility for SPC methods. The type II error represents the probability of concluding the process is in control, when actually it is not; this results from a plot point within the control limits even though the process mean has shifted out of control. The effect on process operations of failing to detect an out-of-control shift would be an increase in non-conforming product and associated costs. 4-4. The statement that a process is in a state of statistical control means that assignable or special causes of variation have been removed; characteristic parameters like the mean, standard deviation, and probability distribution are constant; and process behavior is predictable. One implication is that any improvement in process capability (i.e., in terms of non-conforming product) will require a change in material, equipment, method, etc. 4-5. No. The fact that a process operates in a state of statistical control does not mean that nearly all product meets specifications. It simply means that process behavior (mean and variation) is statistically predictable. We may very well predict that, say, 50% of the product will not meet specification limits! Capability is the term, which refers to the ability to meet product specifications, and a process must be in control in order to calculate capability. 4-6. The logic behind the use of 3-sigma limits on Shewhart control charts is that they give good results in practice. Narrower limits will result in more investigations for assignable causes, and perhaps more false alarms. Wider limits will result in fewer investigations, but perhaps fewer process shifts will be promptly identified. Sometimes probability limits are used - particularly when the underlying distribution of the plotted statistic is known. If the underlying distribution is unknown, care should be exercised in selecting the width of the control limits. Historically, however, 3-sigma limits have been very successful in practice. 4-3 Chapter 4 Exercise Solutions 4-7. Warning limits on control charts are limits that are inside the control limits. When warning limits are used, control limits are referred to as action limits. Warning limits, say at 2-sigma, can be used to increase chart sensitivity and to signal process changes more quickly than the 3-sigma action limits. The Western Electric rule, which addresses this type of shift is to consider a process to be out of control if 2 of 3 plot points are between 2 sigma and 3 sigma of the chart centerline. 4-8. The concept of a rational subgroup is used to maximize the chance for detecting variation between subgroups. Subgroup samples can be structured to identify process shifts. If it is expected that a process will shift and stay at the new level until a corrective action, then sampling consecutive (or nearly) units maximizes the variability between subgroups and minimizes the variability within a subgroup. This maximizes the probability of detecting a shift. 4-9. I would want assignable causes to occur between subgroups and would prefer to select samples as close to consecutive as possible. In most SPC applications, process changes will not be self-correcting, but will require action to return the process to its usual performance level. The probability of detecting a change (and therefore initiating a corrective action) will be maximized by taking observations in a sample as close together as possible. 4-10. This sampling strategy will very likely underestimate the size of the true process variability. Similar raw materials and operating conditions will tend to make any fivepiece sample alike, while variability caused by changes in batches or equipment may remain undetected. An out-of-control signal on the R chart will be interpreted to be the result of differences between cavities. Because true process variability will be underestimated, there will likely be more false alarms on the x chart than there should be. 4-4 Chapter 4 Exercise Solutions 4-11. (a) No. (b) The problem is that the process may shift to an out-of-control state and back to an incontrol state in less than one-half hour. Each subgroup should be a random sample of all parts produced in the last 2½ hours. 4-12. No. The problem is that with a slow, prolonged trend upwards, the sample average will tend to be the value of the 3rd sample --- the highs and lows will average out. Assume that the trend must last 2½ hours in order for a shift of detectable size to occur. Then a better sampling scheme would be to simply select 5 consecutive parts every 2½ hours. 4-13. No. If time order of the data is not preserved, it will be impossible to separate the presence of assignable causes from underlying process variability. 4-14. An operating characteristic curve for a control chart illustrates the tradeoffs between sample size n and the process shift that is to be detected. Generally, larger sample sizes are needed to increase the probability of detecting small changes to the process. If a large shift is to be detected, then smaller sample sizes can be used. 4-15. The costs of sampling, excessive defective units, and searches for assignable causes impact selection of the control chart parameters of sample size n, sampling frequency h, and control limit width. The larger n and h, the larger will be the cost of sampling. This sampling cost must be weighed against the cost of producing non-conforming product. 4-16. Type I and II error probabilities contain information on statistical performance; an ARL results from their selection. ARL is more meaningful in the sense of the operations information that is conveyed and could be considered a measure of the process performance of the sampling plan. 4-5 Chapter 4 Exercise Solutions 4-17. Evidence of runs, trends or cycles? NO. There are no runs of 5 points or cycles. So, we can say that the plot point pattern appears to be random. 4-18. Evidence of runs, trends or cycles? YES, there is one "low - high - low - high" pattern (Samples 13 – 17), which might be part of a cycle. So, we can say that the pattern does not appear random. 4-19. Evidence of runs, trends or cycles? YES, there is a "low - high - low - high - low" wave (all samples), which might be a cycle. So, we can say that the pattern does not appear random. 4-20. Three points exceed the 2-sigma warning limits - points #3, 11, and 20. 4-21. Check: • Any point outside the 3-sigma control limits? NO. • 2 of 3 beyond 2 sigma of centerline? NO. • 4 of 5 at 1 sigma or beyond of centerline? YES. Points #17, 18, 19, and 20 are outside the lower 1-sigma area. • 8 consecutive points on one side of centerline? NO. One out-of-control criteria is satisfied. 4-22. Four points exceed the 2-sigma warning limits - points #6, 12, 16, and 18. 4-23. Check: • Any point outside the 3-sigma control limits? NO. (Point #12 is within the lower 3-sigma control limit.) • 2 of 3 beyond 2 sigma of centerline? YES, points #16, 17, and 18. • 4 of 5 at 1 sigma or beyond of centerline? YES, points #5, 6, 7, 8, and 9. • 8 consecutive points on one side of centerline? NO. Two out-of-control criteria are satisfied. 4-6 Chapter 4 Exercise Solutions 4-24. The pattern in Figure (a) matches the control chart in Figure (2). The pattern in Figure (b) matches the control chart in Figure (4). The pattern in Figure (c) matches the control chart in Figure (5). The pattern in Figure (d) matches the control chart in Figure (1). The pattern in Figure (e) matches the control chart in Figure (3). 4-25 (4-30). Many possible solutions. MTB > Stat > Quality Tools > Cause-and-Effect Cause-and-Effect Diagram for Late Arrival Driv e Family "Turtle" Children/School Route Put out pet Accident Children/Homework Find badge, keys Fix breakfast A rriv e late to O ffice Errands Fix lunch Eat breakfast Carpool Read paper Dress Gas Shower Get up late Activ ities Coffee Stops 4-7 Chapter 4 Exercise Solutions 4-26 (4-31). Many possible solutions. MTB > Stat > Quality Tools > Cause-and-Effect Cause-and-Effect Diagram for Car Accident Driv er Car A sleep Tires Drunk Brakes M isjudgment S uspension Talking on cell phone D istracted S teering S tate of Repair Raining Blocked P oor v isibility Windy W eather Out-of-contr ol car strikes tree Icy /snow -cov ered Road 4-8 Chapter 4 Exercise Solutions 4-27 (4-32). Many possible solutions. MTB > Stat > Quality Tools > Cause-and-Effect Cause-and-Effect Diagram for Damaged Glassware Glassware Glassware Pack aging Deliv ery Serv ice Handling Crushed Not enough padding Strength flaw Dropped Severe transport vibration Dropped Crushed Internal Handling Weak box Broken at start Glassware Damaged Droppped Carelessly packed Manufacturer Handling 4-9 Chapter 4 Exercise Solutions 4-28☺. Many possible solutions. MTB > Stat > Quality Tools > Cause-and-Effect Cause-and-Effect Diagram for Coffee-making Process Personnel Method Machine Worn-out C offee drinkers C leanliness E spresso drinkers Brew temperature Insufficient training A ge of brew Brew method C onsistently Bad C offee Ty pe of filter C offee grind A mount of w ater C offee roast C offee beans Water temperature Env ironment A mount of beans Measurement Water source Material 4-10 . 8 95.1 55. Example of a check sheet to collect data on personal opportunities for improvement. Defect Overeating Being Rude Not meeting commitments Missing class Etc. including defect categories and counts. Many possible solutions. beginning and end of process are shown below. Snooze No Check time Awake Yes 6:30am ? Get out of bed Arrive at work … 4-31☺.7 r he Ot 0 6 4. green is value-added activity. 4-11 .1 g in iss s as Cl e tM No 37 26.9 g in et C m om itm t en 19 13. Yellow is nonvalue-added activity.8 81.0 To reduce total count of defects. Many possible solutions. “Being Rude” represents the greatest opportunity to make an improvement.Chapter 4 Exercise Solutions 4-29☺.3 100. The next step would be to determine the causes of “Being Rude” and to work on eliminating those causes. 1 2 0 2 10 11 4 2 4 6 3 1 9 2 3 4 0 9 2 2 Month/Day 5 6 7 1 0 1 7 10 11 1 0 1 7 9 4 TOTAL 18 21 15 13 16 19 … … … … … 31 TOTAL 1 6 9 76 7 19 2 37 17 19 138 Pareto Chart of Personal Opportunities for Improvement 140 100 120 Count 80 60 60 40 40 20 20 0 Defect g in Be Count Percent Cum % Percent 80 100 de Ru M 76 55. 0027) (1 − 0. Cumulative Probability Cumulative Distribution Function Binomial with n = 5 and p = 0. the probability that at least one of the points is beyond the limits also increases.0027 x P( X <= x ) 0 0.0027) (1 − 0. 4-12 .0027) = 1 − 0.0027.0779 0 ⎝ ⎠ α1 = 1 − Pr{0 of 30 beyond} = 1 − ⎜ Cumulative Distribution Function Binomial with n = 30 and p = 0.986573 m = 10 ⎛10 ⎞ 0 10 ⎟ (0.9733 = 0.0027 x P( X <= x ) 0 0.0027) = 0.0027) (1 − 0. m=5 α1 = Pr{at least 1 out-of-control} = Pr{1 of 5 beyond} + Pr{2 of 5 beyond} + " + Pr{5 of 5 beyond} ⎛5⎞ = 1 − Pr{0 of 5 beyond} = 1 − ⎜ ⎟ (0.Chapter 4 Exercise Solutions 4-32☺.947363 m = 30 ⎛ 30 ⎞ 0 30 ⎟ (0.0134 ⎝0⎠ MTB > Calc > Probability Distributions > Binomial.1025 ⎝0⎠ α1 = 1 − Pr{0 of 50 beyond} = 1 − ⎜ Cumulative Distribution Function Binomial with n = 50 and p = 0.0027 x P( X <= x ) 0 0.0526 ⎝0⎠ α1 = 1 − Pr{0 of 20 beyond} = 1 − ⎜ Cumulative Distribution Function Binomial with n = 20 and p = 0.0027)5 = 1 − 0.0027)0 (1 − 0.0027 x P( X <= x ) 0 0.922093 m = 50 ⎛ 50 ⎞ 0 50 ⎟ (0.0027 x P( X <= x ) 0 0.0027) = 0.0027) = 0.973326 m = 20 ⎛ 20 ⎞ 0 20 ⎟ (0.0267 ⎝0⎠ α1 = 1 − Pr{0 of 10 beyond} = 1 − ⎜ Cumulative Distribution Function Binomial with n = 10 and p = 0.873556 Although the probability that a single point plots beyond the control limits is 0.9866 = 0.0027) (1 − 0. as the number of samples increases (m). 4-13 . sampling frequencies are often designed to increase the likelihood of detecting a special or assignable cause. The lack of independence in the sample statistics will affect the estimates of the process population parameters. the points used to estimate these sample statistics are not independent—they do not reflect a random sample from a population. When the process mean µ and variance σ2 are unknown. they must be estimated by sample means x and standard deviations s.Chapter 4 Exercise Solutions 4-33☺. In fact. However. A2 = 0.00 R chart for Bearing ID (all samples in calculations) X-bar Chart for Bearing ID (all samples in calculations) 41.29 UCL R = D4 R = 2. 3.29 4 31.0 CL = 4. 2.71 LCL R = D3 R = 0(4. D3 = 0 x + x + " + xm 34. An “*” indicates that the description has changed.00 m 24 R + R2 + " + Rm 3 + 4 + " + 2 R= 1 = = 4.5 + 34. it is out of control.96 CL R = R = 4.0 UCL = 9.2 x= 1 2 = = 34.00 R x-bar 35. 5-1.00 − 0.2 + " + 34.0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Sample No.0 12 39. (a) for n = 5. This can be changed in dialog boxes or under Tools>Options>Control Charts and Quality Tools>Estimating Standard Deviation. MINITAB defines some sensitizing rules for control charts differently than the standard rules. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Sample No.71) = 9.96 UCL = 36. 5-1 .577(4. or (2) if a plot point is on or beyond the limits.0 6 33.00 + 0. The MINITAB convention for determining whether a point is out of control is: (1) if a plot point is within the control limits. 4.114.71) = 36.71) = 31. not 8. it is in control. a run of n consecutive points on one side of the center line is defined as 9 points.577.71 m 24 UCL x = x + A2 R = 34. New exercises are denoted with an “☺”. Several exercises in this chapter differ from those in the 4th edition.71 LCL = 31.72 8 CL = 34.00 LCL x = x − A2 R = 34.72 CL x = x = 34.0 2 29.577(4.71) = 0.0 LCL = 0 0 27. D4 = 2. In particular. A second exercise number in parentheses indicates that the exercise number has changed. This can be changed under Tools > Options > Control Charts and Quality Tools > Define Tests.Chapter 5 Exercise Solutions Notes: 1. MINITAB uses pooled standard deviation to estimate standard deviation for control chart limits and capability estimates.0 12 10 15 37.115(4. 5 / 2. R = 4.93 ⎠ ⎣ ⎝ 1.65.0 12 10 UCL = 9.50 31. Assuming an assignable cause is found for these two out-of-control points.CL R = 4. The new process parameter estimates are: x = 33.25.5.06 UCL R = 9.00 x-bar Chart for Bearing ID (samples 12.0 4 2 27. the two samples can be excluded from the control limit calculations.326 = 1.0 CL = 4.CL x = 33.25 8 CL = 33. 0 LCL = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Sample No.93 UCL x = 36.5.0 12 39.0 LCL = 31.29) = 0 + 1 − 0. 12 and Sample No.65.0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Sample No. 15 excluded) 41.52. 15 excluded) R chart for Bearing ID (samples 12.Chapter 5 Exercise Solutions 5-1 (a) continued The process is not in statistical control.65 ⎞ ⎤ = Φ⎜ ⎟ + ⎢1 − Φ ⎜ ⎟⎥ ⎝ 1.99950 = 0. (b) pˆ = Pr{x < LSL} + Pr{x > USL} = Pr{x < 20} + Pr{x > 40} = Pr{x < 20} + [1 − Pr{x < 40}] ⎛ 20 − 33. LCL R = 0. 15.65 ⎞ ⎡ ⎛ 40 − 33. x is beyond the upper control limit for both Sample No. LCL x = 31.06 29.00050 5-2 .93 ⎠ ⎦ = Φ (−7.52 15 37.65 R x-bar 35.07) + 1 − Φ (3.0 UCL =36.0 6 33. σˆ x = R / d 2 = 4. Actual specs are 350 ± 5 V.49 .33 10. (a) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Xbar-R Chart of Ex5-2V Sample M ean 15.00000 10.00000 * USL Sample Mean Sample N StDev(Within) StDev(Overall) Potential (Within) Capability Cp 5.62 50. (b) n = 4.00 -14 0 14 28 5.0 U C L=14.77 5.5 LC L=5.44 4. runs.49 Overall Capability Pp PPL PPU Ppk Cpm -42 Observed Performance PPM < LSL 0.25 / 2. trends. LSLT = –50 USL − LSL +50 − (−50) Cˆ P = = = 5.00 -28 Exp.Chapter 5 Exercise Solutions 5-2.00 PPM > USL 0.059 = 3.88 12. 6σˆ 6(3. Overall Performance PPM < LSL 0.00 PPM Total 0.33. or cycles.03545 3.36 4.49 CPL 6.00 PPM > USL 0.00 PPM Total 0. so the process is capable.5 _ _ X=10. x = 10.23 4. σˆ X = R / d 2 = 6.12282 CPU Cpk CCpk 4.36 5.25 4 0 LC L=0 2 4 6 8 10 Sample 12 14 16 18 20 The process is in statistical control with no out-of-control signals. Within Performance PPM < LSL 0.23 * 42 Exp.0 2 4 6 8 10 Sample 12 14 16 18 20 16 Sample Range U C L=14.00 PPM Total 0.035) MTB > Stat > Quality Tools > Capability Analysis > Normal Process Capability Analysis of Ex5-2V LSL LSL Target USL Within Ov erall Process Data -50. R = 6.25.00 PPM > USL 0. With xi = (observed voltage on unit i – 350) × 10: USLT = +50.0 7.26 12 8 _ R=6.035 .00 5-3 .32500 80 3.34 6. 5 50 LC L=0 0 2 4 6 8 10 Sample 12 14 16 18 20 The process is in statistical control with no out-of-control signals. 5-3.33 3.Chapter 5 Exercise Solutions 5-2 continued (c) MTB > Stat > Basic Statistics > Normality Test Probability Plot of Ex5-2V Normal 99.064 Percent 80 70 60 50 40 30 20 10 5 1 0.3 100 _ R=63.1 0 5 10 Ex5-2V 15 20 A normal probability plot of the transformed output voltage shows the distribution is close to normal.53 Sample M ean 40 20 _ _ X=10.704 0.73 2 4 6 8 10 Sample 12 14 16 18 20 150 Sample Range U C L=134.9 Mean StDev N AD P-Value 99 95 90 10. (a) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Xbar-R Chart of Ex5-3Dia U C L=47.9 0 -20 LC L=-25. 5-4 . trends. runs. or cycles.113 80 0. 35 1.22 .46 1.48 5-5 .00 PPM > USL 0. Cˆ P = 6σˆ x 6(27.5 / 2.Chapter 5 Exercise Solutions 5-3 continued (b) σˆ x = R / d 2 = 63.29384 1.00000 * 100. so the process is capable.17 1.09 1.17 * 90 Exp.326 = 27.22 Overall Capability Pp PPL PPU Ppk Cpm -90 Observed Performance PPM < LSL 0. Within Performance PPM < LSL 24.30009 Potential (Within) Capability Cp 1.09 1.00 -60 Exp. Overall Performance PPM < LSL 5.67 PPM Total 219.90000 100 27.79 PPM Total 574.00000 10.09 -30 0 30 60 1. LSL = –100 USL − LSL +100 − (−100) = = 1.22 CPL CPU Cpk CCpk 25.3 (c) USL = +100.81 PPM > USL 213.30 PPM > USL 549.3) MTB > Stat > Quality Tools > Capability Analysis > Normal Process Capability Analysis of Ex5-3Dia LSL USL Within Ov erall Process Data LSL Target USL Sample Mean Sample N StDev(Within) StDev(Overall) -100.00 PPM Total 0.32 1. Test Failed at points: 22 Test Results for R Chart of Ex5-4Th TEST 1.0012 _ R=0.Chapter 5 Exercise Solutions 5-4.062952 0.0024 U C L=0.002368 0. at points: 15 If graph is updated with new data.0018 0.0640 U C L=0.00092 0.0630 0.062011 1 2 4 6 8 10 12 14 Sample 16 18 20 22 24 1 Sample Range 0.0006 0.00 standard deviations from center line.0620 LC L=0.0635 _ _ X=0. One point more than 3.063893 0.0625 0. the results above may no longer be correct.00 standard deviations from center line. (a) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Xbar-R Chart of Thickness (Ex5-4Th) Sample M ean 0. Test Failed at points: 22 TEST 5. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). One Test Failed * WARNING * * point more than 3. 5-6 .0000 LC L=0 2 4 6 8 10 12 14 Sample 16 18 20 22 24 Test Results for Xbar Chart of Ex5-4Th TEST 1. failing tests on both the x and the R charts. Assuming assignable causes are found.0625 LCL=0. 22) and recalculate control limits.0620 1 2 4 6 8 10 12 14 Sample 16 18 20 22 24 1 0.0635 _ _ X=0. sample 14 is also out-of-control on the x chart. the new control limits are: Xbar-R Chart of Thickness (Ex5-4Th) (Samples 15.064408] 5-7 . Removing all three samples from calculation.06295 ± 3(0.000486 (c) Natural tolerance limits are: x ± 3σˆ x = 0.000486) = [0.0640 1 Sample Mean UCL=0.000823 0.063787 0.0000 LCL=0 2 4 6 8 10 12 14 Sample 16 18 20 22 24 (b) σˆ x = R / d 2 = 0.Chapter 5 Exercise Solutions 5-4 continued The process is out-of-control. With the revised limits.062104 0.0018 0. 0.0006 0.002118 0.0012 _ R=0. remove the out-of-control points (samples 15.061492.693 = 0. 14 removed from control limits calculations) 0.0630 0.062945 0. 22.0024 Sample Range UCL=0.000823 /1. 000486) MTB > Stat > Quality Tools > Capability Analysis > Normal Process Capability Analysis of Thickness (Ex5-4Th_w/o) (Estimated without Samples 14. and excluding samples 14.88 5-8 .00 PPM Total 15151.31 Exp. and 22 from the process capability estimation: USL − LSL +0.0015) = = 1.99 CPU 1.03 CPL 0.0628 0.06450 Sample Mean 0.90 * 0.0632 0.0616 0.03 Overall Capability Pp PPL PPU Ppk Cpm 0.Chapter 5 Exercise Solutions 5-4 continued (d) Assuming that printed circuit board thickness is normally distributed.028 Cˆ P = 6σˆ x 6(0.61 PPM > USL 689.0015 − (−0.52 Exp.33 PPM > USL 1814.90 0.00053 Potential (Within) Capability Cp 1.97 0. 22) LSL USL Within Ov erall Process Data LSL 0.94 0. Overall Performance PPM < LSL 3419.0624 0.0620 0.06295 Sample N 66 StDev(Within) 0. 15. 15. Within Performance PPM < LSL 1467.0636 0.70 PPM Total 2157.06150 Target * USL 0.52 PPM > USL 0.00049 StDev(Overall) 0.07 Cpk 0.99 CCpk 1.0640 0.55 PPM Total 5233.0644 Observed Performance PPM < LSL 15151. 037 Sample M ean 1.0 Sample StDev U C L=1.5 _ _ X=-0.2 3. trends. Estimate” select Sbar as method to estimate standard deviation.0 U C L=0.983 0.0 -0.5 _ _ X=-0. or cycles.5 LC L=0.302 1 2 3 4 5 6 7 8 Sample 9 10 11 12 13 14 15 The process is in statistical control.686 4.0 1 2 3 4 5 6 7 8 Sample 9 10 11 12 13 14 15 The process is in statistical control.5 _ R=3. runs.0 0. with no out-of-control signals.714 0.066 1.5 -1.5 LC L=0. trends.5 -1. Xbar-S Chart of Fill Volume (Ex5-5Vol) U C L=1.Chapter 5 Exercise Solutions 5-5. (a) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-S (Ex5-5Vol) Under “Options.0 -0. 5-9 .990 1 2 3 4 5 6 7 8 Sample 9 10 11 12 13 14 15 Sample Range 6.0 1.0 LC L=-1. (b) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R (Ex5-5Vol) Xbar-R Chart of Fill Volume (Ex5-5Vol) Sample M ean 1. runs. There is no difference in interpretation from the x − s chart. with no out-of-control signals.5 _ S =1.830 1.003 0.0 U C L=5.0 0.003 0. or cycles.0 LC L=-0.043 1 2 3 4 5 6 7 8 Sample 9 10 11 12 13 14 15 2. 0027 (not tabulated in textbook).15 −1 ) = 1.5 CL = 1. Copy results from the session window by holding down the keyboard “Alt” key.0662 (15 − 1) ( 4.010.5 s^2 (Variance) 2. selecting only the variance column.542 LCL = s 2 (n − 1) χ12−(α / 2).066 (15 − 1) ( 31. trends. sample 5 would be within the limits. To obtain sample standard deviations: Stat > Basic Statistics > Store Descriptive Statistics.32 ) = 2.0 1 2 3 4 5 6 7 8 9 Sample 10 11 12 13 14 15 Sample 5 signals out of control below the lower control limit.15−1 ) = 1.330 MINITAB’s control chart options do not include an s2 or variance chart.Chapter 5 Exercise Solutions 5-5 continued (c) Let α = 0. n = 15.5 LCL = 0. and “By Variables” is the sample ID column (Ex5-5Sample). “Variables” is column with sample data (Ex5-5Vol). and there would be no difference in interpretation from either the x − s or the x−R chart.010 / 2. In “Statistics” select “Variance”.n −1 = 1. or cycles. CL = s 2 = 1.0662 (15 − 1) ( χ12−(0. Otherwise there are no runs.0662 (15 − 1) ( χ 0.066.0662 = 1. and then copying & pasting to an empty worksheet column (results in Ex5-5Variance).n −1 = 1.0 0. s = 1. 5-10 .010 / 2). To construct an s2 control chart. Graph > Time Series Plot > Simple Control limits can be added using: Time/Scale > Reference Lines > Y positions Control Chart for Ex5-5Variance UCL = 2.136 2 2 UCL = s 2 (n − 1) χα2 / 2.0 1. If the limits had been calculated using α = 0.136 1.07 ) = 0. Results are displayed in the session window. first calculate the sample standard deviations and then create a time series plot.33 0.542 2. 475 / 2.25 0. runs. trends. σˆ x = R / d 2 = 0.475.268 16. (b) n = 5.268.475 0.4 _ _ X=16.00 0.004 Sample Range 1. or cycles.326 = 0.9940 2 4 6 8 10 Sample 12 14 16 18 20 U C L=1.Chapter 5 Exercise Solutions 5-6.00 LC L=0 2 4 6 8 10 Sample 12 14 16 18 20 The process is in statistical control with no out-of-control signals. x = 16.5420 16.6 Sample M ean U C L=16.50 0.0 LC L=15. R = 0.2 16.204 5-11 .75 _ R=0. (a) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Xbar-R Chart of Net Weight (Ex5-6Wt) 16. 27 0.4 16.6 MTB > Graph > Probability Plot > Single (Ex5-6Wt) Probability Plot of Net Weight (Ex5-6Wt) Normal .75 16.0 16.257 <0.the histogram has one mode.95% CI 99.50 16. 5-12 .2 Ex5-6Wt 16.50 15. and is approximately symmetrical with a bell shape.005 80 70 60 50 40 30 20 10 5 1 0.25 Ex5-6Wt 16.1 15. Points on the normal probability plot generally fall along a straight line.00 Visual examination indicates that fill weights approximate a normal distribution .Chapter 5 Exercise Solutions 5-6 continued (c) MTB > Graph > Histogram > Single (Ex5-6Wt) Histogram of Net Weight (Ex5-6Wt) 20 Frequency 15 10 5 0 15.75 17.2014 100 1.9 Mean StDev N AD P-Value 99 Percent 95 90 16.8 16.00 16. 26800 Sample N 100 StDev(Within) 0.71 * 16.96 (e) ⎛ 15.82 .7} = Φ ⎜ ⎟ = Φ (−2.93 CPU 0.82 Overall Capability Pp PPL PPU Ppk Cpm 15. Overall Performance PPM < LSL 2458.82 CPL 0.5) = = 0.71 0.204 ⎝ ⎠ The MINITAB process capability analysis also reports Exp.0027 0.73 PPM Total 18673.94 0.4 0.0 Exp.2 16.23 PPM > USL 16215.71 CCpk 0.73 PPM Total 18673.41 PPM Total 19902. so the process is not capable of meeting Cˆ P 6σˆ x 6(0. MTB > Stat > Quality Tools > Capability Analysis > Normal Under “Estimate” select Rbar as method to estimate standard deviation.20196 Potential (Within) Capability Cp 0.204) specifications.00 16.268 ⎞ pˆ lower = Pr{x < LSL} = Pr{x < 15.7 − 16.96 5-13 .23 PPM > USL 16215.83 0.00 PPM Total 0. "Overall" Performance PPM < LSL 2458.78) = 0.61 16.6 Exp.70000 Sample Mean 16.70000 Target * USL 16. Process Capability Analysis of Net Weight (Ex5-6Wt) LSL USL Within Ov erall Process Data LSL 15.Chapter 5 Exercise Solutions 5-6 continued (d) USL − LSL +0.20 PPM > USL 17196.71 Cpk 0.20421 StDev(Overall) 0.8 Observed Performance PPM < LSL 0. Within Performance PPM < LSL 2706.5 − (−0.00 PPM > USL 0. 0 1. runs.73 12.5 _ S =2. or cycles. trends.23 15 LC L=0 0 2 4 6 8 10 Sample 12 14 16 18 20 The process is in statistical control with no out-of-control signals. 5-14 .11 2 4 6 8 10 Sample 12 14 16 18 20 60 Sample StDev U C L=52.0 4.0 LC L=0 2 4 6 8 10 Sample 12 14 16 18 20 The process is in statistical control with no out-of-control signals.33 10.Chapter 5 Exercise Solutions 5-7.0 2 4 6 8 10 Sample 12 14 16 18 20 U C L=6. MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-S (Ex5-3Dia) Xbar-S Chart of Deviations from Nominal Diameter (Ex5-3Dia) U C L=46.91 Sample M ean 40 20 _ _ X=10. runs.0 U C L=14.9 0 -20 LC L=-25.0 7.125 Sample StDev 6. MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-S (Ex5-2Vl) Xbar-S Chart of Output Voltage (Ex5-2V) Sample M ean 15. 5-8.703 3. trends. or cycles.92 5.71 45 30 _ S =25.5 0.5 LC L=5.5 _ _ X=10. 01458 Sample M ean 74.00118 74.99 LC L=73.02324 0. (b) The control limits on the x charts in Example 5-3 were calculated using S to estimate σ. (a) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R (Ex5-9ID) Xbar-R Chart of Inner Diameter (Ex5-9ID) U C L=74.036 _ R=0.Chapter 5 Exercise Solutions 5-9☺.00 73. 5-15 . in this exercise R was used to estimate σ.012 0. trends.98777 2 4 6 8 10 12 14 Sample 16 18 20 22 24 U C L=0.04914 Sample Range 0.048 0. They will not always be the same. or cycles. runs.024 0. and in general.01 _ _ X=74. the x control limits based on S will be slightly different than limits based on R.000 LC L=0 2 4 6 8 10 12 14 Sample 16 18 20 22 24 The process is in statistical control with no out-of-control signals. 95} + [1 − Pr{x < 74.59 1.71 CPU 1.63 Cpk 1.05 − 74.009991) specifications. Within Performance PPM < LSL 0.980 73.63 1.95 − 74.950 73.05 − 73.66 Exp. MTB > Stat > Quality Tools > Capability Analysis > Normal Under “Estimate” select Rbar as method to estimate standard deviation. Overall Performance PPM < LSL 0.95000 Target * USL 74.67 CPL 1.00 PPM Total 0.00 Exp.00 PPM > USL 0.995 74.01022 Potential (Within) Capability Cp 1.28 PPM > USL 0.00118 Sample N 125 StDev(Within) 0.02324 / 2.123) + 1 − Φ (4.05} = Pr{x < 73.51 PPM Total 0.965 73.67 1.009991 .00999 StDev(Overall) 0.025 74.040 Observed Performance PPM < LSL 0.05}] ⎛ 73. Process Capability Analysis of Inner Diameter (Ex5-9ID) LSL USL Within Ov erall Process Data LSL 73.00118 ⎞ ⎡ ⎛ 74.16 pˆ = Pr{x < LSL} + Pr{x > USL} = Pr{x < 73.326 = 0.63 CCpk 1.67 Overall Capability Pp PPL PPU Ppk Cpm 1.05000 Sample Mean 74.010 74.00118 ⎞ ⎤ = Φ⎜ ⎟ + ⎢1 − Φ ⎜ ⎟⎥ 0.15 PPM > USL 0. so the process is not capable of meeting USL − LSL 74.009991 ⎝ ⎠ ⎣ ⎝ ⎠⎦ = Φ (−5.95 = = 1.59 * 73.89 PPM Total 1.009991 0.668 Cˆ P = 6σˆ x 6(0.886) = 0 +1−1 =0 5-16 .95} + Pr{x > 74.Chapter 5 Exercise Solutions 5-9 continued (c) σˆ x = R / d 2 = 0. One point more than 3.000 LC L=0 4 8 12 16 20 Sample 24 28 32 36 40 Test Results for Xbar Chart of Ex5-10ID TEST 1.00118 74.01 0. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL).01458 74. 40 TEST 6. 39.99 LC L=73.98777 4 8 12 16 20 Sample 24 28 32 36 40 U C L=0. increasing the process mean. 39 TEST 5. Test Failed at points: 38.024 0.Chapter 5 Exercise Solutions 5-10☺. Test Failed at points: 35. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL).04914 0. 38.00 73. 38. 39. 37.048 Sample Range U C L=74.012 0. 40 The control charts indicate that the process is in control. Test Failed at points: 37.02324 0.00 standard deviations from center line. until the x -value from the 37th sample is plotted.02 1 5 5 _ _ X=74. Since this point and the three subsequent points plot above the upper control limit. 5-17 .036 _ R=0. an assignable cause has likely occurred. MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R (Ex5-10ID) Xbar-R Chart of Inner Diameter (Ex5-10ID) 1 1 Sample M ean 74. LSL +5. σˆ x ≈ 1. B5 = 0.534) = [35.949(10) = 89.669(10) = 16. (e) First.76 5-12* (5-10).068 UCL R = D4 R = 2.016 LCL R = D3 R = 0(4) = 0 (b) natural tolerance limits: x ± 3σˆ x = x ± 3 ( R / d 2 ) = 40 ± 3(4 / 2. ∑ Ri = 200. center the process at 41. n = 10.0 − (−5.533) = 0.276(10) = 2. 50 50 i =1 i =1 n = 6 items/sample.49 LCL x = µ − Aσ x = 80 − 0. µ = 80 in-lb.000005 ⎝ 1.0005%. R = i =1 = =4 m m 50 50 UCL x = x + A2 R = 40 + 0. m = 50 samples (a) 50 ∑ xi 50 ∑ Ri 2000 200 = 40.0) = = 1. σ x = 10 in-lb.433) = 1 − 0.004(4) = 8. ⎝ 1. 5-18 .949.Chapter 5 Exercise Solutions 5-11 (5-9).932 x= i =1 = LCL x = x − A2 R = 40 − 0.264.276 centerline x = µ = 80 UCL x = µ + Aσ x = 80 + 0.69 LCL S = B5σ x = 0.483(4) = 38. so the process is not capable. B6 = 1.579) (d) ⎛ 36 − 40 ⎞ pˆ scrap = Pr{x < LSL} = Pr{x < 36} = Φ ⎜ ⎟ = Φ (−2. Cˆ P = 6σˆ x 6(1.57%. to reduce scrap and rework costs.736] (c) USL .253 . or 0. 44.51 centerline = c4σ x = 0. reduce variability such that the natural process tolerance limits are closer to. not 40.0057 .9727(10) = 9. and A = 0.579 ⎠ or 0.949(10) = 70.483(4) = 41. ∑ xi = 2000. Second.056 .669.579 ⎠ ⎛ 47 − 40 ⎞ pˆ rework = Pr{x > USL} = 1 − Pr{x < USL} = 1 − Φ ⎜ ⎟ = 1 − Φ (4.727 S UCL S = B6σ x = 1. say.999995 = 0. Chapter 5 Exercise Solutions 5-13* (5-11). 50 50 i =1 i =1 n = 4 items/subgroup; ∑ xi = 1000; ∑ Si = 72; m = 50 subgroups (a) 50 x= ∑ xi i =1 m = 1000 = 20 50 50 ∑ Si 72 = 1.44 50 m UCL x = x + A3 S = 20 + 1.628(1.44) = 22.34 S= i =1 = LCL x = x − A3 S = 20 − 1.628(1.44) = 17.66 UCL S = B4 S = 2.266(1.44) = 3.26 LCL S = B3 S = 0(1.44) = 0 (b) ⎛S ⎞ ⎛ 1.44 ⎞ natural process tolerance limits: x ± 3σˆ x = x ± 3 ⎜ ⎟ = 20 ± 3 ⎜ ⎟ = [15.3, 24.7] ⎝ 0.9213 ⎠ ⎝ c4 ⎠ (c) USL - LSL +4.0 − (−4.0) = = 0.85 , so the process is not capable. Cˆ P = 6σˆ x 6(1.44 / 0.9213) (d) ⎛ 23 − 20 ⎞ pˆ rework = Pr{x > USL} = 1 − Pr{x ≤ USL} = 1 − Φ ⎜ ⎟ = 1 − Φ (1.919) = 1 − 0.9725 = 0.0275 ⎝ 1.44 / 0.9213 ⎠ or 2.75%. ⎛ 15 − 20 ⎞ pˆ scrap = Pr{x < LSL} = Φ ⎜ ⎟ = Φ (−3.199) = 0.00069 , or 0.069% ⎝ 1.44 / 0.9213 ⎠ Total = 2.88% + 0.069% = 2.949% (e) ⎛ 23 − 19 ⎞ pˆ rework = 1 − Φ ⎜ ⎟ = 1 − Φ (2.56) = 1 − 0.99477 = 0.00523 , or 0.523% ⎝ 1.44 / 0.9213 ⎠ ⎛ 15 − 19 ⎞ pˆ scrap = Φ ⎜ ⎟ = Φ (−2.56) = 0.00523 , or 0.523% ⎝ 1.44 / 0.9213 ⎠ Total = 0.523% + 0.523% = 1.046% Centering the process would reduce rework, but increase scrap. A cost analysis is needed to make the final decision. An alternative would be to work to improve the process by reducing variability. 5-19 Chapter 5 Exercise Solutions 5-14 (5-12). (a) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Xbar-R Chart of Critical Dimension (Ex5-14ax1, ..., Ex5-14ax5) U C L=154.45 Sample M ean 150 140 _ _ X=130.88 130 120 110 LC L=107.31 2 4 6 8 10 Sample 12 14 16 18 20 U C L=86.40 Sample Range 80 60 _ R=40.86 40 20 0 LC L=0 2 4 6 8 10 Sample 12 14 16 18 20 The process is in statistical control with no out-of-control signals, runs, trends, or cycles. (b) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Under “Options, Estimate” use subgroups 1:20 to calculate control limits. Xbar-R Chart of Critical Dimension (Ex5-14bx1, ..., Ex5-14bx5) 1 Sample M ean 180 1 1 1 1 1 1 1 1 1 160 U C L=154.45 140 _ _ X=130.88 120 LC L=107.31 100 3 6 9 12 15 Sample 18 21 24 27 30 U C L=86.40 Sample Range 80 60 _ R=40.86 40 20 0 LC L=0 3 6 9 12 15 Sample 18 21 24 27 30 Starting at Sample #21, the process average has shifted to above the UCL = 154.45. 5-20 Chapter 5 Exercise Solutions 5-14 continued (c) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Under “Options, Estimate” use subgroups 1:20 to calculate control limits. Xbar-R Chart of Critical Dimension (Ex5-14cx1, ..., Ex5-14cx5) 1 Sample M ean 180 1 1 1 1 1 1 1 1 160 1 U C L=154.45 140 _ _ X=130.88 120 2 6 2 5 LC L=107.31 1 100 4 8 12 16 20 Sample 24 28 32 36 40 U C L=86.40 Sample Range 80 60 _ R=40.86 40 20 0 LC L=0 4 8 12 16 20 Sample 24 28 32 36 40 The adjustment overcompensated for the upward shift. The process average is now between x and the LCL, with a run of ten points below the centerline, and one sample (#36) below the LCL. 5-21 Chapter 5 Exercise Solutions 5-15* (5-13). (a) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Xbar-R Chart of Strength Test (Ex5-15aSt) Sample M ean 85.0 U C L=84.58 82.5 _ _ X=79.53 80.0 77.5 75.0 LC L=74.49 2 4 6 8 10 Sample 12 14 16 18 20 Sample Range 20 U C L=18.49 15 _ R=8.75 10 5 0 LC L=0 2 4 6 8 10 Sample 12 14 16 18 20 Yes, the process is in control—though we should watch for a possible cyclic pattern in the averages. 5-22 Chapter 5 Exercise Solutions 5-15 continued (b) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Under “Options, Estimate” use subgroups 1:20 to calculate control limits. Xbar-R Chart of Strength Test (Ex5-15bSt) 1 Sample M ean 85.0 8 82.5 8 U C L=84.58 8 _ _ X=79.53 80.0 77.5 75.0 LC L=74.49 1 3 6 9 12 15 18 Sample 21 24 27 30 30 33 1 1 Sample Range 1 1 1 1 1 1 20 U C L=18.49 2 _ R=8.75 10 0 LC L=0 3 6 9 12 15 18 Sample 21 24 27 30 33 Test Results for R Chart of Ex5-15bSt TEST Test TEST Test 1. One point more than 3.00 standard deviations from center line. Failed at points: 25, 26, 27, 31, 33, 34, 35 2. 9 points in a row on same side of center line. Failed at points: 32, 33, 34, 35 A strongly cyclic pattern in the averages is now evident, but more importantly, there are several out-of-control points on the range chart. 5-23 Chapter 5 Exercise Solutions 5-16 (5-14). (a) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-S Xbar-S Chart of Strength Test (Ex5-15aSt) Original Data Sample Mean 85.0 UCL=84.64 82.5 _ _ X=79.53 80.0 77.5 75.0 LCL=74.43 2 4 6 8 10 Sample 12 14 16 18 20 Sample StDev 8 UCL=7.468 6 _ S=3.575 4 2 0 LCL=0 2 4 6 8 10 Sample 12 14 16 18 20 Under “Options, Estimate” use subgroups 1:20 to calculate control limits. Xbar-S Chart of Strength Test (Ex5-15bSt) Original plus New Data 1 Sample Mean 85.0 UCL=84.64 82.5 _ _ X=79.53 80.0 77.5 75.0 LCL=74.43 1 3 6 9 12 15 18 Sample 21 24 27 30 1 33 1 1 Sample StDev 10.0 1 1 7.5 1 1 1 1 UCL=7.47 5.0 _ S=3.57 2.5 LCL=0 0.0 3 6 9 12 15 18 Sample 21 24 27 30 33 Test Results for Xbar Chart of Ex5-15bSt TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 24, 31, 34 Test Results for S Chart of Ex5-15bSt TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 22, 25, 26, 27, 31, 33, 34, 35 5-24 Chapter 5 Exercise Solutions 5-16 continued (b) Yes, the s chart detects the change in process variability more quickly than the R chart did, at sample #22 versus sample #24. 5-17 (5-15). nold = 5; xold = 34.00; Rold = 4.7 (a) for nnew = 3 ⎡ d 2(new ) ⎤ ⎡ 1.693 ⎤ UCL x = xold + A2(new ) ⎢ (4.7) = 37.50 ⎥ Rold = 34 + 1.023 ⎢ ⎣ 2.326 ⎥⎦ ⎣⎢ d 2(old) ⎦⎥ ⎡ d 2(new ) ⎤ ⎡ 1.693 ⎤ LCL x = xold − A2(new ) ⎢ ⎥ Rold = 34 − 1.023 ⎢ ⎥ (4.7) = 30.50 ⎣ 2.326 ⎦ ⎢⎣ d 2(old) ⎥⎦ ⎡ d 2(new ) ⎤ ⎡ 1.693 ⎤ UCL R = D4(new) ⎢ ⎥ Rold = 2.574 ⎢ ⎥ (4.7) = 8.81 ⎣ 2.326 ⎦ ⎢⎣ d 2(old) ⎥⎦ ⎡d ⎤ ⎡ 1.693 ⎤ CL R = Rnew = ⎢ 2(new ) ⎥ Rold = ⎢ (4.7) = 3.42 ⎣ 2.326 ⎥⎦ ⎣⎢ d 2(old) ⎦⎥ ⎡d ⎤ ⎡ 1.693 ⎤ LCL R = D3(new) ⎢ 2(new ) ⎥ Rold = 0 ⎢ (4.7) = 0 ⎣ 2.326 ⎥⎦ ⎢⎣ d 2(old) ⎥⎦ (b) The x control limits for n = 5 are “tighter” (31.29, 36.72) than those for n = 3 (30.50, 37.50). This means a 2σ shift in the mean would be detected more quickly with a sample size of n = 5. 5-25 Chapter 5 Exercise Solutions 5-17 continued (c) for n = 8 ⎡ d 2(new ) ⎤ ⎡ 2.847 ⎤ UCL x = xold + A2(new ) ⎢ ⎥ Rold = 34 + 0.373 ⎢ ⎥ (4.7) = 36.15 ⎣ 2.326 ⎦ ⎣⎢ d 2(old) ⎦⎥ ⎡ d 2(new ) ⎤ ⎡ 2.847 ⎤ LCL x = xold − A2(new ) ⎢ (4.7) = 31.85 ⎥ Rold = 34 − 0.373 ⎢ ⎣ 2.326 ⎥⎦ ⎢⎣ d 2(old) ⎥⎦ ⎡ d 2(new ) ⎤ ⎡ 2.847 ⎤ UCL R = D4(new) ⎢ (4.7) = 10.72 ⎥ Rold = 1.864 ⎢ ⎣ 2.326 ⎥⎦ ⎣⎢ d 2(old) ⎦⎥ ⎡d ⎤ ⎡ 2.847 ⎤ CL R = Rnew = ⎢ 2(new ) ⎥ Rold = ⎢ (4.7) = 5.75 ⎣ 2.326 ⎥⎦ ⎢⎣ d 2(old) ⎥⎦ ⎡d ⎤ ⎡ 2.847 ⎤ LCL R = D3(new) ⎢ 2(new ) ⎥ Rold = 0.136 ⎢ ⎥ (4.7) = 0.78 ⎣ 2.326 ⎦ ⎢⎣ d 2(old) ⎥⎦ (d) The x control limits for n = 8 are even "tighter" (31.85, 36.15), increasing the ability of the chart to quickly detect the 2σ shift in process mean. 5-18☺. nold = 5, xold = 74.001, Rold = 0.023, nnew = 3 ⎡ d 2(new ) ⎤ ⎡ 1.693 ⎤ UCL x = xold + A2(new ) ⎢ ⎥ Rold = 74.001 + 1.023 ⎢ ⎥ (0.023) = 74.018 ⎣ 2.326 ⎦ ⎣⎢ d 2(old) ⎦⎥ ⎡ d 2(new ) ⎤ ⎡ 1.693 ⎤ LCL x = xold − A2(new ) ⎢ (0.023) = 73.984 ⎥ Rold = 74.001 − 1.023 ⎢ ⎣ 2.326 ⎥⎦ ⎢⎣ d 2(old) ⎥⎦ ⎡ d 2(new ) ⎤ ⎡ 1.693 ⎤ UCL R = D4(new) ⎢ (0.023) = 0.043 ⎥ Rold = 2.574 ⎢ ⎣ 2.326 ⎥⎦ ⎣⎢ d 2(old) ⎥⎦ ⎡d ⎤ ⎡ 1.693 ⎤ CL R = Rnew = ⎢ 2(new ) ⎥ Rold = ⎢ (0.023) = 0.017 ⎣ 2.326 ⎥⎦ ⎢⎣ d 2(old) ⎥⎦ ⎡d ⎤ ⎡ 1.693 ⎤ LCL R = D3(new) ⎢ 2(new ) ⎥ Rold = 0 ⎢ (0.023) = 0 ⎣ 2.326 ⎥⎦ ⎣⎢ d 2(old) ⎦⎥ 5-26 Chapter 5 Exercise Solutions 5-19 (5-16). 35 35 i =1 i =1 n = 7; ∑ xi = 7805; ∑ Ri = 1200; m = 35 samples (a) 35 x= ∑ xi i =1 m = 7805 = 223 35 35 ∑ Ri 1200 = 34.29 m 35 UCL x = x + A2 R = 223 + 0.419(34.29) = 237.37 R= i =1 = LCL x = x − A2 R = 223 − 0.419(34.29) = 208.63 UCL R = D4 R = 1.924(34.29) = 65.97 LCL R = D3 R = 0.076(34.29) = 2.61 (b) µˆ = x = 223; σˆ x = R / d 2 = 34.29 / 2.704 = 12.68 (c) USL − LSL +35 − (−35) = = 0.92 , the process is not capable of meeting Cˆ P = 6σˆ x 6(12.68) specifications. pˆ = Pr{x > USL} + Pr{x < LSL} = 1 − Pr{x < USL} + Pr{x < LSL} = 1 − Pr{x ≤ 255} + Pr{x ≤ 185} ⎛ 255 − 223 ⎞ ⎛ 185 − 223 ⎞ = 1− Φ ⎜ ⎟+ Φ⎜ ⎟ = 1 − Φ (2.52) + Φ (−3.00) = 1 − 0.99413 + 0.00135 = 0.0072 ⎝ 12.68 ⎠ ⎝ 12.68 ⎠ (d) The process mean should be located at the nominal dimension, 220, to minimize nonconforming units. ⎛ 255 − 220 ⎞ ⎛ 185 − 220 ⎞ pˆ = 1 − Φ ⎜ ⎟+Φ⎜ ⎟ = 1 − Φ (2.76) + Φ (−2.76) = 1 − 0.99711 + 0.00289 = 0.00578 ⎝ 12.68 ⎠ ⎝ 12.68 ⎠ 5-27 114(0.99506 + 0.58) + Φ (−3.155 ⎝ ⎠ ⎝ ⎠ = 1 − 0.00124 5-28 .23) + Φ (−3.577(0.36) = 0 (b) σˆ x = R / d 2 = 0.90 − 26.99938 + 0.155 0.36) = 0.36 m 25 UCL x = x + A2 R = 26.87) = 1 − 0.50.90 − 26.23) 0.50 − 0.40 ⎞ pˆ = 1 − Φ ⎜ ⎟+Φ⎜ ⎟ = 1 − Φ (3.00499 (c) ⎛ 26.50 = 26.36 / 2.00062 = 0.50 25 25 ∑ Ri 9.155 0.50 ⎞ ⎛ 25.Chapter 5 Exercise Solutions 5-20 (5-17). m = 25 samples (a) 25 x= ∑ xi i =1 m = 662.155 ⎝ ⎠ ⎝ ⎠ = 0. 25 25 i =1 i =1 n = 5.155 pˆ = Pr{x > USL} + Pr{x < LSL} = 1 − Pr{x ≤ USL} + Pr{x < LSL} ⎛ 26. ∑ xi = 662.00005 0.90 − 26.36) = 26.50 + 0.40 ⎞ ⎛ 25.50 ⎞ = 1− Φ ⎜ ⎟+Φ⎜ ⎟ = 1 − Φ (2.577(0. ∑ Ri = 9.71 R= i =1 = LCL x = x − A2 R = 26.00.90 − 26.36) = 26.00 = 0.76 LCL R = D3 R = 0(0.29 UCL R = D4 R = 2.326 = 0. 998) 5-29 .427(1. R = 9.79) 1.30.5) = 0 (c) Pr{in control} = Pr{LCL ≤ x ≤ UCL} = Pr{x ≤ UCL} − Pr{x ≤ LCL} ⎛ 22.99 is larger than the width of the tolerance band.326 = 3.60 (b) UCL x = x + A3 S = 20. S = 1.6 5 1.13 LCL S = B3 S = 0(1.8338 6σˆ x 6(3.089(1.Chapter 5 Exercise Solutions 5-21 (5-18).14 − 22 ⎞ ⎛ 17.0 + 1. X ~ N .6 5 ⎝ ⎠ ⎝ ⎠ = 0.9400 = 1. USL − LSL +10 − (−10) Cˆ P = = = 0.5) = 17.5 / 0.20) − Φ (−5.0 − 1.14 LCL x = x − A3 S = 20.5) = 3.84134 = 0.998 and 6σˆ x = 6(3. LSL=90 σˆ x = R / d 2 = 9. So.5.998) = 23. Pr{detect} = 1 − Pr{not detect} = 1 − [Pr{LCL ≤ x ≤ UCL}] = 1 − [Pr{x ≤ UCL} − Pr{x ≤ LCL}] ⎡ ⎛ UCL − µ ⎞ ⎛ LCL − µ x x new ⎟ − Φ ⎜ new = 1 − ⎢Φ ⎜ ⎢ ⎜ ⎟ ⎜ n n σ σ x x ⎠ ⎝ ⎣ ⎝ = 1 − Φ (7) + Φ (1) = 1 − 1 + 0. 2(10) = 20. not all of the output will meet specification. even if the mean is located at the nominal dimension.427(1.57926 5-22 (5-19).86 UCL S = B4 S = 2. n = 5. m = 50 samples (a) σˆ x = S / c4 = 1. 100.57926 − 0 = 0.5) = 22.84134 ⎞⎤ ⎡ ⎛ ⎞ ⎛ ⎞⎤ ⎟ ⎥ = 1 − ⎢Φ ⎜ 209 − 188 ⎟ − Φ ⎜ 191 − 188 ⎟ ⎥ ⎟⎥ ⎢⎣ ⎝ 6 4 ⎠ ⎝ 6 4 ⎠ ⎥⎦ ⎠⎦ 5-23 (5-20).30 / 2. USL=110. x = 104.0. x = 20. n = 5.86 − 22 ⎞ = Φ⎜ ⎟ −Φ⎜ ⎟ = Φ (0. 5) = 0 (c) centerline S = c4σ x = 0. These are standard values.606(2.99 UCL S = B6σ = 2.5) = 1. n = 2.5) = 15.5 .82 UCL R = D2σ = 3.5) = 0 5-30 .5) = 2.22 LCL R = D1σ = 0(2.Chapter 5 Exercise Solutions 5-24* (5-21).128(2.5) = 6.52 LCL S = B5σ = 0(2.121(2.5) = 4.686(2.121(2. µ = 10.7979(2. x (a) centerline x = µ = 10 UCL x = µ + Aσ x = 10 + 2.70 (b) centerline R = d 2σ x = 1.30 LCL x = µ − Aσ x = 10 − 2.5) = 9. σ = 2. x = 20. so the process is not capable of meeting Cˆ P = 6σˆ x 6(1.96 5 ⎠ ⎝ 1.37 UCL R = D4 R = 2.56) = 9. n = 5.56) = 17.56) = 0. (d) Pr{not detect} = Pr{LCL ≤ x ≤ UCL} = Pr{x ≤ UCL} − Pr{x ≤ LCL} ⎛ UCL x − µnew ⎞ ⎛ LCL x − µnew ⎞ ⎛ 22.56.64 LCL R = D3 R = 0(4.05938 − 0 = 0.05938 5-31 .37 − 24 ⎞ = Φ ⎜⎜ ⎟⎟ − Φ ⎜⎜ ⎟⎟ = Φ ⎜ ⎟−Φ⎜ ⎟ ⎝ 1. R = 4. m = 25 samples (a) UCL x = x + A2 R = 20 + 0.85 .577(4.56) = 22.326 = 1.96) specifications.96 5 ⎠ ⎝ σˆ x n ⎠ ⎝ σˆ x n ⎠ = Φ (−1.63 − 24 ⎞ ⎛ 17.96 (c) USL − LSL +5 − (−5) = = 0.56 / 2.Chapter 5 Exercise Solutions 5-25 (5-22).56) + Φ (−7.577(4.114(4.63 LCL x = x − A2 R = 20 − 0.56) = 0 (b) σˆ x = R / d 2 = 4. 00 standard deviations from center line.00 standard deviations from center line. Test Failed at points: 18 No additional subgroups are beyond the control limits.18 30 20 _ R=16.82 450 _ _ X=448.49 430 1 2 4 6 8 10 Sample 12 14 16 18 20 Sample Range 40 UCL=38. MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Sample M ean Xbar-R Chart of TiW Thickness (Ex5-26Th) 460 U C L=460.56 430 1 2 4 6 8 10 Sample 12 14 16 18 20 Sample Range 40 U C L=37.88 Sample Mean 460 _ _ X=449. Excluding subgroup 18 from control limits calculations: Xbar-R Chart of TiW Thickness (Ex5-26Th) Excluding subgroup 18 from calculations UCL=461.98 30 20 _ R=16. One point more than 3. Test Failed at points: 18 The process is out of control on the x chart at subgroup 18.68 450 440 LCL=437.69 440 LC L=436.65 10 0 LC L=0 2 4 6 8 10 Sample 12 14 16 18 20 Test Results for Xbar Chart of Ex5-26Th TEST 1. One point more than 3. so these limits can be used for future production.74 10 0 LCL=0 2 4 6 8 10 Sample 12 14 16 18 20 Test Results for Xbar Chart of Ex5-26Th TEST 1. 5-32 .Chapter 5 Exercise Solutions 5-26☺. 9 Mean StDev N AD P-Value 99 Percent 95 90 448.672 80 70 60 50 40 30 20 10 5 1 0.1 420 430 440 450 Ex5-26Th 460 470 480 A normal probability plot of the TiW thickness measurements shows the distribution is close to normal.13 (c) MTB > Stat > Basic Statistics > Normality Test Probability Plot of TiW Thickness (Ex5-26Th) Normal 99. 5-33 .111 80 0.Chapter 5 Exercise Solutions 5-26 continued (b) Excluding subgroup 18: x = 449.269 0.74 / 2.7 9.68 σˆ x = R / d 2 = 16.059 = 8. 30 440 450 460 470 1.00 PPM Total 0.00000 Sample Mean 448.05 * 480 Exp.92 248. σ x is estimated using R .38 PPM > USL PPM Total 53. LSL = –30 USL − LSL +30 − (−30) = = 1.24.18 CCpk 1.29 Cpk 1.09 1. Within Performance PPM < LSL 194. Cˆ P = 6σˆ x 6(8. or in other words.05 1.17 PPM Total 1154. Overall Performance PPM < LSL 848.13944 Overall Capability Pp PPL PPU Ppk Cpm 420 Observed Performance PPM < LSL 0. 5-34 .23 .24 8.08645 9. Cp = 1.14 1.13) MTB > Stat > Quality Tools > Capability Analysis > Normal Process Capability Analysis of TiW Thickness (Ex5-26Th) LSL USL W ithin O v erall Process Data LSL 420. is estimated from the within-subgroup variation.00000 Target * USL 480.00 430 Exp.24 CPL 1.01 PPM > USL 306.68750 Sample N 80 StDev(Within) StDev(Overall) Potential (Within) Capability Cp 1. so the process is capable.18 CPU 1.Chapter 5 Exercise Solutions 5-26 continued (d) USL = +30.18 The Potential (Within) Capability.00 PPM > USL 0. This is the same result as the manual calculation. 49 430 1 3 6 9 12 15 Sample 18 21 24 27 30 Sample Range 40 UCL=38.18 30 20 _ R=16.Chapter 5 Exercise Solutions 5-27☺.74 10 0 LCL=0 3 6 9 12 15 Sample 18 21 24 27 30 Test Results for Xbar Chart of Ex5-27Th TEST 1.00 standard deviations from center line. Test Failed at points: 18 The process continues to be in a state of statistical control.68 450 440 LCL=437. 5-35 .88 Sample Mean 460 _ _ X=449. One point more than 3. MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Xbar-R Chart of TiW Thickness (Ex5-27Th) Using previous limits with 10 new subgroups UCL=461. 267 ⎢ ⎣ 2.128 ⎤ UCL x = xold + A2(new ) ⎢ (16.17 0 LCL=0 1 2 3 4 5 6 7 8 9 10 Sample The process remains in statistical control.74.059 ⎥⎦ ⎣⎢ d 2(old) ⎦⎥ ⎡ d 2(new ) ⎤ ⎡ 1.059 ⎦ ⎢⎣ d 2(old) ⎥⎦ σˆ new = Rnew d 2(new ) = 9. and enter new parameter values.128 ⎤ LCL x = xold − A2(new ) ⎢ (16. 5-36 .43 430 1 2 3 4 5 6 7 8 9 10 Sample UCL=29. Xbar-R Chart of TiW Thickness (Ex5-28Th) New subgroups with N=2. nnew = 2 ⎡ d 2(new ) ⎤ ⎡ 1.059 ⎥⎦ ⎢⎣ d 2(old) ⎥⎦ ⎡ d 2(new ) ⎤ ⎡ 1.96 Sample Range 30 20 10 _ R=9.128 ⎤ LCL R = D3(new) ⎢ ⎥ Rold = 0 ⎢ ⎥ (16. nold = 4.880 ⎢ ⎣ 2.13 MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Select Xbar-R options.128 ⎤ UCL R = D4(new) ⎢ (16.68 − 1.68 450 440 LCL=432.92 ⎥ Rold = 449.74) = 9. Limits derived from N=4 subgroups Sample Mean 470 UCL=466.44 ⎥ Rold = 449.059 ⎥⎦ ⎣⎢ d 2(old) ⎦⎥ ⎡ d 2(new ) ⎤ ⎡ 1.17 ⎣ 2.93 460 _ _ X=449.Chapter 5 Exercise Solutions 5-28☺.17 1.74) = 432. R old = 16.74) = 0 ⎣ 2. Parameters.128 ⎤ CL R = Rnew = ⎢ ⎥ Rold = ⎢ ⎥ (16.68.68 + 1. xold 449.74) = 466.74) = 29.96 ⎥ Rold = 3.880 ⎢ ⎣ 2.128 = 8.059 ⎦ ⎣⎢ d 2(old) ⎦⎥ ⎡ d 2(new ) ⎤ ⎡ 1. One point more than 3.15 430 1 2 4 6 8 10 Sample 12 14 16 18 20 20 Sample StDev UCL=17.15 430 1 3 6 9 12 15 Sample 18 21 24 27 30 20 Sample StDev UCL=17. Test Failed at points: 18 No additional subgroups are beyond the control limits.70 5 0 LCL=0 3 6 9 12 15 Sample 18 21 24 27 30 The process remains in statistical control. 5-37 .22 Sample Mean 460 _ _ X=449.00 standard deviations from center line. After finding assignable cause.22 Sample Mean 460 _ _ X=449.44 15 10 _ S=7. with prior limits UCL=462. so these limits can be used for future production.Chapter 5 Exercise Solutions 5-29☺. The process is out of control on the x chart at subgroup 18.68 450 440 LCL=437.68 450 440 LCL=437. Xbar-S Chart of Thickness (Ex5-27Th) 10 subgroups of new data. exclude subgroup 18 from control limits calculations: MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-S Xbar-S Chart of Thickness (Ex5-26Th) Excluding subgroup 18 from calculations UCL=462.44 15 10 _ S=7.70 5 0 LCL=0 2 4 6 8 10 Sample 12 14 16 18 20 Xbar-S Chart of Ex5-26Th Test Results for Xbar Chart of Ex5-26Th TEST 1. 42 − 199 ⎞ ⎛ 197.483(5) = 202. 30 30 i =1 i =1 n = 6. Even though the process is centered at nominal.534 = 1.97) The process is not capable of meeting specification.04006 = 0. the variation is large relative to the tolerance.59 − 199 ⎞ ⎟−Φ⎜ ⎟ ⎝ 1. m = 30 samples (a) 30 x= ∑ xi = i =1 m 6000 = 200 30 30 ∑ Ri 150 =5 m 30 UCL x = x + A2 R = 200 + 0.004(5) = 10.97 USL − LSL +5 − (−5) = = 0.97 6 ⎠ ⎝ 1.97 6 ⎠ = Φ (4.85 Cˆ p = 6σˆ x 6(1.483(5) = 197.59 UCL R = D4 R = 2.Chapter 5 Exercise Solutions 5-30 (5-23).02 LCL R = D3 R = 0(5) = 0 (b) σˆ x = R / d 2 = 5 / 2. ∑ xi = 6000.42 R= i =1 = LCL x = x − A2 R = 200 − 0.95994 β − risk = Pr{not detect} = Φ ⎜ 5-38 .75) = 1 − 0. ∑ Ri = 150.25) − Φ (−1. (c) ⎛ 202. 34) ] = 1 − [1 − 0. So.577(3.05.Chapter 5 Exercise Solutions 5-31 (5-24).579 UCL x = x + A2 R = 104 + 0. µ1 = 92 k = ( µ1 − µ0 ) σ = ( 92 − 100 ) 6 = −1.771 UCL R = D4 R = 2.579 / 2.66) − Φ (−0.579) = 101.539 (c) UNTL = x + 3σˆ x = 104 + 3(1.579) = 7.326 = 1.350 LCL R = D3 R = 0(3.579) = 0 (b) Without sample #4.33 Pr{detecting shift on 1st sample} = 1 − Pr{not detecting shift on 1st sample} = 1− β ( ) ( ) = 1 − ⎡Φ ( 3 − (−1.539) = 108. σ = 6.37 ] = 0. n = 4.935 UCL R = D4 R = 2. R = 3.33) 4 ) − Φ ( −3 − (−1. R = 3.05 + 0.566 LCL R = D3 R = 0(3.579) = 106.62 LNTL = x − 3σˆ x = 104 − 3(1.37 5-32 (5-25).577(3.114(3.95) = 106.577(3.329 LCL x = x − A2 R = 104.114(3.38 5-39 .95) = 0 Sample #4 is out of control on the Range chart.95) = 101.95 UCL x = x + A2 R = 104.577(3.33) 4 ) ⎤ ⎣ ⎦ = 1 − ⎡Φ L − k n − Φ − L − k n ⎤ ⎣ ⎦ = 1 − [ Φ (5.95) = 8.05 − 0.065 LCL x = x − A2 R = 104 − 0.539) = 99. σˆ x = R / d 2 = 3. L = 3. µ0 = 100. (a) x = 104. excluding #4 and recalculating: x = 104. 0262 ⎝ 1.0000 + 0.147 LCL R = D 3 R = 0(4.064 ⎠ 5-40 .25) = 1 − 0. first center the process at nominal. m = 30 i =1 (a) m x= ∑ xi i =1 m = 607.997) = 1 − 1.0047 = 0.114(4.60) + Φ (−2.667 ⎠ ⎝ 0.8.064 ⎛ 16 − 20.49 UCL R = D 4 R = 2. ∑ xi = 607.064) = 0.8 30 m UCL x = x + A2 R = 20.0006 = 0. then almost 100% of parts will be within specification.326 = 2.0000 ⎝ 0.0195 ⎝ 2.8) = 10.539 ⎠ (e) To reduce the fraction nonconforming.26 ⎞ pˆ = Pr{x < LSL} = Φ ⎜ ⎟ = Φ (−2.8 = 20.0000 = 0.577(4.539 ⎠ ⎝ 1.9744 + 0.667 . ⎛ 107 − 103 ⎞ ⎛ 99 − 103 ⎞ pˆ = 1 − Φ ⎜ ⎟+ Φ⎜ ⎟ = 1 − Φ (2.03 R= i =1 = LCL x = x − A2 R = 20.539 ⎠ Next work on reducing the variability.577(4.997) + Φ (−5. ⎛ 107 − 103 ⎞ ⎛ 99 − 103 ⎞ pˆ = 1 − Φ ⎜ ⎟+ Φ⎜ ⎟ = 1 − Φ (5. i =1 30 ∑ Ri = 144.8) = 0 (b) σˆ x = R / d 2 = 4.Chapter 5 Exercise Solutions 5-32 continued (d) ⎛ 107 − 104 ⎞ ⎛ 99 − 104 ⎞ pˆ = 1 − Φ ⎜ ⎟+ Φ⎜ ⎟ = 1 − Φ (1. if σˆ x = 0.8) = 17.0094 ⎝ 1.26 − 0.95) + Φ (−3. 30 n = 5.26 30 m ∑ Ri 144 = 4.667 ⎠ 5-33 (5-26).8) = 23.8 / 2.9953 + 0.539 ⎠ ⎝ 1.26 + 0.60) = 1 − 0. 00 standard deviations from center line. Test Failed at points: 12 Process is not in statistical control -.Chapter 5 Exercise Solutions 5-34 (5-27). One point more than 3. 5-41 . Estimate” select Rbar as method to estimate standard deviation. R Chart of Detent (Ex5-34Det) 20 1 15 Sample Range UCL=13.47 5 0 LCL=0 1 2 3 4 5 6 7 8 9 Sample 10 11 12 13 14 15 Test Results for R Chart of Ex5-34Det TEST 1.sample #12 exceeds the upper control limit on the Range chart.67 10 _ R=6. (a) MTB > Stat > Control Charts > Variables Charts for Subgroups > R Under “Options. 42 (d) Assume the cigar lighter detent is normally distributed.64 / 2.0001) 5-42 . Without sample #12: USL − LSL 0. R Chart of Detent (Ex5-34Det) Sample 12 Excluded from Calculations 20 1 Sample Range 15 UCL=11.93 10 _ R=5. Test Failed at points: 12 (c) Without sample #12: σˆ x = R / d 2 = 5. Estimate” omit subgroup 12 and select Rbar.Chapter 5 Exercise Solutions 5-34 continued (b) Excluding Sample Number 12: MTB > Stat > Control Charts > Variables Charts for Subgroups > R Under “Options.00 standard deviations from center line.3220 − 0.38 6σˆ x 6(2.3200 Cˆ P = = = 1.42 × 0. One point more than 3.64 5 0 LCL=0 1 2 3 4 5 6 7 8 9 Sample 10 11 12 13 14 15 Test Results for R Chart of Ex5-34Det TEST 1.326 = 2. 45 5 2.64 LCL=0 2 4 6 8 10 12 14 Sample 16 18 20 22 24 Test Results for Xbar Chart of Ex5-35Det TEST Test TEST Test TEST 1. 22.0 6 -2. 17. 16. 20. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). 3.2 LCL=-3. Failed at points: 1. One point more than 3. Failed at points: 12 2. MTB > Stat > Control Charts > Variables Charts for Subgroups > R Under “Options. Test Failed at points: 2. 9 points in a row on same side of center line. 17. 12. 18. Failed at points: 24. Test Failed at points: 15. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). 2.00 standard deviations from center line. Estimate” use subgroups 1:11 and 13:15.05 1 12 14 Sample 16 18 20 22 24 1 15 UCL=11. 24 Test Results for R Chart of Ex5-35Det TEST Test TEST Test 1. 20.5 6 6 2 2 0. 23 2.5 1 -5. One point more than 3. 13-15 1 1 Sample Mean 5. 13. 25 5-43 .00 standard deviations from center line. 18. Failed at points: 24. 25 5.93 2 10 2 5 0 _ R=5.0 1 1 1 1 1 UCL=3. 20 TEST 6. Xbar-R Chart of Ex5-35Det Limits based on Samples 1-11. 23.Chapter 5 Exercise Solutions 5-35 (5-28). 13. and select Rbar. 9 points in a row on same side of center line. 19.0 2 4 6 8 10 20 Sample Range _ _ X=0. 94 _ _ X=-0. 12. 12 and 13 1 1 Sample Mean 5. 5-44 . the process appears to be unstable. and 13 are out of control on the x chart.0 LCL=-3. 3. the fact that 6 of 15 samples were out of control and eliminated from calculations is an early indication of process instability. 2. While the limits on the above charts may be used to monitor future production. Note that samples 1.0 1 1 1 1 1 1 2. excluding 1. If these samples are removed and the limits recalculated.Chapter 5 Exercise Solutions 5-35 continued We are trying to establish trial control limits from the first 15 samples to monitor future production.5 1 -5.26 1 12 14 Sample 16 18 20 22 24 1 15 1 1 10 1 1 2 2 UCL=9.52 _ R=4. Removing sample 3 gives Xbar-R Chart of Ex5-35Det Limits based on first 15 samples. (a) Given the large number of points after sample 15 beyond both the x and R control limits on the charts above.0 2 4 6 8 10 20 Sample Range UCL=1.5 5 6 6 2 5 2 -2. 2.5 5 0 LCL=0 2 4 6 8 10 12 14 Sample 16 18 20 22 24 Sample 14 is now out of control on the R chart. No additional samples are out of control on the x chart.66 0. sample 3 is also out of control on the x chart. 0 2 4 6 8 10 Sample Range 20 LCL=-3.48 5 LCL=0 0 2 4 6 8 10 12 14 Sample 16 18 20 22 24 With Test 1 only: Test Results for Xbar Chart of Ex5-35Det TEST 1. Test Failed at points: 1.0 -2.Chapter 5 Exercise Solutions 5-35 continued (b) Xbar-R Chart of Detent (Ex5-35Det) All Samples in Calculations 1 1 1 UCL=5. 17 Test Results for R Chart of Ex5-35Det TEST 1.00 standard deviations from center line. One point more than 3. 16.23 0.82 15 10 _ R=7. Test Failed at points: 12 5-45 .55 Sample Mean 5. 12.5 1 -5.5 _ _ X=1. One point more than 3.00 standard deviations from center line.08 1 12 14 Sample 16 18 20 22 24 1 UCL=15.0 2. 13. 16.0 2 4 6 8 10 Sample Range 20 1 12 14 Sample 16 18 20 22 24 1 15 UCL=14.85 5 0 LCL=0 2 4 6 8 10 12 14 Sample 16 18 20 22 24 With Test 1 only: Test Results for Xbar Chart of Ex5-35Det TEST 1. 16. 13. 17 excluded from calculations 1 1 1 1 Sample Mean 5.5 _ _ X=0. Test Failed at points: 12 5-46 . 12. 13. 17.48 10 _ R=6. Test Failed at points: 1. 13.99 0.00 standard deviations from center line. 16. and 17 from calculations: Xbar-R Chart of Detent (Ex5-35Det) Samples 1. 12. One point more than 3.96 1 -5.Chapter 5 Exercise Solutions 5-35 (b) continued Removing samples 1.5 LCL=-2.0 -2. 20 Test Results for R Chart of Ex5-35Det TEST 1.00 standard deviations from center line. One point more than 3.0 UCL=4. 12.94 2. 16.Chapter 5 Exercise Solutions 5-35 continued Sample 20 is now also out of control. One point more than 3. Test Failed at points: 1.78 0. and that the sources of variation need to be identified and removed. with runs of points both above and below the centerline.00 standard deviations from center line. Removing sample 20 from calculations. 12. 13. 20 excluded from calculations 1 1 Sample Mean 5. 17. for a total 7 of the 25 samples. One point more than 3.5 _ _ X=0. 20 Test Results for R Chart of Ex5-35Det TEST 1. 18. 5-47 .0 1 1 1 UCL=4. 17.0 -2.5 1 -5. 13.74 5 LCL=0 0 2 4 6 8 10 12 14 Sample 16 18 20 22 24 With Test 1 only: Test Results for Xbar Chart of Ex5-35Det TEST 1. Test Failed at points: 12 Sample 18 is now out-of-control.66 2. 16. 12.11 1 12 14 Sample 16 18 20 22 24 1 15 UCL=14.00 standard deviations from center line.24 10 _ R=6.0 2 4 6 8 10 Sample Range 20 LCL=-3. Xbar-R Chart of Ex5-35Det Samples 1. This suggests that the process is inherently unstable. 18 R= i =1 = LCL R = D3 R = 0(45. 30 30 i =1 i =1 n = 6.5121 ⎝ 0.978 ( = R / d = (∑ R 20 ) d = (18. ∑ xi = 12.42 + 0.i / mx i =1 σˆ y 10 y 2 i =1 y .Chapter 5 Exercise Solutions 5-36 (5-29).0 30 m σˆ x = R / d 2 = 45.534 = 17.0) = 0 (b) m ∑ xi 12. ∑ Ry . m = 30 (a) m ∑ Ri 1350 = 45.300 σˆ x = Rx / d 2 = ∑ Rx .758 µˆ = x = i =1 = 5-48 .870 = 429.346 5-37 (5-30).326 = 0.09} = 0.09 − z ⎞ Φ −1 ⎜ ⎟ = Φ (0.006.500 ⎠ z = +2.500) + 0.870.978 /10) / 2.i = 18.400 / m ) d = (6.09 − z ⎞ Φ⎜ ⎟ = 0. m y = 10.006) ⎝ 0.09 − z ⎞ ⎜ ⎟ = −2.608 / 20) / 2.09 = 1.608.32 = 0.500 ⎠ ⎛ 0.5121(0.004(45.0 30 m UCL R = D4 R = 2.326 = 0.0 / 2. mx = 20.i = 6. Let z = x − y.006 ⎝ σˆ z ⎠ ⎛ 0. Then σˆ z = σˆ x2 + σˆ y2 = 0.i y 2 2 (b) Want Pr{(x − y) < 0. ∑ Rx . ∑ Ri = 1350.500 ⎛ 0.0) = 90. (a) 20 10 i =1 i =1 n = 5. 751 Cˆ p 6σˆ x 6(17.667 30 m UCL S = B4 S = 2.758 ⎠ (d) To minimize fraction nonconforming the mean should be located at the nominal dimension (440) for a constant variance.9213 = 14.9979 + 0.667) = 30. 30 30 i =1 i =1 n = 4.667 / 0.870. ∑ xi = 12.667) = 0 (b) m ∑ xi 12.63) = 1 − 0.0537 ⎝ 17.870 = 429.758) ⎛ 480 − 429 ⎞ ⎛ 400 − 429 ⎞ pˆ = 1 − Φ ⎜ ⎟+Φ⎜ ⎟ = 1 − Φ (2. ∑ Si = 410.758 ⎠ ⎝ 17.0 30 m σˆ x = S / c4 = 13. m = 30 (a) m ∑ Si 410 = 13.87) + Φ (−1.40 = 400 USL − LSL 480 − 400 = = 0.834 µˆ = x = i =1 = 5-49 .Chapter 5 Exercise Solutions 5-37 continued (c) USL = 440 + 40 = 480. 5-38 (5-31).969 S= i =1 = LCL S = B3 S = 0(13.0516 = 0. LSL = 440 .266(13. 56) ⎟ −Φ⎜ ⎟ = Φ⎜ σx σx ⎝ ⎠ ⎝ ⎠ ⎝ 8 5 ⎠ ⎝ 8 5 ⎠ = 0.9213) 6 ( S c4 ) The process is not capable of meeting specifications.807 8 5-40 (5-33).7678 6σˆ x 6 (1.50 Cˆ P = = = = 0.6658 1 − [Pr{not detected}]3 = 1 − (0.807 ( = µ − k (σ UCL x = µ + kσ x = µ + k σ x LCL x = µ − kσ x x ) ( 4 ) = 111. σ x = 8 Pr{out-of-control signal by at least 3rd plot point} = 1 − Pr{not detected by 3rd sample} = 1 − [Pr{not detected}]3 Pr{not detected} = Pr{LCL x ≤ x ≤ UCL x } = Pr{x ≤ UCL x } − Pr{x ≤ LCL x } ⎛ 104 − 98 ⎞ ⎛ 96 − 98 ⎞ ⎛ UCL x − µ ⎞ ⎛ LCL x − µ ⎞ = Φ⎜ ⎟−Φ⎜ ⎟ = Φ (1.005 / 2 = Z 0. 5-50 .000 0.807 ( 8 4 ) = 88.0025 = 2.2877 = 0. LCL x = 96. k = 3. σ x = 8 ( = µ − 2 (σ UCL x = µ + 2σ x = µ + 2 σ x LCL x = µ − 2σ x x ) ( 4 ) = 108 n ) = 100 − 2 ( 8 4 ) = 92 n = 100 + 2 8 (b) k = Zα / 2 = Z 0. µ = 100. 1 1 1 ARL1 = = = = 2. USL − LSL USL − LSL 202.6658)3 = 0. n = 5. (a) n = 4. centerline x = 100.68) − Φ (−0.9535 − 0.992 1 − β 1 − Pr{not detect} 1 − 0.772 n = 100 + 2.Chapter 5 Exercise Solutions 5-39 (5-32).228 n ) = 100 − 2. UCL x = 104.7049 5-41 (5-34).6658 5-42 (5-35).50-197. µ = 98. 7734.0000 0.Chapter 5 Exercise Solutions 5-43 (5-36).2266.3 1.8000 0.088(10) = 20. LSL = 600 − 20 = 580 (a) USL − LSL USL − LSL 620 − 580 Cˆ P = = = = 1.25.88 LCL S = B5σ x = 0(10) = 0 (b) k = Zα / 2 = Z 0.2 n = 200 + 1.5.9332.75.9974. 0.0 1.0000.9213(10) = 9.0.025 = 1.2000 1.213 UCL S = B6σ x = 2.0000 beta 0.96 (10 4 ) = 190.96 ( = µ − k (σ UCL x = µ + kσ x = µ + k σ x LCL x = µ − kσ x x ) ( 4 ) = 209.5 0.111 6σˆ x 6 (17. 2. σ x = 10 (a) centerline S = c4σ = 0.0013.96 10 5-44 (5-37).5. 3. 0.0 k 5-51 . 0. 1. 0. µ = 200. 0. 0.0668.8 n ) = 200 − 1.05/ 2 = Z 0. 0. β = Φ L − k n − Φ − L − k n ( ) ( ) for k = {0.4000 0. 2.82 / 2.0}. 0. L = 3. 0. USL = 600 + 20 = 620. L = 3 1.6000 0. (b) n = 9.5.0000} Operating Characteristic Curve for n = 9.0 0.8 1.5 2. β = {0. n = 4.0 2. 1. 0.2000 0. 1. n = 9.970 ) 6 ( R d2 ) Process is capable of meeting specifications.5.5 3.0. 324 UCL R = D 4 R = 1. ∑ xi = 2700.59 / 2.059 = 10 Pr{detect shift on 1st sample} = Pr{x < LCL} + Pr{x > UCL} = Pr{x < LCL} + 1 − Pr{x ≤ UCL} ⎛ 785 − 790 ⎞ ⎛ 815 − 790 ⎞ ⎛ LCL − µnew ⎞ ⎛ UCL − µnew ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ ⎟ +1− Φ ⎜ ⎟ = Φ⎜ σx σx ⎝ ⎠ ⎝ ⎠ ⎝ 10 4 ⎠ ⎝ 10 4 ⎠ = Φ (−1) + 1 − Φ (5) = 0.Chapter 5 Exercise Solutions 5-45 (5-38).704 = 1.479) = 1.419) = 2.3 n ) = 600 − 2. µ = 600. 30 30 i =1 i =1 n = 7. α =0.419(4) = 91. σˆ x = R / d 2 = 20.576 (12 9 ) = 589.479 (c) S = c4σˆ x = 0.576 ( = µ − k (σ UCL x = µ + kσ x = µ + k σ x LCL x = µ − kσ x x ) ( 9 ) = 610.576 12 5-47 (5-40).0000 = 0.676 x= i =1 = LCL x = x − A2 R = 90 − 0.118(1.01/ 2 = Z 0.924(4) = 7.005 = 2.419) = 0.696 LCL R = D 3 R = 0.419(4) = 88.7 n = 600 + 2.419 UCL S = 1.304 (b) σˆ x = R / d 2 = 4 / 2.882(1.1587 5-52 .076(4) = 0. R = i =1 = =4 m m 30 30 UCL x = x + A2 R = 90 + 0.671 LCL S = 0. ∑ Ri = 120.01 k = Zα / 2 = Z 0. σ x = 12.1587 + 1 − 1. m = 30 (a) m ∑ xi m ∑ Ri 2700 120 = 90. n = 9.167 5-46 (5-39).9594(1. 424 5-53 .30 ARL1 = 1 − β 1 − Pr{not detect} Pr{detect} 0.970 = 3. k = Zα / 2 = Z 0.005 = 2.0013 + 1 − 0.000 ⎛ LCL − x ⎞ ⎛ UCL − x ⎞ ⎟ +1− Φ ⎜ ⎟ ⎝ σx ⎠ ⎝ σx ⎠ α = Pr{x < LCL} + Pr{x > UCL} = Φ ⎜ ⎛ 357 − 360 ⎞ ⎛ 363 − 360 ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ = Φ (−3) + 1 − Φ (3) = 0.576 3 ) ) ( n = 360 + 2.Chapter 5 Exercise Solutions 5-48 (5-41).9987 = 0.576 9 = 357.91/ 2.5000 = 0.667 6σˆ x 6(3) The process is not capable of producing all items within specification.0000 − 0.01. 1 1 1 1 = = = = 6.1587 5-49 (5-42).01/ 2 = Z 0.576 ( UCL x = x + kσ x = x + k σˆ x ( LCL x = 360 − 2. (c) µnew = 357 ⎛ UCL − µnew Pr{not detect on 1st sample} = Pr{LCL ≤ x ≤ UCL} = Φ ⎜⎜ ⎝ σˆ x n ⎞ ⎛ LCL − µnew ⎟⎟ − Φ ⎜⎜ ⎠ ⎝ σˆ x n ⎞ ⎟⎟ ⎠ ⎛ 363 − 357 ⎞ ⎛ 357 − 357 ⎞ = Φ⎜ ⎟ −Φ⎜ ⎟ = Φ (6) − Φ (0) = 1.576 3 ) 9 = 362.5000 ⎝ 3 9 ⎠ ⎝ 3 9 ⎠ (d) α = 0.0026 ⎝ 3 9 ⎠ ⎝ 3 9 ⎠ (b) USL − LSL +6 − (−6) Cˆ P = = = 0. (a) σˆ x = R / d 2 = 8. Chapter 5 Exercise Solutions 5-50 (5-43).0000 = 0.25) = 0.9213(4) = 3.266(3.000 (b) S = c4σˆ x = 0.0000 5-54 .9772)3 = 1.0000 + 1 − 0.1056 (d) To reduce the fraction nonconforming.059 = 4. (a) σˆ x = R / d 2 = 8.865 UCL S = B4 S = 2.685) = 0 (c) ⎛ LSL − x ⎞ ⎛ USL − x ⎞ pˆ = Pr{x < LSL} + Pr{x > USL} = Φ ⎜ ⎟ +1− Φ ⎜ ⎟ ⎝ σˆ x ⎠ ⎝ σˆ x ⎠ ⎛ 595 − 620 ⎞ ⎛ 625 − 620 ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ 4 4 ⎝ ⎠ ⎝ ⎠ = Φ (−6. (e) Pr{detect on 1st sample} = Pr{x < LCL} + Pr{x > UCL} ⎛ LCL − µnew ⎞ ⎛ UCL − µnew ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ σx σx ⎝ ⎠ ⎝ ⎠ ⎛ 614 − 610 ⎞ ⎛ 626 − 610 ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ ⎝ 4 4 ⎠ ⎝ 4 4 ⎠ = Φ (2) + 1 − Φ (8) = 0.9772 (f) Pr{detect by 3rd sample} = 1 − Pr{not detect by 3rd sample} = 1 − (Pr{not detect})3 = 1 − (1 − 0.8944 = 0.351 LCL S = B3 S = 0(3.685) = 8. try moving the center of the process from its current mean of 620 closer to the nominal dimension of 610. Also consider reducing the process variability.9772 + 1 − 1.236 / 2.25) + 1 − Φ (1. Chapter 5 Exercise Solutions 5-51 (5-44).1006 (d) Pr{detect on 1st sample} = Pr{x < LCL} + Pr{x > UCL} ⎛ LCL − µnew ⎞ ⎛ UCL − µnew ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ σx σx ⎝ ⎠ ⎝ ⎠ ⎛ 703.827 6 ⎠ ⎝ 1.9920)3 = 1.00.827) = 700.8 − 702 ⎞ ⎛ 708.827 ⎠ ⎝ 1.2 − 702 ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ ⎝ 1.41) + 1 − Φ (8.642) + 1 − Φ (1.0503 + 1 − 0.48 LNTL = 706 − 3(1. (a) µˆ = x = 706.738 / 0.827) = 711.827 6 ⎠ = Φ (2.9920 + 1 − 1. σˆ x = S / c4 = 1.827 ⎠ = Φ (−1.52 (c) pˆ = Pr{x < LSL} + Pr{x > USL} ⎛ LSL − x ⎞ ⎛ USL − x ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ ⎝ σˆ x ⎠ ⎝ σˆ x ⎠ ⎛ 703 − 706 ⎞ ⎛ 709 − 706 ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ ⎝ 1.9515 = 1.0000 = 0.9920 (e) Pr{detect by 3rd sample} = 1 − Pr{not detect by 3rd sample} = 1 − (Pr{not detect})3 = 1 − (1 − 0.31) = 0.642) = 0.0000 5-55 .9497 = 0.827 (b) UNTL = x + 3σˆ x = 706 + 3(1. 31) = 0.new ⎝ ⎠ ⎝ ⎠ ⎛ 690 − 693 ⎞ ⎛ 710 − 693 ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ ⎝ 12 4 ⎠ ⎝ 12 4 ⎠ = Φ (−0.661 ⎠ = Φ (−1.22 1 − β 1 − Pr{not detect} Pr{detect} 0.9213 = 8.new x .661 ⎠ ⎝ 8.9896 = 0.3085 + 1 − 0.5) + 1 − Φ (2.0208 (d) Pr{detect on 1st sample} = Pr{x < LCL} + Pr{x > UCL} ⎛ LCL − µnew ⎞ ⎛ UCL − µnew ⎞ = Φ⎜ +1− Φ ⎜ ⎟ ⎟⎟ ⎜ σ ⎟ ⎜ σ x .1251 + 1 − 0.661 4 ⎠ ⎝ 8.Chapter 5 Exercise Solutions 5-52 (5-45).3108 (e) ARL1 = 1 1 1 1 = = = = 3.3108 5-56 .661 (b) pˆ = Pr{x < LSL} + Pr{x > USL} ⎛ LSL − x ⎞ ⎛ USL − x ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ ⎝ σˆ x ⎠ ⎝ σˆ x ⎠ ⎛ 690 − 700 ⎞ ⎛ 720 − 700 ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ ⎝ 8.15) + 1 − Φ (2.31) = 0.661 4 ⎠ = Φ (−2.0104 + 1 − 0.9977 = 0.979 / 0. σˆ x = S / c4 = 7.1355 (c) α = Pr{x < LCL} + Pr{x > UCL} ⎛ LCL − x ⎞ ⎛ UCL − x ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ ⎝ σx ⎠ ⎝ σx ⎠ ⎛ 690 − 700 ⎞ ⎛ 710 − 700 ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ ⎝ 8.31) + 1 − Φ (2.83) = 0.9896 = 0. (a) µˆ = x = 700. 125 16. MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Weight (Ex5-53Wt) Individual Value 16.02 0.0420 2 4 6 8 10 12 14 O bser vation 16 18 20 22 24 M oving Range 0.021055.021055 ⎠ 5-57 . x = 16. MR2 = 0.04 __ M R=0.05 LC L=16.11 16.1052 ⎞ = 100% × Φ ⎜ ⎟ = 100% × Φ (−4.08 16.100 Ex5-53Wt 16.02375 0.1052.00003% ⎝ 0.14 _ X=16.11 0. %underfilled = 100% × Pr{x < 16 oz} ⎛ 16 − 16.397 0. σˆ x = 0.9964) = 0.02044 25 0.1684 16.08 U C L=0.050 16.00 LC L=0 2 4 6 8 10 12 14 O bser vation 16 18 20 22 24 There may be a “sawtooth” pattern developing on the Individuals chart.07760 0.342 Percent 80 70 60 50 40 30 20 10 5 1 16.075 16.06 0.1052 16.150 Visual examination of the normal probability indicates that the assumption of normally distributed coffee can weights is valid.02375 MTB > Stat > Basic Statistics > Normality Test Probability Plot of Weight (Ex5-53Wt) Normal 99 Mean StDev N AD P-Value 95 90 16.Chapter 5 Exercise Solutions 5-53 (5-46).17 U C L=16. MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Hardness (Ex5-54Har) U C L=61.21429 MTB > Stat > Basic Statistics > Normality Test Probability Plot of Hardness (Ex5-54Har) Normal 99 Mean StDev N AD P-Value 95 90 53.712 15 0.217 Percent 80 70 60 50 40 30 20 10 5 1 46 48 50 52 54 Ex5-54Har 56 58 60 Although the observations at the tails are not very close to the straight line.21 2.72 1 2 3 4 5 6 7 8 9 O bser vation 10 11 12 13 14 15 U C L=10.5 5. indicating that it may be reasonable to assume that hardness is normally distributed. MR2 = 3.27 2.84954.0 1 2 3 4 5 6 7 8 9 O bser vation 10 11 12 13 14 15 x = 53.0 7.Chapter 5 Exercise Solutions 5-54(5-47). 5-58 .5 LC L=0 0.465 0.27 50 45 LC L=44.82 Individual Value 60 55 _ X=53. σˆ x = 2. the p-value is greater than 0.0 __ M R=3.50 M oving Range 10.05.2667. (b) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Viscosity (Ex5-55Vis) Individual V alue 3400 U C L=3322.158 5-59 .9 3200 3000 _ X=2928.Chapter 5 Exercise Solutions 5-55 (5-48).9 2800 2600 LC L=2534. (a) MTB > Stat > Basic Statistics > Normality Test Probability Plot of Viscosity (Ex5-55Vis) Normal 99 Mean StDev N AD P-Value 95 90 2929 129. with no out-of-control points.319 0. runs.0 20 0. MR2 = 148.2 120 LC L=0 0 2 4 6 8 10 12 O bser vation 14 16 18 20 The process appears to be in statistical control.9 2 4 6 8 10 12 O bser vation 14 16 18 20 U C L=484.1 M oving Range 480 360 240 __ M R=148. trends. or other patterns. σˆ x = 131.9.346. (c) µˆ = x = 2928.511 Percent 80 70 60 50 40 30 20 10 5 1 2600 2700 2800 2900 3000 Ex5-55Vis 3100 3200 3300 Viscosity measurements do appear to follow a normal distribution. 2 120 LCL=0 0 2 4 6 8 10 12 14 Observation 16 18 20 22 24 All points are inside the control limits.9 2 4 6 8 10 12 14 Observation 16 18 20 22 24 UCL=484. 5-60 . MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Viscosity (Ex5-56Vis) With five next measurements 3400 Individual Value UCL=3322.1 Moving Range 480 360 240 __ MR=148.9 3000 2800 2600 LCL=2534. However all of the new points on the I chart are above the center line.Chapter 5 Exercise Solutions 5-56 (5-49). indicating that a shift in the mean may have occurred.9 3200 _ X=2928. 85 4.85 50 40 LC L=34. MTB > Stat > Basic Statistics > Normality Test Probability Plot of Oxide Thickness (Ex5-57aTh) Normal 99 Mean StDev N AD P-Value 95 90 49.Chapter 5 Exercise Solutions 5-57 (5-50).534 30 0.338 0. (a) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Oxide Thickness (Ex5-57aTh) Individual V alue U C L=65.14 60 _ X=49.79 15 10 __ M R=5.55 30 3 6 9 12 15 18 O bser vation 21 24 27 30 M oving Range 20 U C L=18. 5-61 .480 Percent 80 70 60 50 40 30 20 10 5 1 40 45 50 Ex5-57aTh 55 60 The normality assumption is reasonable.75 5 0 LC L=0 3 6 9 12 15 18 O bser vation 21 24 27 30 The process is in statistical control. 14 _ X=49. several 4 of 5 beyond 1 sigma. 40 TEST 6. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). even without use of the sensitizing rules. Test Failed at points: 35.85 50 40 LCL=34.55 30 4 8 12 16 20 Observation 24 28 32 36 40 Moving Range 20 UCL=18. 39. it is clear that the process is out of control during this period of operation. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). 5-62 . 39. 39. 9 points in a row on same side of center line. 40 We have turned on some of the sensitizing rules in MINITAB to illustrate their use. 40 5. Test Failed at points: 34. However. and several 2 of 3 beyond 2 sigma on the x chart. One point more than 3.75 5 0 LCL=0 4 8 12 16 20 Observation 24 28 32 36 40 Test Results for I Chart of Ex5-57bTh TEST Test TEST Test TEST 1. 38.00 standard deviations from center line. There is a run above the centerline. Failed at points: 38 2. 37.79 15 10 __ MR=5. Failed at points: 38.Chapter 5 Exercise Solutions 5-57 continued (b) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Oxide Thickness (Ex5-57bTh) With 10 new measurements and some sensitizing rules 70 1 Individual Value 5 60 6 6 2 2 UCL=65. Chapter 5 Exercise Solutions 5-57 continued (c) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Oxide Thickness (Ex5-57cTh) 10 + 20 New Measurements.85 50 40 LCL=34.79 15 10 __ MR=5.55 30 1 6 12 18 24 30 Observation 36 48 54 60 1 20 Moving Range 42 UCL=18. 5-63 .14 2 _ X=49.75 5 0 LCL=0 1 6 12 18 24 30 Observation 36 42 48 54 60 The process has been returned to a state of statistical control. with Sensitizing Rules On 70 1 Individual Value 5 60 6 6 2 UCL=65. 88 5 100 80 _ X=73. Test Failed at points: 17 The process is not in control. 5-64 .73 60 LC L=42.59 40 3 6 9 12 15 18 O bser vation 24 27 30 1 40 M oving Range 21 U C L=38. in particular due to the curve of the larger values and three distant values. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). (a) The normality assumption is a little bothersome for the concentration data. Test Failed at points: 11 Test Results for MR Chart of Ex5-58C TEST 1.71 10 0 LC L=0 3 6 9 12 15 18 O bser vation 21 24 27 30 Test Results for I Chart of Ex5-58C TEST 5.Chapter 5 Exercise Solutions 5-58 (5-51).26 30 20 __ M R=11. with two Western Electric rule violations.00 standard deviations from center line. (b) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Concentration (Ex5-58C) Individual V alue U C L=104. One point more than 3. Chapter 5 Exercise Solutions 5-58 continued (c) MTB > Stat > Basic Statistics > Normality Test Probability Plot of ln(Concentration) (Ex5-58lnC) Normal 99 Mean StDev N AD P-Value 95 90 4.9 4.6 4.288 0.327 Percent 80 70 60 50 40 30 20 10 5 1 3.4 Ex5-58lnC 4.1 4.3 4.408 0.5 4.1567 30 0. again due to the curve of the larger values and three distant values.2 4. 5-65 .0 4.7 The normality assumption is still troubling for the natural log of concentration. with the same to Western Electric Rules violations. Test Failed at points: 17 The process is still not in control.8689 3 6 9 12 15 18 O bser vation 21 24 27 30 1 U C L=0.00 LC L=0 3 6 9 12 15 18 O bser vation 21 24 27 30 Test Results for I Chart of Ex5-58lnC TEST 5.00 standard deviations from center line.2884 4.6 4.1577 0.48 0. Test Failed at points: 11 Test Results for MR Chart of Ex5-58lnC TEST 1. 5-66 .2 4.5154 M oving Range 0.0 LC L=3.24 __ M R=0.36 0.12 0. There does not appear to be much difference between the two control charts (actual and natural log). One point more than 3.Chapter 5 Exercise Solutions 5-58 continued (d) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of ln(Concentration) (Ex5-58lnC) Individual Value 4.4 _ X=4. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL).8 U C L=4.7079 5 4. Test Failed at points: 8 The out-of-control signal on the moving range chart indicates a significantly large difference between successive measurements (7 and 8). MTB > Stat > Basic Statistics > Normality Test Probability Plot of Velocity of Light (Ex5-59Vel) Normal 99 Mean StDev N AD P-Value 95 90 909 104.6 100 0 LC L=0 2 4 6 8 10 12 O bser vation 14 16 18 20 I-MR Chart of Ex5-59Vel Test Results for MR Chart of Ex5-59Vel TEST 1.1 1000 _ X=909 750 LC L=582. consider the process to be in a state of statistical process control. use all data for control limits calculations. Since neither of these measurements seems unusual.672 0.Chapter 5 Exercise Solutions 5-59☺.9 20 0.067 Percent 80 70 60 50 40 30 20 10 5 1 600 700 800 900 Ex5-59Vel 1000 1100 1200 Velocity of light measurements are approximately normally distributed. MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Velocity of Light (Ex5-59Vel) Individual V alue 1250 U C L=1235.7 300 200 __ M R=122. There may also be an early indication of less variability in the later measurements. For now.9 500 2 4 6 8 10 12 O bser vation 14 16 18 20 1 M oving Range 400 U C L=400. 5-67 .00 standard deviations from center line. One point more than 3. Failed at points: 36. (a) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR Select I-MR Options. One point more than 3.7 Moving Range 400 300 200 100 2 __ MR=122. which is reflected in the number of observations below the centerline of the moving range chart. 37. 37 The velocity of light in air is not changing. Test Failed at points: 36. Failed at points: 8 2. 38. 9 points in a row on same side of center line. 9 points in a row on same side of center line.Chapter 5 Exercise Solutions 5-60☺. however the method of measuring is producing varying results—this is a chart of the measurement process. meaning the method is producing gradually smaller measurements. 39.9 500 4 8 12 16 20 Observation 24 28 32 36 40 1 UCL=400. There is a distinct downward trend in measurements.6 2 0 LCL=0 4 8 12 16 20 Observation 24 28 32 36 40 I-MR Chart of Ex5-60Vel Test Results for I Chart of Ex5-60Vel TEST 2. 40 Test Results for MR Chart of Ex5-60Vel TEST Test TEST Test 1. (b) Early measurements exhibit more variability than the later measurements. 5-68 .1 1000 _ X=909 2 2 2 750 2 2 LCL=582.00 standard deviations from center line. Estimate to specify which subgroups to use in calculations I-MR Chart of Velocity of Light (Ex5-60Vel) New measurements with old limits Individual Value 1250 UCL=1235. 0 2.Chapter 5 Exercise Solutions 5-61☺.07 5.6 The distribution of the natural-log transformed uniformity measurements is approximately normally distributed. as well as the Anderson-Darling test p-value. Probability Plot of ln(Uniformity) (Ex5-61lnUn) Normal 99 Mean StDev N AD P-Value 95 90 2.626 0. 5-69 . as evidenced by the “S”.546 30 1.093 Percent 80 70 60 50 40 30 20 10 5 1 2.005 Percent 80 70 60 50 40 30 20 10 5 1 0 5 10 15 20 Ex5-61Un 25 30 35 The data are not normally distributed. so a compressive transform such as natural log or square-root may be appropriate.shaped curve to the plot points on a normal probability plot. The data are skewed right.653 0.158 <0. (a) MTB > Stat > Basic Statistics > Normality Test Probability Plot of Uniformity Determinations (Ex5-61Un) Normal 99 Mean StDev N AD P-Value 95 90 15.4 2.8 Ex5-61lnUn 3.2 3.3493 30 0. 720 1.5 2.75 0.586 Individual V alue 3.146 M oving Range 1.5 3.00 0.0 LC L=1.653 2.351 0. 5-70 .0 _ X=2.25 0.5 3 6 9 12 15 18 O bser vation 21 24 27 30 U C L=1.00 LC L=0 3 6 9 12 15 18 O bser vation 21 24 27 30 The etching process appears to be in statistical control.50 __ M R=0.Chapter 5 Exercise Solutions 5-61 continued (b) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of ln (Uniformity) (Ex5-61lnUn) U C L=3. 78 0.174 <0.81 0.87 Purity is not normally distributed.85 0.86 0. 5-71 .Chapter 5 Exercise Solutions 5-62 (5-52).83 0. (a) MTB > Stat > Basic Statistics > Normality Test Probability Plot of Batch Purity (Ex5-62Pur) Normal 99 Mean StDev N AD P-Value 95 90 0.01847 20 1.824 0.79 0.84 Ex5-62Pur 0.82 0.80 0.005 Percent 80 70 60 50 40 30 20 10 5 1 0. 012 0.86459 5 0. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL).024 __ M R=0.01526 0.824 0. (c) all data: µˆ = 0.04987 M oving Range 0.84 _ X=0.78341 0.824 . Test Failed at points: 11.86 6 0.82 6 0. Test Failed at points: 18 TEST 5.048 0. σˆ x = 0. Test Failed at points: 19 TEST 6.0135 without sample 18: µˆ = 0.8216 . σˆ x = 0.00 standard deviations from center line.0133 5-72 .Chapter 5 Exercise Solutions 5-62 continued (b) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Purity (Ex5-62Pur) 1 Individual V alue U C L=0.036 0.000 LC L=0 2 4 6 8 10 12 O bser vation 14 16 18 20 Test Results for I Chart of Ex5-62Pur TEST 1.78 2 4 6 8 10 12 O bser vation 14 16 18 20 U C L=0. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL).80 LC L=0. One point more than 3. 20 The process is not in statistical control. 04 __ M R=0.08 16. control limits for both are essentially the same.05 LC L=16.17 U C L=16.07726 0. MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR Select “Estimate” to change the method of estimating sigma I-MR Chart of Can Weight (Ex5-53Wt) Individual V alue 16.1052 16.06 0.08 U C L=0. 5-73 .Chapter 5 Exercise Solutions 5-63 (5-53).02 0.00 LC L=0 2 4 6 8 10 12 14 O bser vation 16 18 20 22 24 There is no difference between this chart and the one in Exercise 5-53.02365 0.14 _ X=16.11 16.0423 2 4 6 8 10 12 14 O bser vation 16 18 20 22 24 M oving Range 0.1681 16. 27 50 LC L=45.5 0.0 LC L=0 1 2 3 4 5 6 7 8 9 O bser vation 10 11 12 13 14 15 The median moving range method gives slightly tighter control limits for both the Individual and Moving Range charts.0 U C L=9.96 2.Chapter 5 Exercise Solutions 5-64 (5-54).0 __ M R=2. MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR Select “Estimate” to change the method of estimating sigma I-MR Chart of Hardness-Coded (Ex5-54Har) U C L=61. with no practical difference for this set of observations.41 45 1 2 3 4 5 6 7 8 9 O bser vation 10 11 12 13 14 15 M oving Range 10.5 5.13 Individual V alue 60 55 _ X=53.66 7. 5-74 . MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR Select “Estimate” to change the method of estimating sigma I-MR Chart of Polymer Viscosity (Ex5-55Vis) Individual Value 3400 U C L=3337.7 120 0 LC L=0 2 4 6 8 10 12 O bser vation 14 16 18 20 The median moving range method gives slightly wider control limits for both the Individual and Moving Range charts. 5-75 . with no practical meaning for this set of observations.9 2800 2600 LC L=2520.2 M oving Range 480 360 240 __ M R=153.Chapter 5 Exercise Solutions 5-65 (5-55).1 2 4 6 8 10 12 O bser vation 14 16 18 20 U C L=502.7 3200 3000 _ X=2928. 00 standard deviations from center line. 5-76 . 40 5. Failed at points: 38. 39. 39. 9 points in a row on same side of center line.00 standard deviations from center line. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). Test Failed at points: 35. 40 Test Results for MR Chart of Ex5-57cTh TEST 1. One point more than 3. 40 TEST 6.65 5 0 LCL=0 1 6 12 18 24 30 Observation 36 42 48 54 60 Test Results for I Chart of Ex5-57cTh TEST Test TEST Test TEST 1. Test Failed at points: 41 Recall that observations on the Moving Range chart are correlated with those on the Individuals chart—that is.46 15 10 __ MR=5.Chapter 5 Exercise Solutions 5-66 (5-56). Test Failed at points: 34. 39. MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR (a) I-MR Chart of Oxide Thickness (Ex5-57cTh) All 60 Observations--Average Moving Range Method 70 1 Individual Value 5 60 2 6 UCL=66. Remove observation 38 and recalculate control limits. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). the out-of-control signal on the MR chart for observation 41 is reflected by the shift between observations 40 and 41 on the Individuals chart.07 2 6 _ X=51.05 50 40 LCL=36. 37. One point more than 3.03 1 6 12 18 24 30 Observation 36 48 54 60 1 20 Moving Range 42 UCL=18. Failed at points: 38 2. 38. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). Test Failed at points: 41 5-77 . 37. Test Failed at points: 34. 39. 9 points in a row on same side of center line.77 50 40 LCL=35.Chapter 5 Exercise Solutions 5-66 (a) continued Excluding observation 38 from calculations: I-MR Chart of Oxide Thickness (Ex5-57cTh) Less Observation 38 -. One point more than 3.00 standard deviations from center line. 40 TEST 6.58 0 LCL=0 1 6 12 18 24 30 Observation 36 42 48 54 60 Test Results for I Chart of Ex5-57cTh TEST Test TEST Test TEST 1. Test Failed at points: 35. 39. One point more than 3.94 1 6 12 18 24 30 Observation 36 42 48 54 60 1 20 Moving Range UCL=18. 40 5. Failed at points: 38 2. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). 39. 40 Test Results for MR Chart of Ex5-57cTh TEST 1.60 2 6 _ X=50.00 standard deviations from center line.Average Moving Range Method 70 1 Individual Value 5 60 2 6 UCL=65. 38. Failed at points: 38.22 15 10 5 __ MR=5. One point more than 3. 38. Test Failed at points: 41 5-78 . 40 Test Results for MR Chart of Ex5-57cTh TEST 1.05 50 40 LCL=36.83 2 6 _ X=51. Failed at points: 38 2.Median Moving Range Method 70 1 Individual Value 5 60 2 6 UCL=65. 40 5.00 standard deviations from center line. 39. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). Test Failed at points: 34.Chapter 5 Exercise Solutions 5-66 continued (b) I-MR Chart of Oxide Thickness (Ex5-57cTh) All 60 Observations -. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). 39.27 1 6 12 18 24 30 Observation 36 42 48 54 60 1 20 Moving Range UCL=18. Test Failed at points: 35.00 standard deviations from center line.16 15 10 5 __ MR=5.56 0 LCL=0 1 6 12 18 24 30 Observation 36 42 48 54 60 Test Results for I Chart of Ex5-57cTh TEST Test TEST Test TEST 1. 40 TEST 6. 39. 9 points in a row on same side of center line. Failed at points: 38. 37. One point more than 3. 00 15 10 5 __ MR=5. Test Failed at points: 34.Median Moving Range Method 70 1 1 Individual Value 5 60 2 6 2 UCL=64. 40 5. 9 points in a row on same side of center line. Failed at points: 38. One point more than 3. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). 39. 40 Test Results for MR Chart of Ex5-57cTh TEST 1.93 1 6 12 18 24 30 Observation 36 42 48 54 60 1 20 Moving Range UCL=17. Test Failed at points: 35. 39. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). Test Failed at points: 41 (c) The control limits estimated by the median moving range are tighter and detect the shift in process level at an earlier sample.Chapter 5 Exercise Solutions 5-66 (b) continued Excluding observation 38 from calculations: I-MR Chart of Oxide Thickness (Ex5-57cTh) Excluding Observation 38 from Calculations -.77 50 40 LCL=36.20 0 LCL=0 1 6 12 18 24 30 Observation 36 42 48 54 60 Test Results for I Chart of Ex5-57cTh TEST Test TEST Test TEST 1. 39. Failed at points: 33. One point more than 3. 37. 40 TEST 6. 33.00 standard deviations from center line. 5-79 .61 6 _ X=50. 38. 38 2.00 standard deviations from center line. 262 M oving Range 4 3 2 __ M R=1.7979 = 1.157 (b) MTB > Stat > Basic Statistics > Descriptive Statistics Descriptive Statistics: Ex5-67Meas Variable Ex5-67Meas Total Count 25 Mean 10.305 1 0 LC L=0 2 4 6 8 10 12 14 O bser vation 16 18 20 22 24 σˆ x = R / d 2 = 1.128 = 1.305 /1.Chapter 5 Exercise Solutions 5-67 (5-57).342 Median 10.630 σˆ x = S / c4 = 1.549 StDev 1.342 / 0.018 12 _ X=10.549 10 8 LC L=7.682 5-80 .079 2 4 6 8 10 12 14 O bser vation 16 18 20 22 24 U C L=4. (a) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Measurements (Ex5-67Meas) Individual V alue 14 U C L=14. 435 (e) As the span of the moving range is increased.137 (d) Average MR3 Chart: σˆ x = R / d 2 = 2.283 1 LCL=0 0 2 4 6 8 10 12 14 Observation 16 18 20 22 24 σˆ x = R / d 2 = 1.549 10 8 LCL=7.689 = 1.36 / 3.186 / 3.049 /1. and the estimate becomes less reliable. σ gets larger as the span increases.Chapter 5 Exercise Solutions 5-67 continued (c) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Measurements (Ex5-67Meas) Median Moving Range Method--Span = 2 Individual Value 14 UCL=13. 5-81 .693 = 1.210 Average MR4 Chart: σˆ x = R / d 2 = 2.137 2 4 6 8 10 12 14 Observation 16 18 20 22 24 UCL=4.598 / 2. This tends to be true for unstable processes. For this example.961 12 _ X=10. there are fewer observations to estimate the standard deviation.283 /1.192 Moving Range 4 3 2 __ MR=1.406 Average MR20 Chart: σˆ x = R / d 2 = 5.262 Average MR19 Chart: σˆ x = R / d 2 = 5.128 = 1.059 = 1.735 = 1. 8569 5.7566 5.. .0358361 The Individuals and Moving Range charts for the subgroup means are identical.07074 _ R=0.82 _ X=5.03345 0.. For this example. ..04 0. 5-82 .70 LCL=5. Estimate” and choose R-bar method to estimate standard deviation I-MR-R (Between/Within) Chart of Vane Heights (Ex5-68v1. the R chart tells the same story—same data pattern and no out-of-control points.08 UCL=0. Ex5-68v5) Subgroup Mean UCL=5.00 LCL=0 2 4 6 8 10 12 14 16 18 20 Sample Range 0.10 0.00 LCL=0 2 4 6 8 10 Ex5-68Cast 12 14 16 18 20 I-MR-R/S Standard Deviations of Ex5-68v1.0377 0.Chapter 5 Exercise Solutions 5-68 (5-58).6563 MR of Subgroup Mean 2 4 6 8 10 12 14 16 18 20 UCL=0.1233 0.05 __ MR=0. When compared to the s chart for all data...0143831 Between/Within 0. Ex5-68v5 Standard Deviations Between 0. the control schemes are identical.0328230 Within 0..76 5. MTB > Stat > Control Charts > Variables Charts for Subgroups > I-MR-R/S (Between/Within) Select “I-MR-R/S Options. 0611 0.09 _ R=0.1292 Sample Range 0..06 0.. This will lead to a understanding of the greatest source of variability. 7. Test Failed at points: 7 (b) Though the R chart is in control. plot points on the x chart bounce below and above the control limits... 13.. Ex5-69d5) 1 11. 9. . Ex5-69d5 Test Results for Xbar Chart of Ex5-69d1. 17 TEST 5. we might expect that the diameter of a single casting will not change much with location.. between castings or within a casting.03 0. we may want to consider treating the averages as an Individual value and graphing “between/within” range charts.79 U C L=11.82 1 Sample M ean 1 11. .. Since these are high precision castings. One point more than 3. Ex5-69d5 TEST 1.7579 11. . If no assignable cause can be found for these out-of-control points.76 _ _ X=11. (a) MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Xbar-R Chart of Casting Diameter (Ex5-69d1. 5-83 ... 2 out of 3 points more than 2 standard deviations from center line (on one side of CL).73 LC L=11.7931 11.Chapter 5 Exercise Solutions 5-69 (5-59).7226 11.12 0.00 LC L=0 2 4 6 8 10 Sample 12 14 16 18 20 Xbar-R Chart of Ex5-69d1. Test Failed at points: 5.00 standard deviations from center line.70 1 1 2 4 6 8 10 Sample 12 14 16 18 20 U C L=0. 00 2 4 6 8 10 12 14 16 18 20 Sample Range UCL=0.05 0. etc. Estimate” and choose R-bar method to estimate standard deviation MR-R/S (Between/Within) Chart of Casting Diameter (Ex5-69d1.1360 0. . Though the nature of this process leads us to believe that the diameter at any location on a single casting does not change much.Chapter 5 Exercise Solutions 5-69 continued (c) MTB > Stat > Control Charts > Variables Charts for Subgroups > I-MR-R/S (Between/Within) Select “I-MR-R/S Options. (e) The “within” chart is the usual R chart (n > 1).8685 11....05 LCL=0 0....10 _ R=0.. Ex5-69d5 11.10 __ MR=0.0262640 Between/Within 0.1292 0.6472 MR of Subgroup Mean 2 4 6 8 10 12 14 16 18 20 UCL=0. 5-84 . Ex5-69d5 I-MR-R/S Standard Deviations of Ex5-69d1.0349679 Within 0.00 LCL=0 2 4 6 8 10 Sample 12 14 16 18 20 I-MR-R/S (Between/Within) Chart of Ex5-69d1. we should continue to monitor “within” to look for wear.0437327 (d) We are taking several diameter measurements on a single precision casting.0416 0. .0611 0. It describes the measurement variability within a sample (variability in diameter of a single casting).. damage. .9 Subgroup Mean UCL=11. Ex5-69d5 Standard Deviations Between 0.7 LCL=11.. in the wax mold.7579 11.8 _ X=11.. within-wafer variability could be monitored with a standard X − R control chart. (c) Both between-wafer and total process variability could be estimated from measurements at one point on five consecutive wafers. (d) Within-wafer variability can still be monitored with randomly selected test sites. then only one location. a multivariate process control technique could be used. The data from each wafer could also be used to monitor between-wafer variability by maintaining an individuals X and moving range chart for each of the five fixed positions. If the positions are essentially identical. then either five X − R charts or some multivariate technique is needed. However. and (3) sample range is the difference within a wafer ( Rww ) . (2) moving range is the difference between successive wafers. Then a chart for individual x and moving range would provide information on lot-to-lot variability. Alternatively. If it is necessary to separately monitor the variation at each location. no information will be obtained about the pattern of variability within a wafer. needs to be monitored. with one X − R chart.Chapter 5 Exercise Solutions 5-70 (5-60). Individuals X and Moving Range charts should be used for process monitoring. The Minitab “between/within” control charts do this in three graphs: (1) wafer mean ( xww ) is an “individual value”. (a) Both total process variability and the overall process average could be estimated from a single measurement on one wafer from each lot. 5-85 . (b) Assuming that each wafer is processed separately. (e) The simplest scheme would be to randomly select one wafer from each lot and treat the average of all measurements on that wafer as one observation. 074 − 3(0.333 <0.25 Although the p-value is very small.04515 UNTL = x + 3s = 2. The wafer critical dimension is approximately normally distributed.20 2.9 Mean StDev N AD P-Value 99 Percent 95 90 2.939 5-86 .04515) = 1. s = 0.209 LNTL = x − 3s = 2.95 2. with many repeated values.90 1.074. (a) MTB > Stat > Basic Statistics > Normality Test Probability Plot of Critical Dimensions (Ex5-71All) Normal 99.Chapter 5 Exercise Solutions 5-71 (5-61).04515 200 1.074 + 3(0.1 1.05 2. the plot points do fall along a straight line.00 2.15 2. The natural tolerance limits (± 3 sigma above and below mean) are: x = 2.005 80 70 60 50 40 30 20 10 5 1 0.10 Ex5-71All 2.04515) = 2.074 0. .08 _ R=0.00 LCL=0 4 8 12 16 20 24 28 32 Sample (Lot Number-Wafer Order) 36 40 The Range chart is in control.04 0. 5-87 .Chapter 5 Exercise Solutions 5-71 continued (b) To evaluate within-wafer variability..14 Sample Range 0.16 UCL=0.07 0. MTB > Stat > Control Charts > Variables Charts for Subgroups > R R Chart of Critical Dimension Within Wafer (Ex5-71p1.1480 0.02 0.12 0..06 0. construct an R chart for each sample of 5 wafer positions (two wafers per lot number).. indicating that within-wafer variability is also in control.10 0. Ex5-71p5) 0. for a total of 40 subgroups. 08 _ X=2.1603 2. Ex5-71p5) Subgroup Mean Variability between wafers 2.0300946 Between/Within 0.16 UCL=2. ..00 LCL=1.10 0.Chapter 5 Exercise Solutions 5-71 continued (c) To evaluate variability between wafers.0735 6 6 2..08 0.05 __ MR=0.0326 LCL=0 0.00 LCL=0 4 8 12 16 20 Sample 24 28 32 36 40 I-MR-R/S Standard Deviations of Ex5-71p1.1480 _ R=0.. Ex5-71p5 Standard Deviations Between 0..0255911 Within 0. set up Individuals and Moving Range charts where the x statistic is the average wafer measurement and the moving range is calculated between two wafer averages.9868 MR of Subgroup Mean 4 8 12 16 20 24 28 32 36 40 UCL=0.16 UCL=0. indicating that between-wafer variability is also in-control. Estimate” and choose R-bar method to estimate standard deviation I-MR-R/S (Between/Within) Chart of Crit Dim (Ex5-71p1.07 0.. 5-88 . The “within” chart (Range) is not required to evaluate variability between wafers.1066 0. ..0395043 Both “between” control charts (Individuals and Moving Range) are in control. MTB > Stat > Control Charts > Variables Charts for Subgroups > I-MR-R/S (Between/Within) Select “I-MR-R/S Options.00 4 8 12 16 20 24 28 32 36 40 Sample Range 0. 0311891 Between/Within 0.00 LCL=0 2 4 6 8 10 12 14 16 18 20 Sample Range UCL=0. MTB > Stat > Control Charts > Variables Charts for Subgroups > I-MR-R/S (Between/Within) I-MR-R/S (Between/Within) Chart of Ex5-71All Subgroup Mean Lot-to-Lot Variability 2.2 UCL=2. and (3) range within a lot—the Minitab “between/within” control charts.10 0.0459 0.0214 2 4 6 8 10 Ex5-7Lot All 12 14 16 18 20 I-MR-R/S Standard Deviations of Ex5-71All Standard Deviations Between 0. three charts are needed: (1) lot average. (2) moving range between lot averages.1500 0.05 LCL=0. indicating that the lot-to-lot variability is also in-control.0394733 Within 0.Chapter 5 Exercise Solutions 5-71 continued (d) To evaluate lot-to-lot variability.0503081 All three control charts are in control.15 _ R=0. 5-89 .096 0.1706 0.0 LCL=1.9515 2 4 6 8 10 12 14 16 18 20 MR of Subgroup Mean 0.0735 2.08 __ MR=0.16 UCL=0.1 _ X=2.1956 2. 0585 0. it is considered to be in control. 6-1. ∑ Di = 117. This can be changed under Tools > Options > Control Charts and Quality Tools > Define Tests. Test Failed at points: 12 6-1 .0585(1 − 0.0585(1 − 0. 2. m = 20.0585 − 3 = 0. For these solutions.10 0.00 standard deviations from center line.04 0. New exercises are denoted with an “☺”. a run of n consecutive points on one side of the center line is defined as 9 points. In particular.08 _ P=0. If a plot point is on or beyond the control limits.1289 Proportion 0.14 UCL=0. One point more than 3.0585 20(100) UCL p = p + 3 p (1 − p ) 0. Also fewer special cause tests are available for attributes control charts. p = i =1 ∑ Di i =1 mn = 117 = 0. not 8.0585) = 0.06 0.0585) = 0.Chapter 6 Exercise Solutions Notes: 1. m m n = 100.0704 ⇒ 0 n 100 MTB > Stat > Control Charts > Attributes Charts > P P Chart of Nonconforming Assemblies (Ex6-1Num) 0. we follow the MINITAB convention for determining whether a point is out of control.00 LCL=0 2 4 6 8 10 12 Sample 14 16 18 20 Test Results for P Chart of Ex6-1Num TEST 1.16 1 0.12 0.02 0. If a plot point is within the control limits. 3.0585 + 3 = 0. it is considered to be out of control.0585 − 0. MINITAB defines some sensitizing rules for control charts differently than the standard rules.1289 n 100 LCL p = p − 3 p (1 − p ) 0. 10 0. so remove from control limit calculation: m m n = 100. m = 19.0537(1 − 0.00 LCL=0 2 4 6 8 10 12 Sample 14 16 18 20 Test Results for P Chart of Ex6-1Num TEST 1.04 0.12 0.0537 − 0.0537 0. One point more than 3.0537 19(100) UCL p = 0. p = i =1 ∑ Di i =1 mn = 102 = 0.Chapter 6 Exercise Solutions 6-1 continued Sample 12 is out-of-control.0537 + 3 0.0537) = 0.14 UCL=0.16 1 0. Test Failed at points: 12 6-2 .08 _ P=0.0537(1 − 0.0676 ⇒ 0 100 MTB > Stat > Control Charts > Attributes Charts > P P Chart of Nonconforming Assemblies (Ex6-1Num) Sample 12 removed from calculations 0.0537) = 0.06 0.0537 − 3 0. ∑ Di = 102.1213 Proportion 0.1213 100 LCL p = 0.02 0.00 standard deviations from center line. 10 0.0230(1 − 0.0230 + 3 = 0.00 standard deviations from center line.0230 − 3 = 0.0597 0.0230) = 0.08 Proportion 1 0.0230 20(150) UCL p = p + 3 p (1 − p ) 0. ∑ Di = 69. m = 20.0367 ⇒ 0 n 150 MTB > Stat > Control Charts > Attributes Charts > P P Chart of Nonconforming Switches (Ex6-2Num) 1 0.02 LCL=0 0.00 2 4 6 8 10 12 Sample 14 16 18 20 Test Results for P Chart of Ex6-2Num TEST 1.023 0. m m n = 150.Chapter 6 Exercise Solutions 6-2. Test Failed at points: 9.0230) = 0.0230 − 0. p = ∑ Di i =1 mn i =1 = 69 = 0.0597 n 150 LCL p = p − 3 p (1 − p ) 0.0230(1 − 0.06 UCL=0. 17 6-3 .04 _ P=0. One point more than 3. 9.00 standard deviations from center line.02 _ P=0.0473 0.06 1 UCL=0.0163 0.00 LCL=0 2 4 6 8 10 12 Sample 14 16 18 20 Test Results for P Chart of Ex6-2Num TEST 1.10 0.Chapter 6 Exercise Solutions 6-2 continued Re-calculate control limits without samples 9 and 17: MTB > Stat > Control Charts > Attributes Charts > P P Chart of Nonconforming Switches (Ex6-2Num) Samples 9 and 17 excluded from calculations 1 0.04 0. One point more than 3.08 Proportion 1 0. Test Failed at points: 1. 17 6-4 . 0430 150 LCL p = 0. ∑ Di = 36.0141 0.08 Proportion 1 0. 9.0141 17(150) UCL p = 0. 17 excluded from calculations 1 0. 9.0430 0.0141 − 0.0289 ⇒ 0 150 MTB > Stat > Control Charts > Attributes Charts > P P Chart of Nonconforming Switches (Ex6-2Num) Samples 1.04 0. m = 17.00 standard deviations from center line. p = i =1 ∑ Di i =1 mn = 36 = 0.02 _ P=0.0141(1 − 0.06 1 UCL=0.Chapter 6 Exercise Solutions 6-2 continued Also remove sample 1 from control limits calculation: m m n = 150. One point more than 3.0141(1 − 0.10 0.0141) = 0.0141 − 3 0. 17 6-5 . Test Failed at points: 1.0141) = 0.00 LCL=0 2 4 6 8 10 12 Sample 14 16 18 20 Test Results for P Chart of Ex6-2Num TEST 1.0141 + 3 0. 02 0.10 0.06(1 − 0.14 UCL=0.06 − 3 0.06 UCLi = p + 3 p (1 − p ) ni and LCLi = max{0.06) 80 = 0.06 0.04 0.046.06 − 0.06) 80 = 0. p = ∑ Di ∑ ni = 60 1000 = 0. 6-6 . m m m m i =1 i =1 i =1 i =1 m = 10.16 0. ∑ ni = 1000.06 0.08 _ P=0. for n = 80: UCL1 = p + 3 p (1 − p ) n1 = 0. The Fraction Nonconforming for Day 5 should be 0. p − 3 p (1 − p ) ni } As an example.Chapter 6 Exercise Solutions 6-3.12 0.06(1 − 0.0797 ⇒ 0 MTB > Stat > Control Charts > Attributes Charts > P P Chart of Nonconforming Units (Ex6-3Num) 0. ∑ Di = 60. NOTE: There is an error in the table in the textbook.1331 Proportion 0.1397 LCL1 = p − 3 p (1 − p ) n1 = 0.06 + 3 0.00 LCL=0 1 2 3 4 5 6 Sample 7 8 9 10 Tests performed with unequal sample sizes The process appears to be in statistical control. 01 LCL=0 0.0314 ⇒ 0 MTB > Stat > Control Charts > Attributes Charts > P P Chart of Nonconforming Units (Ex6-4Num) 0.0167) 150 = 0.00 2 4 6 8 10 12 Sample 14 16 18 20 The process appears to be in statistical control. ∑ Di = 50.03 0.05 UCL=0. (a) m m i =1 i =1 n = 150.9 Select n = 530. (1 − p) 2 n> L p (1 − 0.0167 + 3 0. (b) Using Equation 6-12.04 0.0167) 150 = 0. m = 20.0167) 2 (3) > 0.0167 UCL = p + 3 p (1 − p ) n = 0.04802 Proportion 0.01667 0.0167 − 0.Chapter 6 Exercise Solutions 6-4. 6-7 .02 _ P=0.0480 LCL = p − 3 p (1 − p ) n = 0.0167(1 − 0. p = ∑ Di mn = 50 20(150) = 0.0167 > 529.0167(1 − 0.0167 − 3 0. 1031 1 1 1 0. 19. Instead.00 standard deviations from center line.1228) 2500 = 0.1228) 2500 = 0. 5.050 1 2 4 6 8 10 12 Sample 14 16 18 20 Test Results for P Chart of Ex6-5Num TEST 1. 16.1228(1 − 0.075 1 1 0.1228 + 3 0. 11. 15. 12.1228(1 − 0.1425 0. (a) UCL = p + 3 p (1 − p ) n = 0.Chapter 6 Exercise Solutions 6-5. 3.1031 MTB > Stat > Control Charts > Attributes Charts > P P Chart of Nonconforming Belts (Ex6-5Num) 0. 17. 2. One point more than 3.1425 LCL = p − 3 p (1 − p ) n = 0.1228 0.150 UCL=0. 6-8 .1228 − 3 0. Test Failed at points: 1.200 1 1 0.125 _ P=0.100 LCL=0.175 1 1 1 Proportion 0. the process should be investigated for causes of the wild swings in p. 20 (b) So many subgroups are out of control (11 of 20) that the data should not be used to establish control limits for future production. 976 ⇒ 0 MTB > Stat > Control Charts > Attributes Charts > NP NP Chart of Number of Nonconforming Units (Ex6-6Num) 1 12 Sample Count 10 UCL=9.00 standard deviations from center line. Test Failed at points: 6 6-9 . One point more than 3. UCL = np + 3 np (1 − p ) = 4 + 3 4(1 − 0.98 8 6 __ NP=4 4 2 LCL=0 0 1 2 3 4 5 6 Ex6-6Day 7 8 9 10 Test Results for NP Chart of Ex6-6Num TEST 1.008) = 4 − 5.976 LCL = np − 3 np (1 − p ) = 4 − 3 4(1 − 0.008) = 9.Chapter 6 Exercise Solutions 6-6. 00 standard deviations from center line.6 continued Recalculate control limits without sample 6: NP Chart of Number of Nonconforming Units (Ex6-6Num) Day 6 excluded from control limits calculations 1 12 Sample Count 10 UCL=8. 6-10 . Test Failed at points: 6 Recommend using control limits from second chart (calculated less sample 6).Chapter 6 Exercise Solutions 6.39 8 6 4 __ NP=3. One point more than 3.11 2 0 LCL=0 1 2 3 4 5 6 Ex6-6Day 7 8 9 10 Test Results for NP Chart of Ex6-6Num TEST 1. 0044 i =1 10 UCL = p + 3 p (1 − p ) n = 0.02 − 0. use the Poisson approximation to the binomial with λ = npnew = 50(0. Pr{detected by 3rd sample} = 1 – Pr{detected after 3rd} = 1 – (1 – 0.0794 LCL = p − 3 p (1 − p ) n = 0.04 < 0.02(1 − 0.0794) | 2} + Pr{D ≤ 50(0) | 2} = 1 – POI(3.0044) 250 = 0. p = UCL = p − 3 p (1 − p ) n = 0. 0. From the 6th sample. 6-11 .02) 50 = 0.0044) 250 = 0.0594 ⇒ 0 Since pnew = 0.0440 = 0.02 − 3 0.135 = 0.00.0044(1 − 0. n = 50 UCL = p + 3 p (1 − p ) n = 0. The data from the shipment do not indicate statistical control.2) + POI(0.624 6-8.1 and n = 50 is "large".0044(1 − 0.02) 50 = 0. n = 250.0044 − 3 0. ∑ pˆ i = 0.2) = 1 – 0.857 + 0.0044 − 0. ( pˆ 6 = 0.0170.020) > 0.278 where POI(⋅) is the cumulative Poisson distribution.278)3 = 0.0170 10 m = 10.02(1 − 0. Pr{detect|shift} = 1 – Pr{not detect|shift} =1–β = 1 – [Pr{D < nUCL | λ} – Pr{D ≤ nLCL | λ}] = 1 – Pr{D < 50(0.0126 ⇒ 0 No. p = 0.0440. the UCL.02 + 3 0.02.0044 + 3 0.Chapter 6 Exercise Solutions 6-7.04) = 2. 000000 0.001179 0.503553 Pr{D ≤ 0|p} 0.10) 64 = 0.000000 0.212 Pr{D ≤ 13|p} 0.10 − 3 0.000000 β 0.2125 LCL = p − 3 p (1 − p ) n = 0.215 0.22 0.962475 0.10 + 3 0.503553 Assuming L = 3 sigma control limits.10.05 0.44154 0.598077 0. n = 64 UCL = p + 3 p (1 − p ) n = 0. p = 0. (1 − p ) 2 n> L p (1 − 0.519279 0.10) 64 = 0.10(1 − 0.994993 0.480098 0.480098 0.10 > 81 6-12 .996172 0.44154 0.2125) | p} − Pr{D ≤ 64(0) | p} = Pr{D < 13.10 − 0.519279 0.10(1 − 0.10 0.6) | p} − Pr{D ≤ 0 | p} p 0.037524 0.000000 0.1125 ⇒ 0 β = Pr{D < nUCL | p} − Pr{D ≤ nLCL | p} = Pr{D < 64(0.000000 0.10) 2 (3) > 0.21 0.Chapter 6 Exercise Solutions 6-9.598077 0.20 0.999999 0. 000 = 0.16) = 5.20 − 0.0.16 UCL = np + 3 np(1 − p ) = 16 + 3 16(1 − 0.00 (a) npnew = 20.5 − 20 ⎞ = 1− Φ ⎜ + Φ⎜ ⎟ ⎜ 20(1 − 0.10) = 81 ⎝δ ⎠ ⎝ 0.970 + 0.20.0 > 15. so use normal approximation to binomial distribution.16 > 47.16) 2 (3) > 0. (1 − p ) 2 n> L p (1 − 0.50.10 = 0. p = 0.16) = 27. assume k = 3 sigma control limits new − p = 0. n = 48 is the minimum sample size for a positive LCL. np = 16.10 ⎠ 6-13 .10)(1 − 0.0873 (b) Assuming L = 3 sigma control limits.2) ⎟⎟ ⎝ ⎠ ⎝ ⎠ = 1 − Φ (1.25 So. desire Pr{detect} = 0. p = 0.875) = 1 − 0. 6-11. Pr{detect shift on 1st sample} = 1 − β = 1 − [Pr{D < UCL | p} − Pr{D ≤ LCL | p}] ⎛ UCL + 1/ 2 − np ⎞ ⎛ LCL − 1/ 2 − np ⎞ = 1− Φ ⎜ +Φ⎜ ⎟ ⎟ ⎜ ⎜ np (1 − p) ⎟⎠ np (1 − p ) ⎟⎠ ⎝ ⎝ ⎛ 27 + 0.10 δ =p new 2 2 ⎛k⎞ ⎛ 3 ⎞ n = ⎜ ⎟ p(1 − p) = ⎜ ⎟ (0.00 LCL = np − 3 np (1 − p ) = 16 − 3 16(1 − 0.875) + Φ (−3.030 Pr{detect by at least 3rd} = 1 – Pr{detected after 3rd} = 1 – (1 – 0.2) ⎟ ⎜ 20(1 − 0.10.5 − 20 ⎞ ⎛ 5 − 0. n = 100.030)3 = 0. p = 16 100 = 0.Chapter 6 Exercise Solutions 6-10. 996] = 0. p = 0.37) = 0. n = 100.20) = 20 > 15.080 < 0.20 0 − 0.161 − 0.07494 (d) Pr{detect shift by at most 4th sample} = 1 – Pr{not detect by 4th} = 1 – (0.080) = 8 − 8.44) − Φ (−7. LCL = 0 (a) np = 100(0.1388 ⇒ 0 (b) p = 0. UCL = 0.08(1 − 0. so use the normal approximation to the binomial.07494 − 0 = 0.004 where POI(⋅) is the cumulative Poisson distribution.8)] = 0 + [1 – 0.14 LCL = np − 3 np (1 − p ) = 8 − 3 8(1 − 0. Pr{type I error} = α = Pr{D < LCL | p} + Pr{D > UCL | p} = Pr{D < LCL | p} + [1 – Pr{D ≤ UCL | p}] = Pr{D < 0 | 8} + [1 – Pr{D ≤ 16 | 8}] = 0 + [1 – POI(16. (c) npnew = 100(0.08(1 − 0.161.Chapter 6 Exercise Solutions 6-12.08) 100 ⎟ ⎜ 0.080) = 8 UCL = np + 3 np(1 − p ) = 8 + 3 8(1 − 0.080) = 16.99997 6-14 .1 and n =100 is large.20 = Φ⎜ −Φ⎜ ⎟ ⎜ 0. so use Poisson approximation to the binomial.07494)4 = 0.08) 100 ⎟⎟ ⎝ ⎠ ⎝ ⎠ = Φ (−1. Pr{type II error} = β = Pr{ pˆ < UCL | pnew } − Pr{ pˆ ≤ LCL | pnew } ⎛ UCL − pnew ⎞ ⎛ LCL − pnew ⎞ = Φ⎜ −Φ⎜ ⎟ ⎜ p (1 − p ) n ⎟ ⎜ p (1 − p ) n ⎟⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ 0.08. 297 + (1 – 0.07) 400 = 0.26.1(1 − 0.1(1 − 0. n = 400 UCL = p + 3 p(1 − p) n = 0.297)(0.07) 400 = 0. (a) p = 0.20 and L = 3 sigma control limits (1 − p ) 2 n> L p (1 − 0.07 − 3 0.1 ⎞ = 1− Φ ⎜ + Φ⎜ ⎟ ⎜ 0.06 2 2 ⎛k⎞ ⎛ 3 ⎞ n = ⎜ ⎟ p(1 − p) = ⎜ ⎟ (0.06 ⎠ 6-15 . δ =pnew − p = 0. k = 3 sigma control limits.10) = > 40.297 (c) Pr{detect on 1st or 2nd sample} = Pr{detect on 1st} + Pr{not on 1st}×Pr{detect on 2nd} = 0.1) 400 ⎟ ⎜ 0.533) + Φ (−4.506 6-14.032 − 0.26 − 0.07 + 3 0.20 > 36 For Pr{detect} = 0.108 − 0.000 = 0.108 LCL = p − 3 p(1 − p ) n = 0.032 (b) npnew = 400(0.07(1 − 0.20)(1 − 0.07.533) = 1 − 0.20) = 400 ⎝δ ⎠ ⎝ 0.703 + 0.1) 400 ⎟⎟ ⎝ ⎠ ⎝ ⎠ = 1 − Φ (0.20) 2 (3) > 0.20 = 0.07(1 − 0.50 after a shift to pnew = 0.Chapter 6 Exercise Solutions 6-13. p = 0.297) = 0. Pr{detect on 1st sample} = 1 − Pr{not detect on 1st sample} = 1− β = 1 − [Pr{ pˆ < UCL | p} − Pr{ pˆ ≤ LCL | p}] ⎛ UCL − p ⎞ ⎛ LCL − p ⎞ = 1− Φ ⎜ +Φ⎜ ⎟ ⎜ p (1 − p) n ⎟ ⎜ p(1 − p ) n ⎟⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎛ 0. so use the normal approximation to the binomial.1 ⎞ 0. 164) = 27.4 15 10 LCL=5. (a) m = 10.4 − 3 16.00 standard deviations from center line.4(1 − 0.29 5 1 2 3 4 5 6 Sample 7 8 9 10 Test Results for NP Chart of Ex6-15Num TEST 1.Chapter 6 Exercise Solutions 6-15.164) = 5. i =1 10 p = ∑ Di i =1 ( mn ) = 164 [10(100)] = 0.4(1 − 0. 10 ∑ Di = 164. One point more than 3.4 UCL = np + 3 np (1 − p ) = 16. Test Failed at points: 3 6-16 . n = 100.51 LCL = np − 3 np (1 − p ) = 16.51 Sample Count 25 20 __ NP=16. np = 16.4 + 3 16.164.292 MTB > Stat > Control Charts > Attributes Charts > NP NP Chart of Number Nonconforming (Ex6-15Num) 1 30 UCL=27. Chapter 6 Exercise Solutions 6-15 continued Recalculate control limits less sample 3: NP Chart of Number Nonconforming (Ex6-15Num) Sample 3 excluded from calculations 1 30 UCL=25.00 standard deviations from center line.13 1 2 3 4 5 6 Sample 7 8 9 10 Test Results for NP Chart of Ex6-15Num TEST 1.42 Sample Count 25 20 __ NP=14.78 15 10 5 LCL=4. Test Failed at points: 3 6-17 . One point more than 3. 16) + POI(0.7255)4 = 0. use the Poisson approximation to the binomial.08)} = 1 − POI(13.Chapter 6 Exercise Solutions 6-15 continued (b) pnew = 0.3) ⎟⎠ ⎝ ⎝ ⎠ = 1 − Φ (−0.16) = 1 − 0.03) 200 = 0.08)} + Pr{D ≤ 200(0) | 200(0.813 6-16.42 + 0. Pr{detect on 1st sample | p} = 1 − Pr{not detect | p} = 1− β = 1 − [Pr{ pˆ < UCL | p} − Pr{ pˆ ≤ LCL | p}] = 1 − Pr{D < nUCL | np} + Pr{D ≤ nLCL | np} = 1 − Pr{D < 200(0.000 = 0.000) = 0.30.50.10 and n is large. (a) UCL p = p + 3 p (1 − p ) n = 0. Pr{detect on 1st} = 1 − Pr{not detect on 1st} = 1− β = 1 − [Pr{D < UCL | p} − Pr{D ≤ LCL | p}] ⎛ UCL + 1/ 2 − np ⎞ ⎛ LCL − 1/ 2 − np ⎞ = 1− Φ ⎜ + Φ⎜ ⎟ ⎟ ⎜ ⎜ np (1 − p ) ⎟⎠ np (1 − p ) ⎟⎠ ⎝ ⎝ ⎛ 25.0362 ⇒ 0 (b) pnew = 0.03(1 − 0.30 is not too far from 0.187) + (0.2745 + 0.13 − 0.3) ⎟⎟ 30(1 − 0. Since (pnew = 0.0662 LCL p = p − 3 p (1 − p ) n = 0. the normal approximation to the binomial can be used.08.03 − 3 0.7544) = 1 − (0.7255 where POI(⋅) is the cumulative Poisson distribution.03 + 3 0.0662) | 200(0. Pr{detect by at least 4th} = 1 – Pr{detect after 4th} = 1 – (1 – 0.03(1 − 0.9943 6-18 .8903) + Φ (−5.03 − 0.08) < 0. and n = 100 > 10. Since p = 0.03) 200 = 0.5 − 30 ⎞ = 1− Φ ⎜ +Φ⎜ ⎟ ⎜ ⎜ 30(1 − 0.5 − 30 ⎞ ⎛ 4. so use the normal approximation to the binomial. np = 400(0.15) ⎟⎟ ⎝ ⎠ ⎝ ⎠ = 1 − Φ (−0.Chapter 6 Exercise Solutions 6-17.583 6-19 . (a) m p = ∑ Di i =1 UCL LCL np np ( mn ) = 1200 [30(400)] = 0. Pr{detect on 1st sample | p} = 1 − Pr{not detect on 1st sample | p} = 1− β = 1 − [Pr{D < UCL | np} − Pr{D ≤ LCL | np}] ⎛ UCL + 1/ 2 − np ⎞ ⎛ LCL − 1/ 2 − np ⎞ = 1− Φ ⎜ + Φ⎜ ⎟ ⎟ ⎜ ⎜ np (1 − p ) ⎟⎠ np (1 − p ) ⎟⎠ ⎝ ⎝ ⎛ 58 + 0.39) = 1 − 0.417 + 0.10.210) + Φ (−5.000 = 0.5 − 60 ⎞ = 1− Φ ⎜ + Φ⎜ ⎟ ⎜ 60(1 − 0.5 − 60 ⎞ ⎛ 22 − 0.15) ⎟ ⎜ 60(1 − 0.10) = 22 (b) npnew = 400 (0.10) = 58 = np − 3 np (1 − p ) = 40 − 3 40(1 − 0.15) = 60 > 15.10) = 40 = np + 3 np (1 − p ) = 40 + 3 40(1 − 0. Chapter 6 Exercise Solutions 6-18. from 6-18(c).20) = 20. 20) − POI(1.000 = 0.000 + 1 − 0.10) = 10. Pr{type II error} = β = Pr{D < nUCL | λ} − Pr{D ≤ nLCL | λ} = Pr{D < 100(0.583 ARL1 = 1/(1 –β) = 1/(0.01) |10} + 1 − Pr{D ≤ 100(0.715 ≅ 2 6-20. (a) UCL = p + 3 p (1 − p ) n 2 ⎛ ⎞ 3 3 ⎛ ⎞ n = p (1 − p ) ⎜ ⎟ = 0. (c) pnew = 0.004 where POI(⋅) is the cumulative Poisson distribution. λ = np = 100(0. 1 – β = 0. NOTE: There is an error in the textbook.10) = 0. β = 0. 6-19. Pr{type I error} = Pr{ pˆ < LCL | p} + Pr{ pˆ > UCL | p} = Pr{D < nLCL | λ} + 1 − Pr{D ≤ nUCL | λ} = Pr{D < 100(0.10) + 1 − POI(19.381) = 1.1) ⎜ ⎟ = 100 ⎝ 0. Using the Poisson approximation to the binomial.1 ⎠ ⎝ UCL − p ⎠ 2 (b) Using the Poisson approximation to the binomial.19) | 20} − Pr{D ≤ 100(0. from 6-17(b). λ = npnew = 100(0.381 ARL1 = 1/(1 –β) = 1/(1 – 0.19 − 0.01) | 20} = POI(18. not 6-18.996 = 0. This is a continuation of Exercise 6-17.583) = 1. 20) = 0.381 − 0.1(1 − 0.381 where POI(⋅) is the cumulative Poisson distribution.616 ≅ 2 6-20 .19) |10} = POI(0.20. Chapter 6 Exercise Solutions 6-21.03 _ P=0.0221) 0. preferred in many cases. 0. (b) There are two approaches for controlling future production. 6-21 . The second approach. would be to construct standardized control limits with control limits at ± 3.01 LCL=0 0.05 UCL=0.0221 ± 3 0. 0.04 0.0216 / ni .02 0.0581] [0.02213 0.05005 0.00 2 4 6 8 10 12 Sample 14 16 18 20 Tests performed with unequal sample sizes Process is in statistical control. 0. (a) For a p chart with variable sample size: p = ∑ i Di ∑ i ni = 83 / 3750 = 0. calculate the exact control limits by p ± 3 p (1 − p ) / ni = 0. 0.0662] [0. UCLi] [0.06 Proportion 0.0221(1 − 0. and to plot Z i = ( pˆ i − 0.0221) ni .0533] [0.07 0. In those cases. The first approach would be to plot pˆ i and use constant limits unless there is a different size sample or a plot point near a control limit.0221 and control limits are at p ± 3 p (1 − p ) / ni [LCLi.0500] ni 100 150 200 250 MTB > Stat > Control Charts > Attributes Charts > P P Chart of Second Visit Required (Ex6-21Sec) 0. 5. MTB > Stat > Control Charts > Attributes Charts > P P Chart of Second Visit Required (Ex6-21Sec) Limits based on average sample size (n=187) 0. and carefully examine points near the control limits.00 LCL=0 2 4 6 8 10 12 Sample 14 16 18 20 Process is in statistical control. Use a sample size of n = 187.Chapter 6 Exercise Solutions 6-22.02219 0. MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex6-21Req Variable Ex6-21Req N 20 Mean 187. however MINITAB accepts only integer values for n.5 Average sample size is 187.06 UCL=0.03 _ P=0.05 Proportion 0.05451 0.04 0.02 0. 6-22 .01 0. zi = ( pˆ i − p ) p (1 − p ) ni = ( pˆ i − 0.0216 / ni MTB > Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Standardized Second Visit Data (Ex6-23zi) UCL=3 3 Individual Value 2 1 _ X=0 0 -1 -2 -3 LCL=-3 2 4 6 8 10 12 Observation 14 16 18 20 Process is in statistical control.Chapter 6 Exercise Solutions 6-23. 6-23 .0221) 0. 0581. CL = 0.01 0.05 0. UCL250 = 0. UCL200 = 0.Chapter 6 Exercise Solutions 6-24.0500 MTB > Graph > Time Series Plot > Multiple Control Chart of Second Visit Data with Limits for Various Sample Sizes (Ex6-24pi) Proportion of Second Visits Required 0.06 0.00 2 4 6 8 10 12 Week 14 16 18 20 6-24 .03 0.02 0.04 0.07 Variable Ex6-24pi Ex6-24n100 Ex6-24n150 Ex6-24n200 Ex6-24n250 Ex6-24C L Ex6-24LCL 0.0662. UCL150 = 0.0221. LCL = 0 UCL100 = 0.0533. 05) = 7. LCL = 0.01 ⎠ > 891 ≥ 892 6-26. UCL = 0.12) = 1 − 1.0399.0809.1. n = 100 ⎛ 1− p ⎞ 2 n>⎜ ⎟L ⎝ p ⎠ ⎛ 1 − 0.8349 np(1 − p) = 20 + 2. Pr{detect shift to 0.8349 np (1 − p ) = 20 − 2. UCL = 0.05 + L 0. use Poisson approximation to binomial. p = CL = 0.01. The np chart is inappropriate for varying sample sizes because the centerline (process center) would change with each ni. n = 400. 6-27.05) = 32.05 + L(0.05) 400 = 0.64 (c) n = 400 is large and p = 0.05 < 0.0109) L = 2.0191 (a) 0. p = CL = 0.36 LCL = np − 2.05(1 − 0.0000 + 0.01 ⎞ 2 >⎜ ⎟3 ⎝ 0.8349 (b) CL = np = 400(0. 6-25 . LCL = 0.0895 where POI(·) is the cumulative Poisson distribution.12) + POI(7.0809 = 0.0500.36 |12} + Pr{D ≤ 7.Chapter 6 Exercise Solutions 6-25.8349 20(1 − 0.05) = 20 UCL = np + 2.64 |12} = 1 − POI(32.0895 = 0.03 on 1st sample} = 1 − Pr{not detect} = 1− β = 1 − [Pr{D < UCL | λ} − Pr{D ≤ LCL | λ}] = 1 − Pr{D < 32.8349 20(1 − 0. 99865 6-29.03 ⎠ 6-26 . (a) UCL = p + L p (1 − p ) n 0.05) 400 L = 4.01 = 0. λ = np = 400(0.15) = 60 > 15.0038 − 0.000 = 0.00) + Φ (−8.0962 − 0.15) 400 ⎟ ⎝ ⎠ ⎝ ⎠ = 1 − Φ (−3.05(1 − 0.01.01 ⎠ > 396 ≥ 397 (b) δ = 0.00135 + 0.15(1 − 0. p = 0. Pr{detect on 1st sample after shift} = 1 − Pr{not detect} = 1− β = 1 − [Pr{ pˆ < UCL | p} − Pr{ pˆ ≤ LCL | p}] ⎛ UCL − p ⎞ ⎛ LCL − p ⎞ = 1− Φ ⎜ +Φ⎜ ⎟ ⎜ p (1 − p ) n ⎟ ⎜ p (1 − p ) n ⎟⎟ ⎝ ⎠ ⎝ ⎠ ⎞ ⎛ ⎞ ⎛ 0. L = 2 (a) ⎛ 1− p ⎞ 2 n>⎜ ⎟L ⎝ p ⎠ ⎛ 1 − 0.03 2 2 ⎛L⎞ ⎛ 2 ⎞ n = ⎜ ⎟ p(1 − p) = ⎜ ⎟ (0.01 ⎞ 2 >⎜ ⎟2 ⎝ 0.15) 400 ⎟ ⎜ 0.0962 = 0.01)(1 − 0.04 – 0. use normal approximation to binomial.01) = 44 ⎝δ ⎠ ⎝ 0.0500 + L 0.19) = 1 − 0.24 (b) p = 15.15(1 − 0.15 0.Chapter 6 Exercise Solutions 6-28.15 = 1− Φ ⎜ +Φ⎜ ⎟ ⎟ ⎜ 0. 6) − POI(0.005) |100(0.002 = 0.04)} + 1 − Pr{D ≤ 100(0.070 where POI(⋅) is the cumulative Poisson distribution.075) |100(0.06)} − Pr{D ≤ 100(0.06) = POI(7. 4) + 1 − POI(7. (a) Pr{type I error} = Pr{ pˆ < LCL | p} + Pr{ pˆ > UCL | p} = Pr{D < nLCL | np} + 1 − Pr{D ≤ nUCL | np} = Pr{D < 100(0. 4) = 0.075) |100(0. 6) = 0.04)} = POI(0. 6-27 .0050) |100(0. (b) Pr{type II error} =β = Pr{D < nUCL | np} − Pr{D ≤ nLCL | np} = Pr{D < 100(0.018 + 1 − 0.742 where POI(⋅) is the cumulative Poisson distribution.948 = 0.Chapter 6 Exercise Solutions 6-30.744 − 0. β = 0.0498 0.0742: ARL1 = 1/(1 –β) = 1/(1 – 0.01 0.0698 0.Chapter 6 Exercise Solutions 6-30 continued (c) β = Pr{D < nUCL | np} − Pr{D ≤ nLCL | np} = Pr{D < 100(0.5 1.6000 0.06 0.4000 0.0000 0.0000 0.742) = 3.9989 0.861 ≅ 4 6-28 .1 0. α = 0.04 0.9489 0.8666 0.7000 Beta 0.0067 0.0025 0.3679 0.0000 Pr{D<=0. UCL=7.1000 0.5|np} 1.07 0.3000 0.0000 0.2 0.03 0.5987 0.8636 0.0000 0.3 p (d) from part (a).05 0.0000 OC Curve for n=100.5 1 2 3 4 5 6 7 8 9 10 12.5 |100 p} − Pr{D ≤ 0.9383 0.0009 0.15 0.xls : worksheet Ex6-30 p 0 0.0183 0.0000 1.0698 0.0008 0.0000 0.15 0.125 0.7415 0.9000 0.0050) |100 p} = Pr{D < 7.070 = 14.0180 0.5.005 0.0000 1.0001 0.08 0.070: ARL0 = 1/α = 1/0. LCL=0.1353 0.2202 0.5|np} 1.0003 0.1 0.6321 0.25 0.4526 0.3238 0.3935 0.0008 0.9306 0.6065 0.0000 0.9881 0.0000 0 0.3239 0.02 0.0000 0.2202 0.0750) |100 p} − Pr{D ≤ 100(0.2000 0.2 0.09 0.29 ≅ 15 from part (b).8599 0.5978 0.25 np 0 0.0000 beta 0.8000 0.0180 0. CL=4.4530 0.5000 0.0000 0.5 |100 p} Excel : workbook Chap06.05 0.5 15 20 25 Pr{D<7.7440 0. n = 100.038 and σˆ p = 0.062 LCL = p − 3 p (1 − p ) n = 0. p = 0.02) 100 = 0. One point more than 3.02 (a) UCL = p + 3 p (1 − p ) n = 0.03 _ P=0.00 standard deviations from center line.01 0.02(1 − 0.04 0.00 LCL=0 1 2 3 4 5 6 Sample 7 8 9 10 Test Results for P Chart of Ex6-31Num TEST 1.09 1 0. LCL = np − k np (1 − p ) > 0 np > k np (1 − p ) ⎛ 1− p ⎞ n > k2 ⎜ ⎟ ⎝ p ⎠ 6-29 .02 0.02 + 3 0.07 UCL=0.062 Proportion 0.05 0.Chapter 6 Exercise Solutions 6-31.06 0.02 0. p = 0.08 0.0191 6-32.02(1 − 0.02 − 3 0.02) 100 ⇒ 0 (b) MTB > Stat > Control Charts > Attributes Charts > P P Chart of Number Nonconforming (Ex6-31Num) 0. Test Failed at points: 4 Sample 4 exceeds the upper control limit. 5 2 1 LCL=0 0 2 4 6 8 10 12 Sample 14 16 18 20 The process is in control.505(1 − 0. n = 150. p = 0.505 − 4. m = 20. results are the same as for the p chart.708 ⇒ 0 MTB > Stat > Control Charts > Attributes Charts > NP NP Chart of Numer of Nonconforming Units (Ex6-4Num) 8 UCL=7.505 UCL = np + 3 np (1 − p ) = 2.213 LCL = np − 3 np (1 − p ) = 2.0167) = 2.0167 CL = np = 150(0.204 7 Sample Count 6 5 4 3 __ NP=2. 6-30 . ∑ D = 50.Chapter 6 Exercise Solutions 6-33.505 + 3 2.0167) = 7. indicating that this data should not be used to establish control limits for future production.8 200 1 1 1 1 1 1 100 2 4 6 8 10 12 Sample 14 16 18 20 Test Results for NP Chart of Ex6-5Num TEST 1. 11. 12. 17. 20 Like the p control chart.77 MTB > Stat > Control Charts > Attributes Charts > NP NP Chart of Number of Nonconforming Belts (Ex6-5Num) 500 1 1 1 400 Sample Count 1 1 UCL=356. 19.23 LCL = np − 3 np (1 − p ) = 307 − 3 307(1 − 0.Chapter 6 Exercise Solutions 6-34. 16. many subgroups are out of control (11 of 20).3 __ NP=307. 5.00 standard deviations from center line. 2.1228) = 356. CL = np = 2500(0. One point more than 3. Test Failed at points: 1. 3. 6-31 .1228) = 257.1 300 LCL=257.1228) = 307 UCL = np + 3 np (1 − p ) = 307 + 3 307(1 − 0. 15. 06) 0.06) 0.Chapter 6 Exercise Solutions 6-35. 6-32 .494 Individual Value 2 1 _ X=0. results are the same as for the p chart.06(1 − 0.040 0 -1 -2 LCL=-2.0564 / ni MTB > Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Standardized Fraction Nonconforming (Ex6-35zi) 3 UCL=2.414 -3 1 2 3 4 5 6 Observation 7 8 9 10 The process is in control.06) / ni = ( pˆ i − 0. p = 0.06 zi = ( pˆ i − 0. 36 = 6. CL = c = 2.36 2 1 0 LCL=0 2 4 6 8 10 12 14 Sample 16 18 20 22 24 Test Results for C Chart of Ex6-36Num TEST 1.36 + 3 2.36 − 3 2. Test Failed at points: 13 No. The plate process does not seem to be in statistical control.Chapter 6 Exercise Solutions 6-36.97 LCL = c − 3 c = 2.969 6 5 4 3 _ C=2. 6-33 .36 ⇒ 0 MTB > Stat > Control Charts > Attributes Charts > C C Chart of Number of Nonconformities on Plate (Ex6-36Num) 9 1 8 Sample Count 7 UCL=6.36 UCL = c + 3 c = 2. One point more than 3.00 standard deviations from center line. 1088.7007 / ni LCLi = u − 3 u ni = 0.2622] [0. 1.1653.2487] [0.7007 + 3 0. 1.0 0.153 0.4 UCL=1. UCLi] [0.7007 − 3 0.7007 / ni ni 18 20 21 22 24 [LCLi.1881. 1.701 0. 1.Chapter 6 Exercise Solutions 6-37.6 0.8 _ U=0.2361] [0.2 1.2133] MTB > Stat > Control Charts > Attributes Charts > U U Chart of Imperfections in Paper Rolls (Ex6-37Imp) 1. 1.1527.2926] [0.1392.7007 UCLi = u + 3 u ni = 0.4 0.2 LCL=0. CL = u = 0.249 Sample Count Per Unit 1.0 2 4 6 8 10 12 Sample 14 16 18 20 Tests performed with unequal sample sizes 6-34 . n = 20.55 = 0.289 Sample Count Per Unit 1.55 = 1.151 0.7007 / 20. however MINITAB accepts only integer values for n.4 UCL=1.6 0. CL = u = 0.1467 MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex6-37Rol Variable Ex6-37Rol N 20 Mean 20.7007 − 3 0.4 0.55 UCL = u + 3 u n = 0.0 0.72 0.7007 / 20.2 1.2 LCL=0.2547 LCL = u − 3 u n = 0. MTB > Stat > Control Charts > Attributes Charts > U U Chart of Imperfections in Paper Rolls (Ex6-37Imp) with average sample size n=20 1.Chapter 6 Exercise Solutions 6-38.7007 + 3 0. Use a sample size of n = 20.8 _ U=0.55.0 2 4 6 8 10 12 Sample 14 16 18 20 6-35 . and carefully examine points near the control limits.550 Average sample size is 20.7007. 004 0 -1 LCL=-1. zi = (ui − u ) u ni = (ui − 0.Chapter 6 Exercise Solutions 6-39.7007) 0.7007 / ni MTB > Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Standardized Paper Roll Imperfections (Ex6-39zi) 2 UCL=1.898 Individual Value 1 _ X=-0.906 -2 2 4 6 8 10 12 Observation 14 16 18 20 6-36 . 5 UCL = c + 3 c = 1.5 + 3 1. c chart based on # of nonconformities per cassette deck CL = c = 1.5 1 0 LCL=0 2 4 6 8 10 Sample 12 14 16 18 Process is in statistical control.17 LCL ⇒ 0 MTB > Stat > Control Charts > Attributes Charts > C C Chart of Cassette Deck Nonconformities (Ex6-40Num) UCL=5. 6-37 . Use these limits to control future production.5 = 5.174 5 Sample Count 4 3 2 _ C=1.Chapter 6 Exercise Solutions 6-40. 38 15 10 _ C=8. UCL = c + 3 c = 8. CL = c = 8. 22 6-38 .59. 11.59 ⇒ 0 MTB > Stat > Control Charts > Attributes Charts > C C Chart of Number of Nonconformities (Ex6-41Num) per 1000 meters telephone cable 25 1 20 1 1 Sample Count UCL=17.59 + 3 8.00 standard deviations from center line.Chapter 6 Exercise Solutions 6-41. LCL = c − 3 c = 8.59 5 0 LCL=0 2 4 6 8 10 12 Sample 14 16 18 20 22 Test Results for C Chart of Ex6-41Num TEST 1.59 = 17.59 − 3 8. Test Failed at points: 10.384. One point more than 3. 17 5 LCL=0 0 2 4 6 8 10 12 Sample 14 16 18 20 22 Test Results for C Chart of Ex6-41Num TEST 1. 11 and 22. One point more than 3.17. UCL = c + 3 c = 6. then re-calculate the control limits. three subgroups exceed the UCL. 11. Test Failed at points: 10.62.Chapter 6 Exercise Solutions 6-41 continued Process is not in statistical control.62 10 _ C=6.00 standard deviations from center line.17 + 3 6. Subgroup 15 will then be out of control and should also be excluded. 15. 15. 11. LCL ⇒ 0 C Chart of Number of Nonconformities (Ex6-41Num) Samples 10. Exclude subgroups 10. 22 excluded from calculations 25 1 Sample Count 20 1 1 1 15 UCL=13. 22 6-39 .17 = 13. CL = c = 6. CL = nc = 4(1. total nonconformities 111 = = 15.43 UCL = nc + 3 nc = 15.35 LCL = u − 3 u n = 6 − 3 6 1 ⇒ 0 6-43. (b) The sample is n =1 new inspection units.42 + 3 15.65 The plot point. is the average number of nonconformities found in 2500m.42 − 3 15.Chapter 6 Exercise Solutions 6-42. CL = nc = 2. A u chart of average nonconformities per inspection unit is appropriate. A c chart of the total number of nonconformities per inspection unit is appropriate. and since n = 1. (a) The new inspection unit is n = 4 cassette decks.21 LCL = nc − 3 nc = 15.35 LCL = nc − 3 nc = 6 − 3 6 ⇒ 0 (b) The sample is n =1 new inspection units.5) = 6 UCL = nc + 3 nc = 6 + 3 6 = 13.5(6. A c chart of the total number of nonconformities per inspection unit is appropriate. is the total number of nonconformities found while inspecting a sample 2500m in length.42 CL = u = total inspection units (18 × 1000) / 2500 UCL = u + 3 u n = 15. total nonconformities 27 = = 6. A u chart of average nonconformities per inspection unit is appropriate.20 LCL = u − 3 u n = 15.17) = 15. this is the same as the total number of nonconformities. 6-40 .5 of the old unit. (a) The new inspection unit is n = 2500/1000 = 2.64 The plot point.43 + 3 15.00 CL = u = total inspection units (18 / 4) UCL = u + 3 u n = 6 + 3 6 1 = 13. u .42 /1 = 3.43 = 27. c .43 = 3.42 /1 = 27.43 − 3 15. since this chart was established for average nonconformities per unit.211 ⇒ 0 MTB > Stat > Control Charts > Attributes Charts > U U Chart of Manual Transmission Subassemblies (Ex6-44Num) 1.6 _ U=0.396 LCL = u − 3 u n = 0.422 + 3 0.422 4 = 1. (a) A u chart of average number of nonconformities per unit is appropriate.Chapter 6 Exercise Solutions 6-44. the same control limits may be used for future production with the new sample size.75 16 = 0. (If this was a c chart for total nonconformities in the sample.2 1. with n = 4 transmissions in each inspection.396 Sample Count Per Unit 1.422 UCL = u + 3 u n = 0. CL = u = ∑ ui m = ( ∑ xi / n ) m = (27 / 4) 16 = 6. However.) 6-41 .422 − 3 0.4 0.0 0. the control limits would need revision.422 0.0 LCL=0 2 4 6 8 10 Sample 12 14 16 (b) The process is in statistical control.8 0.2 0.422 4 = −0. (c) The new sample is n = 8/4 = 2 inspection units.4 UCL=1. Chapter 6 Exercise Solutions 6-45. (a) CL = c = 4 UCL = c + 3 c = 4 + 3 4 = 10 LCL = c − 3 c = 4 − 3 4 ⇒ 0 (b) c = 4.0774. c = 16 Pr{x ≤ 21| c = 16} = 0. n = 4 CL = u = c / n = 4 / 4 = 1 UCL = u + 3 u n = 1 + 3 1/ 4 = 2. Use the cumulative Poisson tables. n = 4 CL = u = c / n = 16 / 4 = 4 UCL = u + 3 u n = 4 + 3 4 / 4 = 7 LCL = u − 3 u n = 4 − 3 4 / 4 = 1 6-42 . (a) CL = c = 9 UCL = c + 3 c = 9 + 3 9 = 18 LCL = c − 3 c = 9 − 3 9 = 0 (b) c = 16.9108.5 LCL = u − 3 u n = 1 − 3 1/ 4 ⇒ 0 6-46. UCL = 21 Pr{x ≤ 10 | c = 16} = 0. LCL = 10 6-47. 954 0.4 − 1.6 − 1.645 1.054 6 3 = 3. the normal distribution gives: UCL = u + z0.980 u n = 6 + 2.972 0.6 = 2.975 c = 7.025 c = 7.Chapter 6 Exercise Solutions 6-48.095 6-49.0 and n = 3. and LCL = x/n = 7/10 = 0. the normal distribution gives: UCL = u + z0.95 u n = 1. the normal distribution gives UCL = c + z0. As a comparison. As a comparison.030 0.6 + 1.6} 0.980 and Pr{D < LCL} = 0.20 6-50.054 6 3 = 8.70.4 10 = 2.983 UCL = x/n = 27/3 = 9. u chart with u = 6.6 = 13.952 UCL = x/n = 20/10 = 2.062 0.784 6-43 .020 u n = 6 − 2.4(10) = 14: x 7 8 19 20 Pr{D ≤ x | c = 14} 0.020.4 + 1. and LCL = x/n = 9/3 = 3. Using the cumulative Poisson distribution: x 2 3 12 13 Pr{D ≤ x | c = 7.055 0.923 0. Using the cumulative Poisson distribution with c = u n = 1.032 0.015 0.05 u n = 1.00.645 1. UCL = 13 and LCL = 2.96 7.016 LCL = u + z0.019 0. c = u × n = 18.00 LCL = c − z0.905 LCL = u + z0.96 7.976 for the c chart. As a comparison.4 10 = 0. From the cumulative Poisson tables: x 9 10 26 27 Pr{D ≤ x | c = 18} 0. Find limits such that Pr{D ≤ UCL} = 0. UCLi = 7 + 3 7 / ni .995.005)(0.0} = 0.94 Sample Count Per Unit 14 12 10 8 _ U=7 6 4 2 0 LCL=0 1 2 3 4 5 6 Ex6-51Day 7 8 9 10 Tests performed with unequal sample sizes The process is in statistical control.0. 6-52. u chart with control limits based on each sample size: u = 7. (a) From the cumulative Poisson table. LCLi = 7 − 3 7 / ni MTB > Stat > Control Charts > Attributes Charts > U U Chart of Total Number of Imperfections (Ex6-51Imp) 16 UCL=14. (b) Pr{two consecutive out-of-control points} = (0. So set UCL = 6.00003 6-44 .Chapter 6 Exercise Solutions 6-51.005) = 0. Pr{x ≤ 6 | c = 2. 5 = 0.949 14 0.10. (a) Plot the number of nonconformities per water heater on a c chart. the normal distribution gives UCL = c + z0.5 .030 13 0.25 UCL = c + 3 c = 5.5 = 3.22 |10.88 8.02 6-54.12 LCL ⇒ 0 Plot the results after inspection of each water heater.5} + [1 − Pr{D ≤ 20.25 + 3 5.25) = 10.5 + 1. From the cumulative Poisson distribution: x Pr{D ≤ x | c = 8.5) ] = 0.97 c = 8.973 LCL = 3 and UCL = 13.78 |10.88 8.10. CL = c = ∑ D m = 924 /176 = 5.5 = 20. (b) Let new inspection unit n = 2 water heaters CL = nc = 2(5.997 ] = 0. For comparison.22 LCL = nc − 3 nc = 10.Chapter 6 Exercise Solutions 6-53.98 LCL = c + z0.5} 3 0. c = 850 /100 = 8.5}] = POI(0.5 + 3 10.25 = 12.5 − 3 10.78 (c) Pr{type I error} = Pr{D < LCL | c} + Pr{D > UCL | c} = Pr{D < 0.03 c = 8.5 = 13.5 UCL = nc + 3 nc = 10. approximately 8/day.5 − 1.5) + [1 − POI(20. A c chart with one inspection unit equal to 50 manufacturing units is appropriate.000 + [1 − 0.003 6-45 . 00 An UCL = 9 will give a probability of 0. u = 4. 2.99 1. 0.24 0.43 0.Chapter 6 Exercise Solutions 6-55.mtw Ex6-55X 0 1 2 3 4 5 6 7 8 9 10 11 Ex6-55alpha 0.533) ] = 1 − 0.723] (b) α = Pr{D < LCL | c} + Pr{D > UCL | c} = Pr{D < 0 | 0. Desire α = 0. Use the cumulative Poisson distribution to determine the UCL: MTB : worksheet Chap06.533 ± 3 0. 6-46 . when in fact it is.533 = [0. c = 16 / 30 = 0.533}] = 0 + [1 − POI(2.95 0. 6-56. ∑ Di = 16 in 30 refrigerators.72 | 0.99 of concluding the process is in control.99.533} + [1 − Pr{D ≤ 2.98 0.00 1.89 0.02 0.533 (a) 3-sigma limits are c ± 3 c = 0.0 average number of nonconformities/unit. Use a c chart for nonconformities with an inspection unit n = 1 refrigerator.983 = 0.017 where POI(⋅) is the cumulative Poisson distribution.79 0.63 0.09 0. 533} + [1 − Pr{D ≤ 1. 2) − POI(0.372 ≈ 2 1 − β 1 − 0.1004 where POI(⋅) is the cumulative Poisson distribution.533 (a) c ± 2 c = 0.1353 = 0.1.Chapter 6 Exercise Solutions 6-56 continued (c) β = Pr{not detecting shift} = Pr{D < UCL | c} − Pr{D ≤ LCL | c} = Pr{D < 2. (c) β = Pr{D < UCL | c} − Pr{D ≤ LCL | c} = Pr{D < 1.993] (b) α = Pr{D < LCL | c } + Pr{D > UCL | c } = Pr{D < 0 | 0. 2) = 0.18 ≈ 2 1 − β 1 − 0. c = 0.533)] = 1 − 0.993 | 2} − Pr{D ≤ 0 | 2} = POI(1.8996 = 0. (d) ARL1 = 1 1 = = 2.0} − Pr{D ≤ 0 | 2. 0.533 = [0.993 | 0.271 6-47 .541 6-57.72 | 2.5414 where POI(⋅) is the cumulative Poisson distribution.533}] = 0 + [1 − POI(1.0} = POI(2.406 − 0. 2) = 0.271 where POI(⋅) is the cumulative Poisson distribution.135 = 0.533 + 2 0.6767 − 0. 2) − POI(0. (d) ARL1 = 1 1 = = 1. L: sigma control limits nc − L nc > 0 nc > L nc n > L2 c 6-48 .5 UCL = c + 3 c = 4.75(6) = 4. 1 inspection unit = 10 radios.75 nonconformities/clock CL = c = u × n = 0. u = 0.86 LCL ⇒ 0 6-61. 4)] = 1 − 0. 6-60. u = average # nonconformities/calculator = 2 (a) c chart with c = u × n = 2(2) = 4 nonconformities/inspection unit CL = c = 4 UCL = c + k c = 4 + 3 4 = 10 LCL = c − k c = 4 − 3 4 ⇒ 0 (b) Type I error = α = Pr{D < LCL | c } + Pr{D > UCL | c } = Pr{D < 0 | 4} + [1 − Pr{D ≤ 10 | 4}] = 0 + [1 − POI(10.708 LCL ⇒ 0 6-59.997 = 0.5 = 10. 1 inspection unit = 6 clocks.5 + 3 4. u = 0. c: nonconformities per unit.Chapter 6 Exercise Solutions 6-58.5 average nonconformities/radio CL = c = u × n = 0.003 where POI(⋅) is the cumulative Poisson distribution.5(10) = 5 UCL = c + 3 c = 5 + 3 5 = 11. 005 Percent 80 70 60 50 40 30 20 10 5 1 -30 -20 -10 0 10 20 Ex6-62Bet 30 40 50 There is a huge curve in the plot points.806 0. (a) MTB > Graphs > Probability Plot > Single Probability Plot of Days-Between-Homicides (Ex6-62Bet) Normal .5635 28 0. 6-49 .95% CI 99 Mean StDev N AD P-Value 95 90 1.238 0.572 <0.04 28 1.Chapter 6 Exercise Solutions 6-62.760 Percent 80 70 60 50 40 30 20 10 5 1 0 1 2 Ex6-62t27 3 4 The 0. (b) Probability Plot of Transformed "Days-between-Homicides" (Ex6-62t27) Normal .95% CI 99 Mean StDev N AD P-Value 95 90 12. indicating that the normal distribution assumption is not reasonable.2777th root transformation makes the data more closely resemble a sample from a normal distribution.25 12. 0 0.4789 28 0.695 0.0 1.5 3.5 The 0.223 0.246 0.5 UCL=3. It is not very different from the transformed data in (b). (d) MTB > Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Transformed Homicide Data (0.5 LCL=0.Chapter 6 Exercise Solutions 6-62 continued (c) Probability Plot of Transformed "Days-betwee-Homicides" (Ex6-62t25) Normal .0 3 6 9 12 15 18 Observation 21 24 27 6-50 .95% CI 99 Mean StDev N AD P-Value 95 90 1.0 0.2777 root) (Ex6-62t27) 3.5 2.25th root transformation makes the data more closely resemble a sample from a normal distribution.0 3.5 1.366 3.5 2.0 Individual Value 2.806 1.0 Ex6-62t25 2.5 1.0 _ X=1.807 Percent 80 70 60 50 40 30 20 10 5 1 0. mean and LCL. wrong package labeling).5 1. One could track the number of orders filled incorrectly (wrong parts.5 LCL=0.Chapter 6 Exercise Solutions 6-62 continued (e) I Chart of Transformed Homicide Data (0. There is no difference in interpretation. too few/many parts.0 _ X=1. errors in tax preparation.5 2. (f) The “process” is stable. the mean time between may get longer (or shorter) with plot points above the upper (or below the lower) control limit. wrong part labeling.0 3 6 9 12 15 18 Observation 21 24 27 Both Individuals charts are similar. law. (hopefully caught internally with a verification step).695 1.25 root) (Ex6-62t25) UCL=3. packaged incorrectly (wrong material. Consider a product distribution center (or any warehouse) with processes for filling and shipping orders. say in population. 6-63. meaning that the days-between-homicides is approximately constant. If a change is made. Or consider an accounting firm—errors in statements.0 Individual Value 2. which affects the rate at which homicides occur. etc. etc. 6-51 . etc. policy. invoiced incorrectly..0 0.). There are endless possibilities for collection of attributes data from nonmanufacturing processes. workforce. with an identical pattern of points relative to the UCL.025 3.365 0. If time-between-events data (say failure time) is being sought for internally generated data. One point more than 3. 6-52 .86 20 15 LCL=15. A u chart which monitors the average number of CAT scans per NYRSB is appropriate. Collection of “start” time data may be facilitated by serializing or date coding product. 6-65☺. The variable NYRSB can be thought of as an “inspection unit”.94 30 25 _ U=25. the conditions of use and the definition of “failure” may not be consistently applied.00 standard deviations from center line. it may be hard to determine with sufficient accuracy—both the “start” and the “end”. However. it can usually be obtained reliably and consistently. representing an identical “area of opportunity” for each “sample”. The “process characteristic” to be controlled is the rate of CAT scans. if you’re looking for data on time-between-events that must be obtained from external sources (for example.Chapter 6 Exercise Solutions 6-64. There are ways to address these difficulties. Also. Test Failed at points: 15 The rate of monthly CAT scans is out of control. time-to-field failures).77 94 94 94 94 94 94 94 94 94 94 94 94 95 95 95 M EB A R PR A Y UN JUL UG EP C T OV EC A N EB A R A J F J S F D A J A M N O M M Ex6-65MON Tests performed with unequal sample sizes Test Results for U Chart of Ex6-65NSCANB TEST 1. MTB > Stat > Control Charts > Attributes Charts > U U Chart of CAT Scans (Ex6-65NSCANB) Sample Count Per Unit 40 1 35 UCL=35. Chapter 6 Exercise Solutions 6-66☺. (a) MTB > Stat > Control Charts > Attributes Charts > U U Chart of Number of Office Visits (Ex6-66aNVIS) Phase 1 Sample Count Per Unit 2500 UCL=2476. A u chart which monitors the average number of office visits per NYRSB is appropriate.5 2400 _ U=2303.5 2100 JAN94 FEB94 MAR94 APR94 MAY94 Ex6-66aMON JUN94 JUL94 AUG94 Tests performed with unequal sample sizes The chart is in statistical control 6-53 . The variable NYRSE can be thought of as an “inspection unit”. representing an identical “area of opportunity” for each “sample”. The “process characteristic” to be controlled is the rate of office visits.0 2300 2200 LCL=2129. 14. 13.0 2200 2100 LCL=2141. 15 The phase 2 data appears to have shifted up from phase 1. 10.0 4 4 4 4 4 4 4 4 5 5 4 5 4 4 4 N9 EB9 A R9 PR9 A Y9 UN9 UL9 UG9 EP9 CT 9 V 9 EC9 A N9 EB9 AR9 A J F M J J F M J S A M A NO D O Ex6-66MON Tests performed with unequal sample sizes Test Results for U Chart of Ex6-66NVIS TEST 1. One point more than 3.00 standard deviations from center line. 11. The 2nd phase is not in statistical control relative to the 1st phase. 6-54 . Test Failed at points: 9. 12.Chapter 6 Exercise Solutions 6-66 continued (b) U Chart of Number of Office Visits (Ex6-66NVIS) Phase 1 Limits 1 2800 Sample Count Per Unit 1 2700 2600 2500 1 1 1 1 1 UCL=2465.0 2400 2300 _ U=2303. are in statistical control. 6-55 . separated from the Phase 1 data.5 2600 2500 LCL=2450.5 2700 _ U=2623.6 2400 9 10 11 12 Sample 13 14 15 Tests performed with unequal sample sizes The Phase 2 data.Chapter 6 Exercise Solutions 6-66 continued (c) U Chart of Number of Office Visits (Ex6-66NVIS) Phase 2 Sample Count Per Unit 2800 UCL=2796. 36 Cˆ pl = 3σˆ 3(1.000 ± 0.93) ( ) Cˆ pk = min Cˆ pl .023. n = 5.023 2.010) µˆ − LSL 74.965 = = 1.Chapter 7 Exercise Solutions Note: Several exercises in this chapter differ from those in the 4th edition.20 Cˆ pl = 3σˆ 3(0. In Exercise 5-1.93) µˆ − LSL 33.65 = = 1.010) ( ) Cˆ pk = min Cˆ pl .65 − 20 = = 2.001 = = 1. samples 12 and 15 are out of control. µˆ = x = 74. A second exercise number in parentheses indicates that the exercise number has changed. and the new process parameters are used in the process capability analysis. LSL = 20 USL − LSL 40 − 20 = = 1.93 USL = 40.965.001. σˆ = R d 2 = 0.326 = 0.035 − 73. µˆ = x = 33.010 SL = 74.035] USL − LSL 74.001 − 73.10 7-1 . R = 0. 74.035 − 74.5.10 Cˆ pu = 3σˆ 3(1.93) USL − µˆ 40 − 33.17 Cˆ p 6σˆ 6(0. An “*” indicates that the description has changed.65.13 7-2.73 Cˆ p 6σˆ 6(1. σˆ = R d 2 = 1. R = 4.010) USL − µˆ 74. New exercises are denoted with an “☺”. Cˆ pu = 1. 7-1. Cˆ pu = 1.035 = [73.13 Cˆ pu = 3σˆ 3(0.965 = = 1. 059 = 3. the mean µ lies within approximately the middle fourth of the specification band. ξˆ = µˆ − T 10. this PCR does not tell where the mean is located within the specification band.34 pk pu pl This is an extremely capable process. indicating that the process is not centered and is not achieving potential capability.Chapter 7 Exercise Solutions 7-3. σˆ x = R d 2 = 6.54 pk = = 0.04 Cˆ pkm = Cˆ 1. Note that the Cpk is less than Cp.04) µˆ − LSL x 10.54 2 1+V 1 + (−3.412 1 + ξˆ 2 7-2 . However.25 2. Rx = 6.375 = = 4. USL x − µˆ 50 − 10.375 = = −3.375 − 0 = = 3.375.25. T − x 0 − 10.375 − (−50) = = 6.48 Cˆ p = 6σˆ x 6(3. µˆ = x = 10.4128) 2 V= Since Cpm is greater than 4/3.4128 S 3.41 σˆ 3. LSL x = [(350 − 5) − 350] × 10 = −50 xi = (obsi − 350) ×10 USL x − LSL x 50 − (−50) = = 5.62 Cˆ pl = 3σˆ x 3(3. Cˆ ) = 4.48 Cˆ pm = = = 1.04) Cˆ = min(Cˆ .34 Cˆ pu = 3σˆ x 3(3.43 1 + 3.04 USL x = [(350 + 5) − 350] ×10 = 50.04) The process produces product that uses approximately 18% of the total specification band.04 Cˆ p 5. with an estimated percent defective much less than 1 ppb. 00273 Cˆ 1.35 3σˆ 3(0. USL − µˆ 0.01 − 0.09 pk = = 1.09 pk pl pu This process is not considered capable.00273 Cˆ pkm = Cˆ 1.399 S 0. the mean µ lies within approximately the middle third of the specification band.00109 − 0 = = 0.00273 .01 Cˆ p = = = 1.01) Cˆ pl = = = 1.00273) The process produces product that uses approximately 82% of the total specification band.00273) Cˆ = min(Cˆ .399 σˆ 0.09 3σˆ 3(0.13 1+V 2 1 + (−0.01 + 0.399) 2 V= Since Cpm is greater than 1. Cˆ ) = 1. tolerances: 0 ± 0. ξˆ = µˆ − T 0.00109 − (−0. σˆ x = 0. failing to meet the minimally acceptable definition of capable Cpk ≥ 1. R = 0.22 6σˆ 6(0.00273) µˆ − LSL 0.00109 = = −0. n = 5.00635.399 + 1+ ξ 7-3 . x = 0.00109 Cˆ pu = = = 1.33 T − x 0 − 0.01 2 2 ˆ 1 0.00109.01 USL − LSL 0.Chapter 7 Exercise Solutions 7-4.22 p ˆ C pm = = = 1. 117 ⎭⎥⎦ ⎩ ⎩ = Φ (−13.117) Cˆ = min(Cˆ .9400 = 1. Cˆ ) = 1.05 0.953) + [1 − Φ (8.Chapter 7 Exercise Solutions 7-5.000000 7-4 . µˆ = x = 100.48 3σˆ x 3(1.05.000000 + [1 − 1.117 (a) USL − LSL (95 + 10) − (95 − 10) = = 2.117) (b) µˆ − LSL x 100 − (95 − 10) Cˆ pl = = = 4.429) + [1 − Φ (4.49 pk pl pu (c) pˆ Actual = Pr{x < LSL} + Pr{x > USL} = Pr{x < LSL} + [1 − Pr{x ≤ USL}] LSL − µˆ ⎫ ⎡ USL − µˆ ⎫⎤ ⎧ ⎧ = Pr ⎨ z < ⎬ + ⎢1 − Pr ⎨ z ≤ ⎬⎥ σˆ ⎭ ⎣ σˆ ⎩ ⎩ ⎭⎦ 85 − 100 ⎫ ⎡ ⎧ ⎧ 105 − 100 ⎫⎤ = Pr ⎨ z < ⎬ + ⎢1 − Pr ⎨ z ≤ ⎬ 1.999996] = 0.117 ⎭ ⎣ 1.98 Potential: Cˆ p = 6σˆ 6(1.117) USL x − µˆ (95 + 10) − 100 Actual: Cˆ pu = = = 1. σˆ x = s c4 = 1.476) ] = 0.000000] = 0.117 ⎭⎥⎦ ⎩ ⎩ = Φ (−8.953) ] = 0.49 3σˆ x 3(1. s = 1.117 ⎭ ⎣ 1.000004 85 − 95 ⎫ ⎡ ⎧ ⎧ 105 − 95 ⎫⎤ pˆ Potential = Pr ⎨ z < ⎬ + ⎢1 − Pr ⎨ z ≤ ⎬ 1.0000 + [1 − 0. 37 pk pl pu (c) The current fraction nonconforming is: pˆ Actual = Pr{x < LSL} + Pr{x > USL} = Pr{x < LSL} + [1 − Pr{x ≤ USL}] LSL − µˆ ⎫ ⎡ USL − µˆ ⎫ ⎤ ⎧ ⎧ = Pr ⎨ z < ⎬ + ⎢1 − Pr ⎨ z ≤ ⎬⎥ σˆ ⎭ ⎣ σˆ ⎩ ⎩ ⎭⎦ 208 − 199 ⎫⎤ ⎧ 192 − 199 ⎫ ⎡ ⎧ = Pr ⎨ z < ⎬ + ⎢1 − Pr ⎨ z ≤ ⎬ 1. LSL = 200 – 8 = 192 (a) USL − LSL 208 − 192 = = 1.Chapter 7 Exercise Solutions 7-6☺. R = 3. µˆ = x = 199.70) The process produces product that uses approximately 64% of the total specification band.5 2. σˆ x = R d 2 = 3.70 USL = 200 + 8 = 208. (b) USL − µˆ 208 − 199 Cˆ pu = = = 1.70 ⎭⎥⎦ ⎩ ⎩ = Φ (−4.70) Cˆ = min(Cˆ .0000026 7-5 .37 Actual: Cˆ pl = 3σˆ 3(1.0000013 = 0.5.0000191 + [1 − 1] = 0. n = 4.059 = 1.0000191 If the process mean could be centered at the specification target.1176) + [1 − Φ (5.76 3σˆ 3(1.70 ⎭ ⎩ = 2 × 0. the fraction nonconforming would be: ⎧ 192 − 200 ⎫ pˆ Potential = 2 × Pr ⎨ z < ⎬ 1.70) µˆ − LSL 199 − 192 = = 1. Cˆ ) = 1.2941) ] = 0.70 ⎭ ⎣ 1.57 Potential: Cˆ p = 6σˆ 6(1. Cpm. σˆ x = R d 2 = 2.216) (b) USL − µˆ 45 − 39.75 = = = 0.5. 2 (d) The current fraction nonconforming is: pˆ Actual = Pr{x < LSL} + Pr{x > USL} = Pr{x < LSL} + [1 − Pr{x ≤ USL}] LSL − µˆ ⎫ ⎡ USL − µˆ ⎫⎤ ⎧ ⎧ = Pr ⎨ z < ⎬ + ⎢1 − Pr ⎨ z ≤ ⎬⎥ σˆ ⎭ ⎣ σˆ ⎩ ⎩ ⎭⎦ 35 − 39.135) 2 The closeness of estimates for Cp.7.74 2 1+V 1 + (−0.128 = 2.025348 7-6 .39170)] = 0.80 3σˆ 3(2. LSL = 40 – 5 = 35 (a) USL − LSL 45 − 35 = = 0. R = 2.991615] = 0.7 ⎫⎤ ⎧ ⎧ = Pr ⎨ z < ⎬ + ⎢1 − Pr ⎨ z ≤ ⎬ 2.71 = 0.216 USL = 40 + 5 = 45.216 ⎭⎥⎦ ⎩ ⎩ = Φ (−2.7 ⎫ ⎡ 45 − 39.216 Cˆ p 0. and Cpkm indicate that the process mean is very close to the specification target.216) Cˆ = min(Cˆ .216 ⎭ ⎣ 2.75 Potential: Cˆ p = 6σˆ 6(2.7 Cˆ pu = = = 0.216) µˆ − LSL 39. Cˆ ) = 0.7 − 35 = = 0.71 Actual: Cˆ pl = 3σˆ 3(2. µˆ = x = 39. Cpk.12094) + [1 − Φ (2.71 pk pl pu (c) V= Cˆ pm x − T 39.135) 2 Cˆ pkm = Cˆ pk = 0.5 1.70 1+V 1 + (−0.135 s 2.0169634 + [1 − 0.Chapter 7 Exercise Solutions 7-7☺.7 − 40 = = −0. n = 2. 02382 7-8 (7-6).13 ⎭ ⎩ ⎩ = Φ (−3.13 (a) USL − LSL 2(8) Potential: Cˆ p = = = 1.13 ⎭ 2.13) (b) µˆ − LSL 75 − (80 − 8) Cˆ pl = = = 0.756) + 1 − Φ (3.000172 7-7 . S = 2.47 pk pl pu (c) Let µˆ = 80 pˆ Potential = Pr{x < LSL} + Pr{x > USL} LSL − µˆ ⎫ USL − µˆ ⎫ ⎧ ⎧ = Pr ⎨ z < ⎬ + 1 − Pr ⎨ z ≤ ⎬ σˆ ⎭ σˆ ⎩ ⎩ ⎭ 72 − 80 ⎫ 88 − 80 ⎫ ⎧ ⎧ = Pr ⎨ z < ⎬ + 1 − Pr ⎨ z ≤ ⎬ 2.9400 = 2.13) Cˆ = min(Cˆ .13) USL − µˆ 80 + 8 − 75 = = 2.756) = 0.01191 = 0. σˆ = Sˆ c4 = 2 0.26} = 2 × 0. µˆ = 75.999914 = 0.000086 + 1 − 0.47 3σˆ 3(2.216 ⎭ ⎩ = 2 × Pr{z < −2.03 Actual: Cˆ pu = 3σˆ 3(2.Chapter 7 Exercise Solutions 7-7 (d) continued If the process mean could be centered at the specification target. the fraction nonconforming would be: 35 − 40 ⎫ ⎧ pˆ Potential = 2 × Pr ⎨ z < ⎬ 2. Cˆ ) = 0.25 6σˆ 6(2. 13) + 1 − Φ (3.191 USL − LSL (100 + 10) − (100 − 10) = = 1.00087 + 1 − 0.191) µˆ − LSL x 100 − (100 − 10) Cˆ pl = = = 1. Cˆ pu ) = 1.045 = = = 1.191 Cˆ p 1.9400 = 3. Cˆ ) = 1.00174 Process B µˆ = xB = 105. Assume n = 5 Process A µˆ = x A = 100.13) = 0.Chapter 7 Exercise Solutions 7-9 (7-7).9400 = 1.045 Cˆ p = 6σˆ 6(3.699 3σˆ x 3(1.045 3σˆ x 3(3.045 3σˆ x 3(3.191) Cˆ = min(Cˆ . sB = 1.133 6(1.191) USL x − µˆ (100 + 10) − 100 Cˆ pu = = = 1.191 ⎭ 3.566 Cˆ pu = 3σˆ x 3(1. σˆ A = s A c4 = 3 0. s A = 3.99913 = 0.045 2 1+V 1 + (0) 2 pˆ = Pr{x < LSL} + Pr{x > USL} = Pr{x < LSL} + 1 − Pr{x ≤ USL} LSL − µˆ ⎫ USL − µˆ ⎫ ⎧ ⎧ = Pr ⎨ z < ⎬ + 1 − Pr ⎨ z ≤ ⎬ σˆ ⎭ σˆ ⎩ ⎩ ⎭ 90 − 100 ⎫ ⎧ ⎧ 110 − 100 ⎫ = Pr ⎨ z < ⎬ + 1 − Pr ⎨ z ≤ ⎬ 3.064 Cˆ p = USL − LSL 6σˆ = (100 + 10) − (100 − 10) = 3.064) Cˆ pk = min(Cˆ pl .064) USL x − µˆ x (100 + 10) − 105 = = 1.064) µˆ − LSL x 105 − (100 − 10) Cˆ pl = x = = 4.045 pk pl V= Cˆ pm pu x − T 100 − 100 = =0 s 3.566 7-8 .191 ⎭ ⎩ ⎩ = Φ (−3. σˆ B = sB c4 = 1 0. 064) 2 = 4.064 ⎭ ⎩ ⎩ = Φ (−14.000001 instead of Process A with estimated fallout 0.064 Cˆ p 3.758 Process B will result in fewer defective assemblies. A = 1.699) 2 90 − 105 ⎫ ⎧ ⎧ 110 − 105 ⎫ pˆ = Pr ⎨ z < ⎬ + 1 − Pr ⎨ z ≤ ⎬ 1.133 Cˆ pm = = = 0.064 ⎭ 1.001726.Chapter 7 Exercise Solutions 7-9 continued x − T 100 − 105 V= = = −4. For the parts Cˆ pk .699) = 0.098) + 1 − Φ (4.999999 = 0.699 s 1.000000 + 1 − 0. σˆ B = 20σˆ 2 = 20(1.652 1+V 2 1 + (−4.271 Process B: µˆ B = 20(105) = 2100. B indicates that more parts from Process B are within ( ) ( ) specification than from Process A.045 < 1. Process A: µˆ A = 20(100) = 2000. σˆ A = 20σˆ 2 = 20(3.566 = Cˆ pk .191) 2 = 14.000001 Prefer to use Process B with estimated process fallout of 0. 7-9 . 7-10 (7-8). 1350 7-12☺.000 Ex7-9Wt 1.0225) = 0.9968 0.985 kg µˆ − LSL 0.985 − 0.556) = 0.19 C pl = 3σˆ 3(0.950 0. MTB > Stat > Basic Statistics > Normality Test Probability Plot of 1-kg Containers (Ex7-9Wt) Normal 99 Mean StDev N AD P-Value 95 90 0.0200 − 0.9975.Chapter 7 Exercise Solutions 7-11 (7-9).9975 = 0.9975 ⎫ ⎧ ⎧ pˆ = Pr ⎨ z < ⎬ = Pr ⎨ z < ⎬ = Φ (−0.323 0. p84 = 1. LSL = 0.0225 6σˆ = 6(0.975 1.985 = = 0.492 Percent 80 70 60 50 40 30 20 10 5 1 0.289105 0.0200 σˆ = p84 − p50 = 1. x ≈ p50 = 0.0225) LSL − µˆ ⎫ 0.025 1.0225 σˆ ⎭ ⎩ ⎩ ⎭ 7-10 .050 A normal probability plot of the 1-kg container weights shows the distribution is close to normal.02167 15 0.9975 − 0. 98 19.99986 = 0.99 20.00905 − 19.05514 7-11 .009242 25 0.02 A normal probability plot of computer disk heights shows the distribution is close to normal.Chapter 7 Exercise Solutions 7-13☺.515 0.00905 σˆ = p84 − p50 = 20.01 20.00 0.) Probability Plot of Disk Height (Ex7-13Ht) Normal 99 95 90 84 Percent 80 Mean StDev N AD P-Value 20. MTB > Stat > Basic Statistics > Normality Test (Add percentile lines at Y values 50 and 84 to estimate µ and σ.99986 10 5 1 19.99986 p84 = 20.00905 20 20.00919) = 0.00 Disk Height. mm 20. x ≈ p50 = 19.00919 6σˆ = 6(0.174 70 60 50 40 30 50 19. 27 10 10 15 20 Reimbursement Cycle Time.2 = 4.27 − 13.401 0.27 σˆ = p84 − p50 = 17.097 30 0.Chapter 7 Exercise Solutions 7-14☺.07 6σˆ = 6(4. Days 25 A normal probability plot of reimbursement cycle times shows the distribution is close to normal.07) = 24. MTB > Stat > Basic Statistics > Normality Test (Add percentile lines at Y values 50 and 84 to estimate µ and σ.42 7-12 .340 70 60 50 40 30 50 20 1 13. x ≈ p50 = 13.2 4.2 p84 = 17.) Probability Plot of Cycle Time (Ex7-14CT) Normal 99 95 90 84 Percent 80 Mean StDev N AD P-Value 13.2 5 5 17. 2 6σˆ = 6(12.2 ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ = 1 − Φ (1.27 40 0.) Probability Plot of Response Time (Ex7-15Resp) Normal 99 95 90 84 Percent 80 Mean StDev N AD P-Value 98.78 ⎫ pˆ = Pr ⎨ z > ⎬ = 1 − Pr ⎨ z < ⎬ = 1 − Pr ⎨ z < ⎬ σˆ σˆ 12.041017 7-13 .Chapter 7 Exercise Solutions 7-15☺.78 = = 0.2 (b) USL = 2 hrs = 120 mins USL − µˆ 120 − 98. MTB > Stat > Basic Statistics > Normality Test (Add percentile lines at Y values 50 and 84 to estimate µ and σ. (a) x ≈ p50 = 98.58 C pu = 3σˆ 3(12.98 − 98.463 0.243 70 60 50 40 30 50 20 1 98.2) USL − µˆ ⎫ USL − µˆ ⎫ ⎧ ⎧ ⎧ 120 − 98.98 10 90 100 110 Response Time.958983 = 0.98 σˆ = p84 − p50 = 110.739) = 1 − 0. minutes 120 130 A normal probability plot of response times shows the distribution is close to normal.78 12.78 5 70 80 110.2) = 73.78 = 12.78 p84 = 110. 27 = 2.217 70 60 50 40 30 50 20 1 53.96 − 53.69) = 16. x ≈ p50 = 53.712 15 0.96 10 56 58 60 A normal probability plot of hardness data shows the distribution is close to normal.27 5 46 48 50 52 54 Hardness 55.69 6σˆ = 6(2. MTB > Stat > Basic Statistics > Normality Test (Add percentile lines at Y values 50 and 84 to estimate µ and σ.96 σˆ = p84 − p50 = 55.27 p84 = 55.465 0.27 2.) Probability Plot of Hardness Data (Ex5-59Har) Normal 99 95 90 84 Percent 80 Mean StDev N AD P-Value 53.Chapter 7 Exercise Solutions 7-16 (7-10).14 7-14 . Chapter 7 Exercise Solutions 7-17 (7-11). so it is not appropriate to estimate capability. MTB > Stat > Basic Statistics > Normality Test Probability Plot of Failure Times (Ex7-17FT) Normal 99 Mean StDev N AD P-Value 95 90 1919 507.1 10 0.587 Percent 80 70 60 50 40 30 20 10 5 1 1000 1500 2000 Ex7-17FT 2500 3000 The plot shows that the data is not normally distributed. 7-15 .272 0. 26 ≤ C p 49 The company cannot demonstrate that the PCR exceeds 1.80 ≤ C p ≤ 1.40 2 χα2/ 2.24 = 12.11 25 − 1 25 − 1 0.n −1 = χ 0. 1. n = 25.9303 χ12−α . n −1 0. 7-19 (7-13).36 ≤ C p ≤ 1.49 49 = 1.52 33.40 39.52 ⎠ 1 − α = 0.n −1 ˆ ˆ Cp ≤ Cp ≤ Cp n −1 n −1 1.52 ⎝ 1.Chapter 7 Exercise Solutions 7-18 (7-12).49 χ2 Cˆ p 1−α .52 p 1 − α = 0.05 χ2 = χ2 1−α / 2.975. LSL = 75.n −1 ≤ C p n −1 1.95.33 2 ⎛ 1.5) (b) α = 0.33 at a 95% confidence level.42 This confidence interval is wide enough that the process may either be capable (ppm = 27) or far from it (ppm ≈ 16.11 12.n −1 = χ 0.5 (a) USL − LSL 85 − 75 = = 1.11 Cˆ p = 6σˆ 6(1.49 7-16 .24 = 39.9303 = 1. USL = 85.395).36 χ12−α / 2.33 ⎞ χ = 49 ⎜ ⎟ = 37.025.12 2 1−α .88 α = 0.n −1 χα2/ 2.95 2 = 33. n = 50 Cˆ = 1.52 χ12−α . S = 1. x = 97.63 ⎢1 − 1. LSL = 90 (a) USL x − µˆ x 100 − 97 Cˆ pu = = = 0. Cˆ pu ) = 0.63) 2(30 − 1) ⎦ 9(30)(0.05 zα / 2 = z0. S = 1.4287 ≤ C pk ≤ 0.63 (b) α = 0.6.6) Cˆ pk = min(Cˆ pl .63) 2(30 − 1) ⎦ ⎣ ⎣ 0.96 ⎥ ≤ C pk ≤ 0.Chapter 7 Exercise Solutions 7-20 (7-14).8313 7-17 .6) µˆ − LSL x 97 − 90 Cˆ pl = x = = 1. n = 30.960 ⎡ 1 1 ⎤ ⎥ ≤ C pk ≤ Cˆ pk + Cˆ pk ⎢1 − zα / 2 2 2(n − 1) ⎥ 9nCˆ pk ⎢⎣ ⎦ ⎡ 1 1 ⎤ ⎢1 + zα / 2 ⎥ + 2 2(n − 1) ⎥ 9nCˆ pk ⎢⎣ ⎦ ⎡ ⎤ ⎡ ⎤ 1 1 1 1 + + 0. USL = 100.025 = 1.63 3σˆ x 3(1.96 ⎥ 2 2 9(30)(0.46 3σˆ x 3(1.63 ⎢1 + 1. 05.97 Cˆ pl = x 3σˆ x 3(60) Cˆ pk = min(Cˆ pl .025 = 1.42) 2(50 − 1) ⎦ ⎣ ⎣ 0.42 (b) α = 0. but it is not too far off.96. x = 2275. from Ex.Chapter 7 Exercise Solutions 7-21 (7-15).5443 7-22 (7-16).63 ⎢1 − 1.42 ⎢1 − 1.42 ⎢1 + 1.63 ⎢1+1. 7-20.2957 ≤ C pk ≤ 0.42) 2(50 − 1) ⎦ 9(50)(0.63. 7-23 (7-17).96 ⎥ 2 2 9(50)(0.960 ⎡ 1 1 ⎤ ⎥ ≤ C pk ≤ Cˆ pk Cˆ pk ⎢1 − zα / 2 + 2 2(n − 1) ⎥ 9nCˆ pk ⎢⎣ ⎦ ⎡ 1 1 ⎤ ⎢1 + zα / 2 ⎥ + 2 2(n − 1) ⎥ 9nCˆ pk ⎢⎣ ⎦ ⎡ ⎤ ⎡ ⎤ 1 1 1 1 0. n = 50 (a) USL x − µˆ x 2350 − 2275 = = 0. σˆ I = 3. nominal = 2225. zα / 2 = z0.79 The approximation yields a narrower confidence interval. n = 30 ⎡ ⎡ 1 ⎤ 1 ⎤ Cˆ pk ⎢1 − zα / 2 ⎥ ≤ C pk ≤ Cˆ pk ⎢1 + zα / 2 ⎥ 2(n − 1) ⎦ 2(n − 1) ⎦ ⎣ ⎣ ⎡ ⎤ ⎡ ⎤ 1 1 0. LSL = 2100.96 ⎥ 2(30 − 1) ⎦ 2(30 − 1) ⎦ ⎣ ⎣ 0. USL = 2350. zα / 2 = 1. σ OI = 0.96 + + ⎥ ≤ C pk ≤ 0. Cˆ pk = 0.47 ≤ C pk ≤ 0. σˆ Total = 5 2 2 2 σˆ Total = σˆ Meas + σˆ Process 2 2 σˆ Process = σˆ Total − σˆ Meas = 52 − 32 = 4 7-18 . Cˆ pu ) = 0. s = 60.96 ⎥ ≤ C pk ≤ 0.42 Cˆ pu = 3σˆ x 3(60) µˆ − LSL x 2275 − 2100 = = 0. 07 25 _ _ X=21. R = 2.Chapter 7 Exercise Solutions 7-24 (7-18).15 Sample Range 8 6 4 _ R=2. (a) n = 2.8. 15. The new gauge is more repeatable than the old one.8. 12.482 MTB > Stat > Control Charts > Variables Charts for Subgroups > X-bar R Xbar-R Chart of Part Measurements (Ex7-24All) 30 1 Sample M ean U C L=27. x = 21.482) P = ×100 = ×100 = 49. and the x chart has a few out-of-control parts. 20 The R chart is in control.53 20 U C L=9. (b) specs: 25 ± 15 6σˆ Gauge 6(2. One point more than 3. Test Failed at points: 8.8 2 0 LC L=0 2 4 6 8 10 Sample 12 14 16 18 20 Test Results for Xbar Chart of Ex7-24All TEST 1.6% T USL − LSL 2(15) 7-19 . σˆ Gauge = 2.00 standard deviations from center line.8 20 1 15 2 4 6 8 1 1 10 Sample 12 14 16 18 LC L=16. σˆ Gauge = R d 2 = 2.Chapter 7 Exercise Solutions 7-25 (7-19).0 _ _ X=98. 3 The x chart has a couple out-of-control points.3 1.5% 4.921 4.00 standard deviations from center line.359 × 100 = 62.693 = 1.5 96.359) = = = 0.5 0.272 T USL − LSL 115 − 85 7-20 .2 97.3592 = 2. LSL = 100 – 15 = 85 6σˆ Gauge P 6(1. MTB > Stat > Control Charts > Variables Charts for Subgroups > X-bar R Xbar-R Chart of Part Mesaurements (Ex7-25All) 102.2.5 3.3.0 LC L=0 1 2 3 4 5 6 7 8 9 10 Sample Test Results for Xbar Chart of Ex7-25All TEST 1.5 99.717 2 2 2 σˆ Product = σˆ Total − σˆ Gauge = 4.695 (c) σˆ Gauge σˆ Total × 100 = 1.717 − 1. This indicates that the operator is not having difficulty making consistent measurements.717 (d) USL = 100 + 15 = 115.359 2 σˆ Total = 4. and the R chart is in control.0 LC L=95.3 1.0 1 U C L=100. R = 2. One point more than 3. Test Failed at points: 2. (b) x = 98.872 σˆ Product = 1.0 _ R=2.553 Sample M ean 100.847 1 1 2 3 4 5 6 7 8 9 10 Sample Sample Range 6.0 U C L=5. 7% T USL − LSL 60 − 40 7-21 .693 = 1.17 n = 2 operators d 2 = 1.128 σˆ Reproducibility = Rx d 2 = 0.00 1.17 1.03.87. x2 = 49.191 (c) specs: 50 ± 10 6σˆ Gauge P 6(1.xls : worksheet Ex7-26 x1 = 50. (a) Excel : workbook Chap07.Chapter 7 Exercise Solutions 7-26 (7-20).191) = ×100 = × 100 = 35.30 R = 2. R2 = 2.128 = 0. R1 = 1.151 (b) 2 2 2 ˆ2 ˆ2 σˆ Measurement Error = σ Repeatability + σ Reproducibility = 1.181 + 0.418 σˆ Measurement Error = 1.151 = 1.181 Rx = 0.693 σˆ Repeatability = R d 2 = 2.00 n = 3 repeat measurements d 2 = 1.70. 0 1 2 3 4 5 6 7 8 Sample 9 10 11 12 13 14 15 Test Results for R Chart of Ex7-27All TEST 1.0 1 1 2 3 4 5 6 7 8 Sample 9 10 Sample Range 6. Test Failed at points: 11.128 = 1.5 3.0 U C L=23.0 17. 12 Out-of-control points on R chart indicate operator difficulty with using gage.010 4.0 _ R=1.5 LC L=17.533 1.0 11 12 1 1 13 14 15 U C L=5.82 1 15. One point more than 3. (a) σˆ Gauge = R d 2 = 1.359 Gauge capability: 6σˆ = 8.5 LC L=0 0. 7-22 .533 1.5 _ _ X=20.Chapter 7 Exercise Solutions 7-27 (7-21).00 standard deviations from center line.58 22.154 (b) MTB > Stat > Control Charts > Variables Charts for Subgroups > X-bar R Xbar-R Chart of Part Measurements (Ex7-27All) 1 1 Sample M ean 25.7 20. For the reduced model: ANOVA: Ex7-28Reading versus Ex7-28Part.712 − 0.712 2 σˆ Operator = = 0.500 0.1399 0.592 S = 0. 12. 17. 6.0149 = pn 20(2) MSP − MSP×O 62.0149 -0. 20 Ex7-28Op Values 1.Chapter 7 Exercise Solutions 7-28☺.2513 0. Since the Part × Operator term is not significant (α = 0.995825 1 2 3 4 9. 16. 10.861 R-Sq(adj) = 90. 2. 13.000 0.84 0.2798 = on 3(2) 2 σˆ Part×Operator = The manual calculations match the MINITAB results.74% Variance component 10.72 P 0. R-Sq = 95.2798 0.8832 Error term 3 3 Expected Mean Square for Each Term (using unrestricted model) (3) + 6 (1) (3) + 40 (2) (3) 7-23 . we can fit a reduced model without that term.617 1.10).992 Total 119 1274. 3 Analysis of Variance for Ex7-28Reading Source DF SS MS Ex7-28Part 19 1185.9917 Error term 3 3 4 Expected Mean Square for Each Term (using unrestricted model) (4) + 2 (3) + 6 (1) (4) + 2 (3) + 40 (2) (4) + 2 (3) (4) 2 σˆ Repeatability = MS Error = 0.308 Ex7-28Part*Ex7-28Op 38 27.33% Source Ex7-28Part Ex7-28Op Ex7-28Part*Ex7-28Op Error F 87. 1. Note the Part × Operator variance component is negative.992 = = −0. Ex7-28Op Factor Ex7-28Part Ex7-28Op Type random random Factor Ex7-28Part 18.391 Ex7-28Op 2 2.050 0.712 Error 60 59.391 − 0. 5.712 2 σˆ Part = = 10.0106 0. 7. 15. select “Display expected mean squares and variance components” ANOVA: Ex7-28Reading versus Ex7-28Part. Levels 20 3 3.425 62. 4. MTB > Stat > ANOVA > Balanced ANOVA In Results. 11. 19. 2.992 MSP×O − MSE 0.308 − 0.1400 ⇒ 0 n 2 MSO − MSP×O 1. 8. 14.65 1.173 0. Ex7-28Op … 1 2 3 Source Ex7-28Part Ex7-28Op Error Variance component 10. 6470 1.6186 Part-To-Part 3.91 100.3908 1.8832 0.10. and total gauge variability may also be found using: MTB > Stat > Quality Tools > Gage Study > Gage R&R Study (Crossed) Gage R&R Study .9454 n = = 0.861 Two-Way ANOVA Table Without Interaction Source Ex7-28Part Ex7-28Op Repeatability Total DF 19 2 98 119 SS 1185.2106 Total Variation 3.7118 0.93977 5.8832 = 0.00 7-24 .Chapter 7 Exercise Solutions (a) 2 2 σˆ Reproducibility = σˆ Operator = 0.8938 0.02 7.8380 0.000 0.0106 2 2 σˆ Repeatability = σˆ Error = 0.000 0.6724 Repeatability 0.3083 0. Estimates of variance components.3083 0.94541 5.1451 %Contribution (of VarComp) 8.05 59.92 0.6447 1.50 1274.8832 F 70. reproducibility.10 91.20176 19.173 0.10310 0.10310 0.33842 20.3908 1.0305 Number of Distinct Categories = 4 %Study Var (%SV) 28.8832 (b) 2 2 2 σˆ Gauge = σˆ Reproducibility + σˆ Repeatability = 0.7178 P 0.62 27.10 0.59 MS 62.8938 σˆ Gauge = 0.55 1274. repeatability.98 100.2513 11.43 2.32 28.0106 10.0106 + 0.59 MS 62.1050 /T = P USL-LSL 60 − 6 This gauge is borderline capable since the estimate of P/T ratio just exceeds 0.ANOVA Method Two-Way ANOVA Table With Interaction Source Ex7-28Part Ex7-28Op Ex7-28Part * Ex7-28Op Repeatability Total DF 19 2 38 60 119 SS 1185.09 95.4814 P 0.62 86.9917 F 87.09 3.232 Gage R&R Source Total Gage R&R Repeatability Reproducibility Ex7-28Op Part-To-Part Total Variation VarComp 0.00 Study Var Source StdDev (SD) (6 * SD) Total Gage R&R 0.6186 Ex7-28Op 0.0106 0.9454 (c) 6 × σˆ Gauge 6 × 0.43 2.15 3.6386 Reproducibility 0. 39 20 LCL=20.757 30 25 2 _ R=1.15 20 LCL=0 0 1 2 Ex7-28Op Xbar Chart by Ex7-28Op 1 2 3 Ex7-28Op * Ex7-28Part Interaction 3 30 30 25 UCL=24.23 Average Sample Mean 8 Ex7-28Op 1 2 3 25 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 5 16 17 18 19 20 Ex7-28Part 7-25 .Chapter 7 Exercise Solutions 7-28 continued Visual representations of variability and stability are also provided: Gage R&R (ANOVA) for Ex7-28Reading Reported by : Tolerance: M isc: G age name: Date of study : Components of Variation Ex7-28Reading by Ex7-28Part 100 % Contribution 30 Percent % Study Var 25 50 20 0 Gage R&R Repeat Reprod 1 Part-to-Part 2 3 4 5 R Chart by Ex7-28Op Sample Range 4 1 2 6 7 9 10 11 12 13 14 15 16 17 18 19 20 Ex7-28Part Ex7-28Reading by Ex7-28Op 3 UCL=3.55 _ _ X=22. this is also the number of distinct categories of parts that the gauge is able to distinguish) DR = 23.1451 n= SNR 2 ρˆ P 2(0. but larger than a value of two (or less) that indicates inadequate gauge capability. exceeding the minimum recommendation of four.2513.9198 m = 1 + ρˆ P = 1 + 0. the gauge is capable.9198 2 σˆ Total 11.Chapter 7 Exercise Solutions 7-29☺.000) = 1 − 0. σˆ Total = 11. µ = µ1 + µ2 + µ3 = 100 + 75 + 75 = 250 σ = σ 12 + σ 22 + σ 33 = 42 + 42 + 22 = 6 Pr{x > 262} = 1 − Pr{x ≤ 262} 262 − µ ⎫ ⎧ = 1 − Pr ⎨ z ≤ ⎬ σ ⎭ ⎩ 262 − 250 ⎫ ⎧ = 1 − Pr ⎨ z ≤ ⎬ 6 ⎩ ⎭ = 1 − Φ (2.9772 = 0.94 DR 1 − ρˆ P 1 − 0.79 indicates that fewer than five distinct levels can be reliably obtained from the measurements.94.0228 7-26 .79 1 − ρˆ P 1 − 0. This is near the AIAG-recommended value of five levels or more. (Also note that the MINITAB Gage R&R output indicates “Number of Distinct Categories = 4”. 2 2 σˆ Part = 10.9198 SNR = 4.1451 ρˆ P = 2 σˆ Part 10.9198) = = 4. By this measure.2513 = = 0.9198 = 23. 7-30 (7-22). 0(3.08(10)(20)(3) = 48 2 σˆ Weight ≅ d 2 ⎡⎣ µˆW2 σˆ L2σˆT2 + µˆ L2σˆW2 σˆT2 + µˆT2σˆW2 σˆ L2 ⎤⎦ = 0.9 µ y = µ1 − µ2 = 20 − 19.25 ⎠ = Φ (−0.4 σ y2 = σ 12 + σ 22 = 0.01) + 4.6 = 0.6.1 or y = x1 − x2 > 0.Chapter 7 Exercise Solutions 7-31 (7-23).1} + 1 − Pr{ y ≤ 0.01)(0.4 ⎞ = Φ⎜ ⎟ +1− Φ ⎜ ⎟ ⎝ 0.32 ).082 ⎡⎣10 2 (0. 0.42 ) Nonconformities will occur if y = x1 − x2 < 0.12 ) + 202 (0. x2 ~ N (19.9 − 0.0)(4.4330 7-32 (7-24).1} + Pr{ y > 0.2743 + 1 − 0. Volume = L × H × W ≅ µ L µ H µW + ( L − µ L ) µ H µW + ( H − µ H ) µ L µW + (W − µW ) µ L µ H µˆ Volume ≅ µ L µ H µW = 6.0) = 72.01)(0.9} = Pr{ y < 0.8413 = 0.12 ) + 32 (0.6) + 1 − Φ (1.04252 7-27 .1 − 0.00181 σˆ Weight ≅ 0.32 + 0.02 (0.00) = 0.22 )(0.4 ⎞ ⎛ 0.32 )(0.32 ) ⎤⎦ = 0.01) + 3.50 Pr{Nonconformities} = Pr{ y < LSL} + Pr{ y > USL} = Pr{ y < 0.02 (0.01) = 0.0061 7-33 (7-25).0 2 σ Volume ≅ µ L2σ H2 σ W2 + µ H2 σ L2σ W2 + µW2 σ L2σ H2 = 6.02 (0.42 = 0. x1 ~ N (20.2 2 )(0. 0.25 ⎠ ⎝ 0. Weight = d × W × L × T ≅ d [ µW µ L µT + (W − µW ) µ L µT + ( L − µ L ) µW µT + (T − µT ) µW µ L ] µˆ Weight ≅ d [ µW µ L µT ] = 0.01)(0.9} ⎛ 0.25 σ y = 0. 05(3.1282) ] (0.1026 26 ⎜⎝ 4 2 3 2 ⎟⎠ 2 ⎣ 26 ⎦ s = (3 + 0.05)(0. 2 ≤ x ≤ 4 26 4 4 4⎞ 1 ⎛5 3 ⎡1 ⎤ E ( x) = µ x = ∫ xf ( x)dx = ∫ x ⎢ (5 x − 2) ⎥ dx = ⎜ x − x 2 ⎟ = 3.05µ x )(0.Chapter 7 Exercise Solutions 7-34 (7-26).05)σ x2 = 2 [3 + 0.1026 − (3.05( µ x )] = [3 + 0.05 x)2 ⎤ ⎥ =⎢ ∂x ⎢⎣ ⎥⎦ 2 σ x2 µx = 2(3 + 0.05 x) 2 and f ( x) = σ x2 = E ( x 2 ) − [ E ( x)] = 10.1001 7-35 (7-27).1282) 2 = 0.9629 2 ⎡ ∂ g ( x) ⎤ σ ≅⎢ ⎣ ∂ x ⎥⎦ 2 2 2 s σ x2 µx ⎡ ∂ (3 + 0.3170 2 µ s ≅ g ( x) = [3 + 0.3170) = 0.1282 2⎟ 26 ⎜⎝ 3 2 2 ⎣ 26 ⎦ ⎠ 4 4 4 ⎞ 1 ⎛5 2 ⎡1 ⎤ E ( x 2 ) = ∫ x 2 f ( x)dx = ∫ x 2 ⎢ (5 x − 2) ⎥ dx = ⎜ x 4 − x3 ⎟ = 10.1282)] = 9. I = E ( R1 + R2 ) µI ≅ µE (µR + µR ) 1 σ I2 ≅ σ 2 µE σ 2 + σ R2 2 ( R (µR + µR ) (µR + µR ) 1 2 E + 2 1 1 2 ) 2 7-28 . 1 (5 x − 2).05(3. x1 ~ N ( µ1 .006 ⎧⎪ 0.3002 ) µ y = µ1 − µ2 σ y = σ 12 + σ 22 = 0.4 = 5.002236 Pr{positive clearance} = 1 − Pr{interference} = 1 − Pr{ y < 0} ⎛ 0 − 0.346 7-37 (7-29).000005 σ y = 0.4002 + 0.989 +⎜ = +⎜ = 299 ⎟ ⎟ 2 ⎝ α ⎠ 4 2 ⎝ 0.683) = 1 − 0.3002 = 0.004.010. 0.09 − µ y = −2.0012 = 0.0036 = 0.01 γ = 0.989 1 ⎛ 2 − α ⎞ χ1−γ .09] = 1.0022 + 0.01 ⎠ 4 2 n≅ 7-29 .512) − 0.4002 ).5(−2. ID ~ N (2.0022 ) and OD ~ N (2.006 2 2 σ y2 = σ ID + σ OD = 0.010 − 2.004 = 0.0012 ) Interference occurs if y = ID – OD < 0 µ y = µID − µOD = 2.5 µ y = −[0. 0. 0.000005 ⎠ = 1 − Φ (−2.4 1 ⎛ 2 − 0. 0.20.006 ⎞ = 1− Φ ⎜ ⎟ ⎝ 0.9964 7-38 (7-30).80 2 χ12−γ .Chapter 7 Exercise Solutions 7-36 (7-28).01 ⎞ 5.09} = 0.512 0.09 − µ y ⎫⎪ −1 Pr ⎨ z < ⎬ = Φ (0.006) σ y ⎭⎪ ⎩⎪ 0. α = 0.5 Pr{ y < 0. x2 ~ N ( µ2 .4 = χ 0. n = 20.838(1) = 83.55 7-40 (7-32). γ = 0.10. α = 0.208 UTL = x + KS = 350 + 2. a natural tolerance interval would be the smallest to largest observations. x ~ N (350.10.355(10) = 323.95 . x ~ N (85.162 7-41 (7-33).208(10) = 372.05 ⎞ 7. γ = 0.05. x ~ N (300. 7-30 .4 1 ⎛ 2 − 0.4 = χ 0. n = 10.10.779 +⎜ = +⎜ = 77 ⎟ ⎟ 2 ⎝ α ⎠ 4 2 ⎝ 0. n = 25.838 x − KS = 85 − 1.95 . γ = 0.102 ).4 = 7. one-sided From Appendix VIII: K = 2.779 1 ⎛ 2 − α ⎞ χ1−γ .05 ⎠ 4 2 n≅ After the data are collected.Chapter 7 Exercise Solutions 7-39 (7-31).355 UTL = x + KS = 300 + 2. α = 0.08 7-42 (7-34).05 γ = 0.90 2 χ12−γ .12 ). α = 0. α = 0. one-sided From Appendix VIII: K = 1. one-sided From Appendix VIII: K = 2.102 ).90 . n −1 S ( n = 0.95. 7-44 (7-36).0003) = [0.00032 ( ) (a) α = 0. the interval relates to individual observations (random variables).1265] Part (a) is a tolerance interval on individual thickness observations.1264 ± 2. part (b) is a confidence interval on mean thickness.445(0. and two-sided From Appendix VII: K = 2.1263.95 log(1 − γ ) log(1 − 0.05. 7-31 . 0.n −1 = t0.1257.05. tα / 2. 0.445 TI on x : x ± KS = 0. In part (a).1264.023 CI on x : x ± tα / 2. 0. x ~ N 0.025.1264 ± 2.95) n= = = 59 log(1 − α ) log(1 − 0.Chapter 7 Exercise Solutions 7-43 (7-35).023 0.39 = 2. while in part (b) the interval refers to a parameter of a distribution (an unknown constant).05) The largest observation would be the nonparametric upper tolerance limit. γ = 0.1271] (b) α = 0. α = 0. γ = 0.0003 ) 40 = [0.05. A number in parentheses gives the exercise number from the 4th edition.4345 No. µ0 = 1050.5. (b) σˆ = MR2 d 2 = 38. k = 0. New exercises are denoted with an “☺”. The estimate used for σ is much smaller than that from the data.128 = 34. σ = 25. The point at which the assignable cause occurred can be determined by counting the number of increasing plot points. 8-1 . std dev = 25.5. The assignable cause occurred after observation 10 – 3 = 7. K = (δ/2)σ = (1/2)25 = 12. h = 5 1250 Cumulative Sum 1000 750 500 250 UCL=125 0 0 LCL=-125 2 4 6 8 10 12 Sample 14 16 18 20 The process signals out of control at observation 10. 8-1.Chapter 8 Exercise Solutions Several exercises in this chapter differ from those in the 4th edition.8421/1. H = 5σ = 5(25) = 125 (a) MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Molecular Weight (Ex8-1mole) target = 1050. An “*” following the exercise number indicates that the description has changed. δ = 1σ. std dev = 25.5. h = 5 50 Cumulative Sum 40 30 20 10 UCL=5 0 0 LCL=-5 2 4 6 8 10 12 Sample 14 16 18 20 The process signals out of control at observation 10. 8-2 . k = 0. MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Standardized Molecular Weight (Ex8-2std) target = 1050.Chapter 8 Exercise Solutions 8-2. The assignable cause occurred after observation 10 – 3 = 7. 5 MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Molecular Weight (Ex8-1mole) FIR=H/2 = 62. k = 0.5 std dev units) 1250 Cumulative Sum 1000 750 500 250 UCL=125 0 0 LCL=-125 2 4 6 8 10 12 Ex8-1Obs 14 16 18 20 For example.Chapter 8 Exercise Solutions 8-3.5. the same as the CUSUM without a FIR feature. K = 12. the process signals out of control at observation 10. (a) µ0 = 1050.5) + 62.5 FIR = H/2 = 62. 8-3 .5 (or 2. C1+ = max ⎡⎣0. H/2 = 125/2 = 62.5/25 = 2.1045 − (1050 + 12. σ = 25. in std dev units = 62. xi − ( µ0 − K ) + C0+ ⎤⎦ = max [ 0.5.5] = 45 Using the tabular CUSUM.5. h = 5. 3 1100 1000 -3.5SL=0 2 4 6 8 10 12 Observation 14 16 18 20 Using 3. there are no out-of-control signals.8 Individual Value 1200 _ X=1116. 8-4 .5SL=141.5-sigma limits +3. However there does appear to be a trend up from observations 6 through 12—this is the situation detected by the cumulative sum.Chapter 8 Exercise Solutions 8-3 continued (b) MTB > Stat > Control Charts > Variables Charts for Individuals > I-MR I-MR Chart of Molecular Weight (Ex8-1mole) with 3.8 40 0 -3.8 2 4 6 8 10 12 Observation 14 16 18 20 160 Moving Range +3.5SL=1236.6 120 80 __ MR=38.5σ limits on the Individuals chart.5SL=995. µ0 = 8.05.128 = 0.77.5 = −0.2 LCL=-0.1 0.05 is probably not reasonable.2385 Cumulative Sum 0.3 2 4 6 8 10 12 14 Sample 16 18 20 22 24 There are no out-of-control signals.936) − 1 = 742. h = 4.05) = 0.02.5. so σ = 0.936)] + 2(−0.48 8-5 . H = hσ = 4.128 = 0.2385 -0.05.0 0 -0. σ = 0.964 2(−0.5 exp[−2(−0. b = h + 1. ∆ + = δ * − k = 0 − 0.5)(5.936 δ * = 0.77 0.5)(5.2 0.02.02.5 = −0.2385 (a) MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Can Weight (Ex8-4can) target = 8.5.Chapter 8 Exercise Solutions 8-4.5) 2 1 1 1 2 = + = = 0.3 UCL=0.0186957 /1. k = 0.166 = 4.0166 .0027 = 371. h = 4.77 (0.166 = 5.77. (b) σˆ = MR2 1. k = 1/ 2.77 + 1.1 -0. ∆ − = −δ * − k = −0 − 0. and h=4. In Exercise 8-4: µ0 = 8. σ = 0. k=1/2.964 ARL+0 = ARL−0 ≅ ARL0 = 1/ 0.0027 + − ARL0 ARL0 ARL0 742. b = h + 1.1 -0.05.01 UCL=0. k = 0. h = 8.01.4005 -0. H = hσ = 8.4005 0.3 LCL=-0.25 = −0.176)] + 2(−0.0 -0.01 + 1.05) = 0.02.6771 2(−0.176 δ * = 0.25 exp[−2(−0.0027 + − ARL0 ARL0 ARL0 741. In Exercise 8-5: µ0 = 8.4 0. ∆ − = −δ * − k = −0 − 0.176) − 1 = 741.02.02. 8-6 .25.25.01 (0.6771 ARL+0 = ARL−0 ≅ ARL0 = 1 0. k = 0.2 0.166 = 9.1 0 0. ∆ + = δ * − k = 0 − 0.25)(9.5 2 4 6 8 10 12 14 Sample 16 18 20 22 24 There are no out-of-control signals.84 The theoretical performance of these two CUSUM schemes is the same for Exercises 8-4 and 8-5. µ0 = 8.25)(9.01.25. k = 0.3 Cumulative Sum 0.4 -0. σ = 0.Chapter 8 Exercise Solutions 8-5. h = 8.4005 MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Can Weight (Ex8-4can) target = 8. h = 8. σ = 0.2 -0.166 = 8.0027 = 370.25 = −0.25.05.25) 2 1 1 1 2 = + = = 0. 2 0. k = 0. k = 1/2.77 (0.77.2 LCL=-0.1 -0. σ = 0. h = 4.2385 FIR = H/2. h = 4.2385 Cumulative Sum 0.1 0. 8-7 . Process was out of control at process start-up.77 0.3 UCL=0.385 MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Ex8-4can FIR = 2.00.0 0 -0.Chapter 8 Exercise Solutions 8-6.2385 -0.05.05) = 0. H = h σ = 4. FIR in # of standard deviations = h/2 = 4. µ0 = 8.00.77/2 = 2.3 2 4 6 8 10 12 14 Sample 16 18 20 22 24 The process signals out of control at observation 20.385 std dev. target = 8.5. The assignable cause occurred after observation 12 – 10 = 2.8 Cumulative Sum 50 25 0 0 -25 -50 LCL=-60.7215 /1. σˆ = 12.5. k = 1/ 2.8 1 8 16 24 32 40 48 Sample 56 64 72 80 Test Results for CUSUM Chart of Ex8-7temp TEST. 8-8 . 13 The process signals out of control at observation 12.16 (b) µ0 = 950. h = 5 75 UCL=60. Test Failed at points: 12. h = 5 MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Temperature Readings (Ex8-7temp) target = 950. (a) σˆ = MR2 d 2 = 13.128 = 12. One point beyond control limits. k = 0.16.Chapter 8 Exercise Solutions 8-7. 19. 17. 20. 32 The process signals out of control on the lower side at sample 3 and on the upper side at sample 12. 22. h = 5 MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Bath Concentrations (Ex8-8con) target = 175.Chapter 8 Exercise Solutions 8-8. k = 1/2. 15. std dev = 5. 21. 28. 25. One point beyond control limits. 26. 29. 27. 30.35 /1. 24.1 0 -100 -200 3 6 9 12 15 18 Sample 21 24 27 30 Test Results for CUSUM Chart of Ex8-8con TEST.1 0 LCL=-28.128 = 5. 18.35) (b) µ0 = 175. Assignable causes occurred after startup (sample 3 – 3 = 0) and after sample 8 (12 – 4). 16. 8-9 . Test Failed at points: 12.629. h = 5 400 Cumulative Sum 300 200 100 UCL=28. 31. σˆ = 5. 14. k = 1/ 2. (a) σˆ = MR2 d 2 = 6. 23.629 (from a Moving Range chart with CL = 6. 13.629. 25. σˆ = 5.949. Test Failed at points: 16.7 -100 -200 -300 -400 -500 4 8 12 16 20 Sample 24 28 32 36 Test Results for CUSUM Chart of Ex8-9vis TEST. One point beyond control limits.25. h = 8. to give longer in-control ARLs with shorter out-of-control ARLs.01 UCL=47. std dev = 5.Chapter 8 Exercise Solutions 8-9. Assignable causes occurred after startup (sample 2 – 2) and after sample 9 (16 – 7).01 MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Viscosity Measurements (Ex8-9vis) target = 3200. k = 0. 17. should be balanced by a larger control limit. 18 The process signals out of control on the lower side at sample 2 and on the upper side at sample 16. k = 0.949.25.71/1. 8-10 . (a) σˆ = MR2 d 2 = 6.71) (b) µ0 = 3200.949 (from a Moving Range chart with CL = 6.7 0 0 Cumulative Sum LCL=-47.01. (c) Selecting a smaller shift to detect. h = 8. k = 0.128 = 5. h = 8. One point beyond control limits.5.. σ = 0. Exm5-1x5 TEST.8 0.250 -0.4 UCL=0. µ0 = 1. n = 5.. k = 0.0626 δ = 1. 41.0313.250 0. The x -R chart shown in Figure 5-4 signals out of control at sample 43. H = hσ x = 0.5. and remains above the upper limit.4 4 8 12 16 20 24 Sample 28 32 36 40 44 Test Results for CUSUM Chart of Exm5-1x1. k = δ 2 = 0.2 1. 44.14 5 = 0. . at sample 40 versus sample 43.50.2 LCL=-0. Exm5-1x5) target = 1.14. h = 4. 42.2504 MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Flow Width Data (Exm5-1x1...6 0... 43.14.0 Cumulative Sum 0. K = kσ x = 0.0 -0.Chapter 8 Exercise Solutions 8-10*. σ x = σ n = 0.2 0 0. h = 4 1.50. . This CUSUM detects the shift in process mean earlier. std dev = 0. 45 The CUSUM chart signals out of control at sample 40. 8-11 . Test Failed at points: 40. 74 1188 5.29 1064 0.57 1 1.26 1139 3. Vi = | yi | − 0.71 1238 7.Chapter 8 Exercise Solutions 8-11.07 1055 0.78 11 OOC 7 33.76 3.61 1163 4.38 1146 3.xls : worksheet Ex8-11 mu0 = sigma = delta = k= h= 1050 25 1 sigma 0.40 1128 3.36 1050 0 -2.56 -0.52 3. Process variability is increasing.88 13 OOC 7 one-sided lower cusum Si.9 10 OOC 7 30.07 1037 -0.65 4 0 0 0 0 1.848 2 0 0 0.84 3. 8-12 .36 1087 1.22 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The process is out of control after observation 10 – 3 = 7.68 3.04 3.52 5.55 7 OOC 7 21.66 9 OOC 7 26.5 5 Obs.44 5 OOC 7 14.90 1151 4.56 3.61 1146 3.84 3.15 2 0.989 1 1.2 -1.26 1167 4.21 1095 1.94 3 0.35 6 OOC 7 16.738 2 5.52 -0.05 1169 4.12 2.498 3 OOC 7 8.86 1 0. i No FIR xi 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 yi vi 1045 -0.84 one-sided upper cusum Si+ N+ OOC? When? 0 0 0 0 0 0 0 0 0 0.2 -1.13 1125 3 2.50 1125 3 2.N.48 1.631 1 2.349 ( ) Excel file: workbook Chap08.822 0.68 1.049 4 OOC 7 11.54 12 OOC 7 36.8 1.OOC? When? 0 0.52 4.49 1008 -1.56 8 OOC 7 23. 0199 2.62 22.56 10.946 2 150 -4.34 20 OOC 0 one-sided lower cusum Si.6646 2.4410 3.47 13 OOC 0 192 3.68 6 OOC 0 158 -3.59 14 OOC 0 186 1.74 15 OOC 0 197 3.9540 1.6294 (from Exercise 8-8) 1 sigma 0.31 26.32 12 OOC 0 175 0.5987 1.0199 2.27 23.OOC? When? 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1.3093 2.5 5 one-sided upper cusum xi yi vi Si+ N+ OOC? When? 0 160 -2.2435 0.19 4 OOC 0 153 -3.9081 3.2633 3. Vi = | yi | − 0.7106 0.65 20.36 20.55 16 OOC 0 190 2.65 23.6646 2.129 3 OOC 0 151 -4.05 23 10 OOC 0 179 0.06 22.46 31 19 OOC 0 182 1.56 11 OOC 0 184 1.8 7 OOC 0 162 -2.9540 1. Process variability has been increasing since start-up.62 17.0000 -2.xls : worksheet Ex8-12 mu0 = sigma = delta = k= h= i No FIR 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 175 5.84 31.37 17 OOC 0 189 2.822 1 158 -3.5528 3.18 15.7764 1.N.32 1.04 18 OOC 0 185 1. 8-13 .Chapter 8 Exercise Solutions 8-12.4869 2.822 0.3 8 OOC 0 186 1.9081 3.45 9 OOC 0 195 3.31 13 5 OOC 0 154 -3.7304 3.349 ( ) Excel file : workbook Chap08.32 28.86 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … The process was last in control at period 2 – 2 = 0.62 3.68 7.0199 2.00 19.16 30. 2)(9.3 ∆ − = −δ * − k = −0.7)(9.166 = 8 + 1.166) − 1 ARL−1 = = 381.166)] + 2(0.Chapter 8 Exercise Solutions 8-13.2 = −0.166 exp[−2(0.166 exp[−2(−0. two-sided cusum with k = 0.2 b = h + 1. 767 ARL1 = 1/ 0.005 + − ARL0 ARL0 ARL0 430.2 and h = 8 In control ARL performance: δ* = 0 ∆ + = δ * − k = 0 − 0.005 = 215.023 381.7)(9.5 − 0.2)(9.3) 2 exp[−2(−0.556 2(−0.3)(9.166) − 1 ARL+0 = ARL−0 ≅ = 430.7 b = h + 1.2 ∆ − = −δ * − k = −0 − 0.556 ARL0 = 1/ 0.023 2(0.23 Out of control ARL Performance: δ * = 0.3)(9.2) 2 1 1 1 2 = + = = 0.040 = 25.166)] + 2(−0.166 = 9.2 = 0.5 − 0.166)] + 2(−0. Standardized.040 + − ARL1 ARL1 ARL1 25.166 = 9.02 8-14 .166 = 8 + 1.2 = −0.2 = −0. 767 2(−0.7) 2 1 1 1 1 1 = + = + = 0.166) − 1 ARL+1 = = 25.5 ∆ + = δ * − k = 0. 8-15 . The upper CUSUM is used to detect upward shifts in the level of the process.95238. µ0 = 3150.95238) = 2.5.Chapter 8 Exercise Solutions 8-14.762 MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Viscosity Measurements (Ex8-9vis) target = 3150 1200 Cumulative Sum 1000 800 600 400 200 UCL=30 0 LCL=-30 0 4 8 12 16 20 Sample 24 28 32 36 MINITAB displays both the upper and lower sides of a CUSUM chart on the same graph.976. The assignable cause occurred at start-up (2 – 2). h = 5 K = ks = 0. The process signals out of control on the upper side at sample 2. s = 5. there is no option to display a single-sided chart.95238) = 29. k = 0. H = hs = 5(5.5 (5. of either a trend/drift or a shit in measurements. The CUSUM chart reflects a consistent run above the target value 734. k = 0.35 H = hσˆ = 5(108.5. The outof-control signals should lead us to investigate and determine the assignable cause. with a centerline at x = 909 .7) = 54.6 /1.6) µ0 = 734. h = 5 K = kσˆ = 0. starting at about sample 18.Chapter 8 Exercise Solutions 8-15☺. from virtually the first sample.5(108.7) = 543. There is a distinct signal on both charts. σˆ = MR2 d 2 = 122.5.7 (from a Moving Range chart with CL = 122.5.128 = 108.5 MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Light Velocity (Ex5-60Vel) target = 734. 8-16 . displayed a distinct downward trend in measurements.5 4000 Cumulative Sum 3000 2000 1000 UCL=544 0 0 LCL=-544 4 8 12 16 20 24 Sample 28 32 36 40 The Individuals I-MR chart. 1.1.7.5. L = 2.8 750 _ _ X=734.5.2 4 8 12 16 20 24 Sample 28 32 36 40 The EWMA chart reflects the consistent trend above the target value.7 MTB > Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Light Velocity (Ex5-60Vel) lambda = 0. and also indicates the slight downward trend starting at about sample 22.Chapter 8 Exercise Solutions 8-16☺. 8-17 .5 700 -2.7 900 EWMA 850 800 +2. CL = µ0 = 734. 734. λ = 0.7SL=801. L = 2. σ = 108.7SL=667. 7SL=1065. CL = µ0 = 1050. L = 2.7. λ = 0.11] (c) λ = 0. (a) λ = 0.11.2.2 (2 − 0.51 MTB > Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Molecular Weight (Ex8-1mole) lambda = 0.5] As λ increases.31.49.4) = [8.4 _ _ X=1050 1060 1040 -2.69] (b) λ = 0.7 1140 1120 EWMA 1100 1080 +2.6 2 4 6 8 10 12 Sample 14 16 18 20 Process exceeds upper control limit at sample 10.7SL=1034. σ = 25. L = 3 limits = µ0 ± Lσ λ (2 − λ ) = 10 ± 3(1) 0.1) = [9. 8-18 .1.4 (2 − 0.Chapter 8 Exercise Solutions 8-17 (8-15).5. UCL = 1065.1 (2 − 0.1. L = 2. the width of the control limits also increases. 8-18 (8-16).1. the same as the CUSUM chart.2) = [9. L = 3 limits = µ0 ± Lσ λ (2 − λ ) = 10 ± 3(1) 0.10.4. L = 3 limits = µ0 ± Lσ λ (2 − λ ) = 10 ± 3(1) 0. LCL = 1034. 02 8. Assume σ = 0.08 UCL=8. L = 3.9700 7.05.2. UCL = 8. LCL = 7.97 MTB > Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Can Weight (Ex8-4can) lambda = 0.04 _ _ X=8.0700 8. λ = 0.96 2 4 6 8 10 12 14 Sample 16 18 20 22 24 The process is in control.02 8.06 EWMA 8.Chapter 8 Exercise Solutions 8-19 (8-17).2. CL = µ0 = 8.02. L = 3 8.98 LCL=7. 8-19 .07.00 7. 05 8.1. L = 2.04 EWMA 8. There is not much difference between the control charts.00 7.7. LCL = 7. 8-20 .03 _ _ X=8.02 8.1.02.7SL=8.02 8.7SL=7.05087 8. Assume σ = 0. L = 2. λ = 0.99 MTB > Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Can Weight (Ex8-4can) lambda = 0.05.98913 2 4 6 8 10 12 14 Sample 16 18 20 22 24 The process is in control.Chapter 8 Exercise Solutions 8-20 (8-18).01 8. UCL = 8. CL = µ0 = 8.7 +2.99 -2.05. 1. 8-21 . UCL = 957. L = 2.7. One point beyond control limits. CL = µ0 = 950.53 EWMA 955 _ _ X=950 950 945 -2.47.Chapter 8 Exercise Solutions 8-21 (8-19).53. σˆ = 12.7SL=957.16 .1.7 960 +2. 12 and 13. MTB > Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Temperature Readings (Ex8-7temp) lambda = 0. 13 Process is out of control at samples 8 (beyond upper limit. but not flagged on chart). L = 2.7SL=942. Test Failed at points: 12.47 1 8 16 24 32 40 48 Sample 56 64 72 80 Test Results for EWMA Chart of Ex8-7temp TEST. LCL = 942. λ = 0. 4. One point beyond control limits. as compared to the chart in the Exercise 21 (with the smaller λ) which signaled out of control at earlier samples. σˆ = 12. λ = 0.76 930 1 8 16 24 32 40 48 Sample 56 64 72 80 Test Results for EWMA Chart of Ex8-7temp TEST. MTB > Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Temperature Readings ( Ex8-7temp) lambda = 0.24.Chapter 8 Exercise Solutions 8-22 (8-20). CL = µ0 = 950.4. UCL = 968.76. LCL = 931. L = 3 970 UCL=968. 8-22 . the process is out of control at observation 70.24 EWMA 960 _ _ X=950 950 940 LCL=931. Test Failed at points: 70 With the larger λ. L = 3.16 . L = 2.634 .30 _ _ X=175 175 -2. L = 2.30. The process average of µˆ = 183. λ = 0.6SL=172. UCL = 177.6SL=177.594 is too far from the process target of µ0 = 175 for the process variability. LCL = 172.Chapter 8 Exercise Solutions 8-23 (8-21). CL = µ0 = 175.70. σˆ = 5. MTB > Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Bath Concentrations (Ex8-8con) lambda = 0.05. The data is grouped into three increasing levels.6.6 190 EWMA 185 180 +2. 8-23 .05.70 170 3 6 9 12 15 18 Sample 21 24 27 30 Process is out of control. Chapter 8 Exercise Solutions 8-24☺.1 t.23 5.5 10 r .4-3).9 26 .25th-root transformed data showed no out-of-control signals. L = 2.ep.2 25 28 29 n.7 24 28 .7SL=6.p ay y .4 .ec.and 0.27 17. 8-24 .n .Sep ep . L = 2. Individuals control charts of 0.n.n.5 10.b.0 +2.ecJa F e F e M M a A M Ma Ma J Ju Ju Ju Ju O N N D D S S Ex6-62Day In Exercise 6-62.25 12.Oc Oc O c ct.1. The EWMA chart also does not signal out of control.y .7 16 16 22 25 l -6 l.4 t.Ju Ju Ju Ju l.8 19 .0 _ _ X=12.8 l.2777th.5 -2.0 20 23 25 r .7 MTB > Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Homicide Data (Ex6-62Bet) lambda = 0.0 7.5 EWMA 15.b.7 20. λ = 0. a properly designed EWMA chart is very robust to the assumption of normally distributed data. As mentioned in the text (Section 8.1.a r .9 22 24 t.ov o v .un n.7SL=18. 68 _ _ X=3200 3200 -2.7 3205 +2. µ0 = 3200.1. 8-25 . sigma=5. L=2.7SL=3196. λ = 0.7 MTB > Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Viscosity (Ex8-9vis) Target=3200. σˆ = 5. L = 2.Chapter 8 Exercise Solutions 8-25 (8-22).7SL=3203.95 (from Exercise 8-9).95. lambda=0.1.32 EWMA 3195 3190 3185 3180 3175 4 8 12 16 20 24 Ex5-60Meas 28 32 36 The process is out of control from the first sample. 6 _ _ X=1050 1050 LCL=1019. 12. 15.4 MTB > Stat > Control Charts > Time-Weighted Charts > Moving Average Moving Average Chart of Molecular Weight (Ex8-1mole) w = 6. LCL = 1019. µ0 = 1050. 20 Process is out of control at observation 10. 18.6. 11.Chapter 8 Exercise Solutions 8-26 (8-23). CL = 1050. target value = 1050. Test Failed at points: 10.4 1000 2 4 6 8 10 12 Sample 14 16 18 20 Test Results for Moving Average Chart of Ex8-1mole TEST. std dev = 25 1200 Moving Average 1150 1100 UCL=1080. 14. UCL = 1080. 19. 17. One point beyond control limits. 16. σ = 25. 8-26 . w = 6. the same result as for Exercise 8-1. 13. Chapter 8 Exercise Solutions 8-27 (24). w = 5, µ0 = 8.02, σ = 0.05, CL = 8.02, UCL = 8.087, LCL = 7.953 MTB > Stat > Control Charts > Time-Weighted Charts > Moving Average Moving Average Chart of Can Weight (Ex8-4can) w = 5, process target = 8.02, std dev = 0.05 8.20 Moving Average 8.15 8.10 UCL=8.0871 8.05 _ _ X=8.02 8.00 LCL=7.9529 7.95 7.90 2 4 6 8 10 12 14 Sample 16 18 20 22 24 The process is in control, the same result as for Exercise 8-4. 8-27 Chapter 8 Exercise Solutions 8-28☺. w=5 MTB > Stat > Control Charts > Time-Weighted Charts > Moving Average Moving Average Chart of Homicide Data (Ex6-62Bet) w = 5, target and std dev estimated from data UCL=25.31 25 Moving Average 20 15 _ _ X=12.25 10 5 0 LCL=-0.81 3 6 9 12 15 Sample 18 21 24 27 Because these plot points are an average of five observations, the nonnormality of the individual observations should be of less concern. The approximate normality of the averages is a consequence of the Central Limit Theorem. 8-28 Chapter 8 Exercise Solutions 8-29 (8-25). Assume that t is so large that the starting value Z 0 = x has no effect. ∞ ⎡ ∞ ⎤ E ( Z t ) = E[λ xt + (1 − λ )( Z t −1 )] = E ⎢λ ∑ (1 − λ ) j xt − j ⎥ = λ ∑ (1 − λ ) j E ( xt − j ) j =0 ⎣ j =0 ⎦ ∞ Since E ( xt − j ) = µ and λ ∑ (1-λ ) j = 1 , E ( Z t ) = µ j =0 8-30 (8-26). ⎡ ∞ ⎤ var( Z t ) = var ⎢ λ ∑ (1 − λ ) j xt − j ⎥ ⎣ j =0 ⎦ ∞ ⎡ ⎤ = ⎢λ 2 ∑ (1 − λ ) 2 j ⎥ ⎡⎣ var( xt − j ) ⎤⎦ ⎣ j =0 ⎦ λ ⎛σ 2 ⎞ = ⎜ ⎟ 2−λ ⎝ n ⎠ 8-31 (8-27). For the EWMA chart, the steady-state control limits are x ± 3σ λ . (2 − λ )n ⎛ 2 ⎞ ⎜ ⎟ ⎝ w + 1 ⎠ = x ± 3σ 1 = x ± 3σ , Substituting λ = 2/(w + 1), x ± 3σ 2 ⎞ wn ⎛ wn ⎜2− ⎟n w +1 ⎠ ⎝ which are the same as the limits for the MA chart. 8-32 (8-28). 1 w−1 w −1 . In the ∑ j= w j =0 2 EWMA, the weight given to a sample mean j periods ago is λ(1 - λ)j , so the average age ∞ 1− λ . By equating average ages: is λ ∑ (1 − λ ) j j = The average age of the data in a w-period moving average is j =0 λ 1− λ w −1 λ 2 2 λ= w +1 = 8-29 Chapter 8 Exercise Solutions 8-33 (8-29). For n > 1, Control limits = µ0 ± 3 ⎛ σ ⎞ 3σ ⎜ ⎟ = µ0 ± w⎝ n⎠ wn 8-34 (8-30). x chart: CL = 10, UCL = 16, LCL = 4 UCL = CL + kσ x 16 = 10 − kσ x kσ x = 6 EWMA chart: UCL = CL + lσ λ [(2 − λ )n] = CL + l σ n 0.1 (2 − 0.1) = 10 + 6(0.2294) = 11.3765 LCL = 10 − 6(0.2294) = 8.6236 8-35 (8-31). λ = 0.4 For EWMA, steady-state limits are ± Lσ λ (2 − λ ) For Shewhart, steady-state limits are ± kσ kσ = Lσ λ (2 − λ ) k = L 0.4 (2 − 0.4) k = 0.5 L 8-30 Chapter 8 Exercise Solutions 8-36 (8-32). The two alternatives to plot a CUSUM chart with transformed data are: 1. Transform the data, target (if given), and standard deviation (if given), then use these results in the CUSUM Chart dialog box, or 2. Transform the target (if given) and standard deviation (if given), then use the Box-Cox tab under CUSUM Options to transform the data. The solution below uses alternative #2. From Example 6-6, transform time-between-failures (Y) data to approximately normal distribution with X = Y 0.2777. TY = 700, TX = 700 0.2777 = 6.167, k = 0.5, h = 5 MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Transformed Failure Data (Ex8-37trans) X = Y^0.277, target - 6.167, k = 0.5, h = 5 UCL=10.46 Cumulative Sum 10 5 0 0 -5 -10 LCL=-10.46 2 4 6 8 10 12 Sample 14 16 18 20 A one-sided lower CUSUM is needed to detect an increase in failure rate, or equivalently a decrease in the time-between-failures. Evaluate the lower CUSUM on the MINITAB chart to assess stability. The process is in control. 8-31 Chapter 8 Exercise Solutions 8-37 (8-33). µ0 = 700, h = 5, k = 0.5, estimate σ using the average moving range MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM, also CUSUM options > Estimate > Average Moving Range CUSUM Chart of Valve Failure Data (Ex8-37fail) Target=700, h=5, k=0.5 4000 UCL=3530 3000 Cumulative Sum 2000 1000 0 0 -1000 -2000 -3000 LCL=-3530 -4000 2 4 6 8 10 12 Ex8-37No 14 16 18 20 A one-sided lower CUSUM is needed to detect an increase in failure rate. Evaluate the lower CUSUM on the MINITAB chart to assess stability. The process is in control. Though the data are not normal, the CUSUM works fairly well for monitoring the process; this chart is very similar to the one constructed with the transformed data. 8-32 Chapter 8 Exercise Solutions 8-38 (8-34). µ0 = TX = 700 0.2777 = 6.167, λ = 0.1, L = 2.7 MTB > Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Transformed Failure Data (Ex8-37trans) X = Y^0.2777, target = 6.167, lambda = 0.1, L = 2.7 7.5 +2.7SL=7.453 7.0 EWMA 6.5 _ _ X=6.167 6.0 5.5 5.0 -2.7SL=4.881 2 4 6 8 10 12 Sample 14 16 18 20 Valve failure times are in control. 8-39 (8-35). The standard (two-sided) EWMA can be modified to form a one-sided statistic in much the same way a CUSUM can be made into a one-sided statistic. The standard (two-sided) EWMA is zi = λ xi + (1 − λ ) zi−1 Assume that the center line is at µ0. Then a one-sided upper EWMA is zi+ = max ⎡⎣ µ0 , λ xi + (1 − λ ) zi −1 ⎤⎦ , and the one-sided lower EWMA is zi− = min ⎡⎣ µ0 , λ xi + (1 − λ ) zi −1 ⎤⎦ . 8-33 Chapter 9 Exercise Solutions Note: Many of the exercises in this chapter were solved using Microsoft Excel 2002, not MINITAB. The solutions, with formulas, charts, etc., are in Chap09.xls. 9-1. σˆ A = 2.530, nA = 15, µˆ A = 101.40 σˆ B = 2.297, nB = 9, µˆ B = 60.444 σˆ C = 1.815, nC = 18, µˆ C = 75.333 σˆ D = 1.875, nD = 18, µˆ D = 50.111 Standard deviations are approximately the same, so the DNOM chart can be used. R = 3.8, σˆ = 2.245, n = 3 x chart: CL = 0.55, UCL = 4.44, LCL = −3.34 R chart: CL = 3.8, UCL = D4 R = 2.574 (3.8) = 9.78, LCL = 0 Stat > Control Charts > Variables Charts for Subgroups > Xbar-R Chart Xbar-R Chart of Measurements (Ex9-1Xi) U C L=4.438 Sample M ean 4 2 _ _ X=0.55 0 -2 LC L=-3.338 -4 2 4 6 8 10 Sample 12 14 16 18 20 Sample Range 10.0 U C L=9.78 7.5 5.0 _ R=3.8 2.5 0.0 LC L=0 2 4 6 8 10 Sample 12 14 16 18 20 Process is in control, with no samples beyond the control limits or unusual plot patterns. 9-1 Chapter 9 Exercise Solutions 9-2. Since the standard deviations are not the same, use a standardized x and R charts. Calculations for standardized values are in: Excel : workbook Chap09.xls : worksheet : Ex9-2. n = 4, D3 = 0, D4 = 2.282, A2 = 0.729; RA = 19.3, RB = 44.8, RC = 278.2 Graph > Time Series Plot > Simple Control Chart of Standardized Xbar (Ex9-2Xsi) 1.5 1.0 Ex9-2Xsi +A2 = 0.729 0.5 0 0.0 -0.5 -A2 = -0.729 -1.0 Ex9-2Samp Ex9-2Part 2 A 4 A 6 A 8 B 10 B 12 C 14 C 16 C 18 C 20 C Control Chart of Standardized R (Ex9-2Rsi) 2.5 D4 = 2.282 Ex9-2Rsi 2.0 1.5 1.006 1.0 0.5 0.0 Ex9-2Samp Ex9-2Part D3 = 0 2 A 4 A 6 A 8 B 10 B 12 C 14 C 16 C 18 C 20 C Process is out of control at Sample 16 on the x chart. 9-2 c1385 = 26. c1130 = 64. In a short production run situation.25 64. j − µ0.00 12. is the same for each part type.25.67. c4610 = 4. j ) / σ j (where j represents the part type) would be used to calculate each plot statistic.Chapter 9 Exercise Solutions 9-3. UCL = +3. Note: In the textbook.63 4. The plot statistic is Z i = ( ci − c ) c. in standard deviation units.13 Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Standardized Total # of Defects (Ex9-4Zi) 3 UCL=3 Individual Value 2 1 _ X=0 0 -1 -2 -3 LCL=-3 4 8 12 16 20 24 Observation 28 32 36 40 Process is in control.63. 9-3 .67 26. The standardized variable ( yi . with CL = 0. c1261 = 12.00. 9-4. Set up a standardized c chart for defect counts.13 c1055 = 13.67 50. the 4th part on Day 246 should be “1385” not “1395”. LCL = −3. The chart would be designed so that δ. a standardized CUSUM could be used to detect smaller deviations from the target value. Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Rx9-4Def Rx9-4Def 1055 1130 1261 1385 4610 8611 13.67. c8611 = 50. 00 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Sam ple Group Range Control Chart 8 Range 6 Ex9-5Rmax Ex9-5RUCL 4 Ex9-5RLCL 2 0 1 3 5 7 9 11 13 15 17 19 Sam ple There is no situation where one single head gives the maximum or minimum value of x six times in a row.xls : Worksheet Ex9-5 Grand Avg = Avg R = s= n= A2 = D3 = D4 = Xbar UCL = Xbar LCL = R UCL = R LCL = 52.574 55.00 Ex9-5Xbmax Xbar 55. Excel : Workbook Chap09.Chapter 9 Exercise Solutions 9-5.00 Ex9-5XbLCL 49.00 57. so the process is out-of-control.023 0 2.596 6.00 Ex9-5XbUCL 51.379 50. 9-4 .00 47.988 2.00 Ex9-5Xbmin 53.00 45. not just a specific one.017 0. The assignable cause affects all heads.000 Group Xbar Control Chart 61. There are many values of x max and x min that are outside the control limits.00 59.338 4 heads 3 units 1. Chapter 9 Exercise Solutions 9-6. a process change may have caused output of this head to be different from the others.xls : Worksheet Ex9-6 Group Control Chart for Xbar 65 60 Xbar Ex9-5Xbmax Ex9-5Xbmin 55 Ex9-5XbUCL Ex9-5XbLCL 50 45 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 Sample Group Control Chart for Range 7 6 Range 5 Ex9-5Rmax 4 Ex9-5RUCL 3 Ex9-5RLCL 2 1 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 Sample The last four samples from Head 4 are the maximum of all heads. Excel : Workbook Chap09. 9-5 . 050 0.267 58.988 2. 6 5 Ex9-7aMRmax 4 Ex9-7aMRUCL 3 Ex9-7aMRLCL 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Sample See the discussion in Exercise 9-5.248 7.727 47.000 Group Control Chart for Individual Obs. Obs. Obs. 65 Ex9-7aXmax 60 Ex9-7aXmin 55 Ex9-7aXbUCL 50 Ex9-7aXbLCL 45 40 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Sample Group Control Chart for Moving Range 8 7 Individ.Chapter 9 Exercise Solutions 9-7.xls : Worksheet Ex9-7A Grand Avg = Avg MR = s= n= d2 = D3 = D4 = Xbar UCL = Xbar LCL = R UCL = R LCL = 52.128 0 3. 70 Individ. 9-6 . (a) Excel : Workbook Chap09.158 4 heads 2 units 1. 9-7 .988 2.158 4 heads 2 units 1.248 7. Obs. 65 Individ. indicating a potential process change. 60 Ex9-7bXmax Ex9-7bXmin 55 Ex9-7bXbUCL Ex9-7bXbLCL 50 45 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 Sample Group Control Chart for Moving Range 8 7 MR 6 5 Ex9-7bMRmax 4 Ex9-7bMRUCL 3 Ex9-7bMRLCL 2 1 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 Sample The last four samples from Head 4 remain the maximum of all heads.050 0.Chapter 9 Exercise Solutions 9-7 continued (b) Excel : Workbook Chap09.267 58.128 0 3.xls : Worksheet Ex9-7b Grand Avg = Avg MR = s= n= d2 = D3 = D4 = Xbar UCL = Xbar LCL = R UCL = R LCL = 52.727 47.000 Group Control Chart for Individual Obs. Chapter 9 Exercise Solutions 9-7 continued (c) Stat > Control Charts > Variables Charts for Subgroups > Xbar-S Chart Note: Use “Sbar” as the method for estimating standard deviation.159 Sample M ean 56 54 _ _ X=52.. Ex9-7X4) U C L=56..948 2 1 0 LC L=0 2 4 6 8 10 Sample 12 14 16 18 20 Failure to recognize the multiple stream nature of the process had led to control charts that fail to identify the out-of-control conditions in this process.415 Sample StDev 4 3 _ S =1. 9-8 .816 2 4 6 8 10 Sample 12 14 16 18 20 U C L=4..988 52 50 LC L=49. Xbar-S Chart of Head Measurements (Ex9-7X1. . Xbar-S Chart of Head Measurements (Ex9-7X1..988 52 50 LC L=49. .948 1.00 standard deviations from center line... Ex9-7X4) U C L=56.0 LC L=0 3 6 9 12 15 Sample 18 21 24 27 30 Test Results for S Chart of Ex9-7X1.8 27 1 30 1 U C L=4. . Test Failed at points: 27.. One point more than 3. 29 Only the S chart gives any indication of out-of-control process.415 3.6 2.. Ex9-7X4 TEST 1..Chapter 9 Exercise Solutions 9-7 continued (d) Stat > Control Charts > Variables Charts for Subgroups > Xbar-S Chart Note: Use “Sbar” as the method for estimating standard deviation.4 _ S =1. 9-9 .816 3 6 9 12 15 Sample 18 21 24 Sample StDev 4.159 Sample M ean 56 54 _ _ X=52.2 0. Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex9-8Xbar.552 − 0.55025.00227 / 2..004 0.002 0.00227 0. σˆ = R / d 2 = 0. Ex9-8R Variable Ex9-8Xbar Ex9-8R Mean 0..83 PCR Stat > Control Charts > Variables Charts for Subgroups > R Chart R Chart of Range Values ( Ex9-8R.001 LCL=0 0. .55025 0.Chapter 9 Exercise Solutions 9-8.002270 n=5 x = 0.326 = 0.000976)] = 6.00227. Ex9-8Rdum4) 0.004800 Sample Range 0.548 ) [6(0.000976 n = ( USL-LSL ) 6σˆ = ( 0. 9-10 .003 _ R=0.. R = 0.005 UCL=0.000 2 4 6 8 10 12 Sample 14 16 18 20 The process variability. as shown on the R chart is in control. Chapter 9 Exercise Solutions 9-8 continued (a) 3-sigma limits: δ = 0.01, Zδ = Z 0.01 = 2.33 ( LCL = LSL + ( Z UCL = USL − Zδ − 3 δ −3 ) ( n ) σˆ = (0.550 − 0.020) + ( 2.33 − 3 n σˆ = (0.550 + 0.020) − 2.33 − 3 ) 20 ) (0.000976) = 0.5316 20 (0.000976) = 0.5684 Graph > Time Series Plot > Simple Note: Reference lines have been used set to the control limit values. Control Chart of Xbar Values (Ex9-8Xbar) 0.57 UCL = 0.5684 Ex9-8Xbar 0.56 0.55 0.54 LCL = 0.5316 0.53 2 4 6 8 10 12 Ex9-8Samp 14 16 18 20 The process mean falls within the limits that define 1% fraction nonconforming. Notice that the control chart does not have a centerline. Since this type of control scheme allows the process mean to vary over the interval—with the assumption that the overall process performance is not appreciably affected—a centerline is not needed. 9-11 Chapter 9 Exercise Solutions 9-8 continued (b) γ = 0.01, Zγ = Z 0.01 = 2.33 1 − β = 0.90, Z β = z0.10 = 1.28 ( LCL = LSL + ( Z UCL = USL − Zγ + Z β γ + Zβ ) ( n ) σˆ = (0.550 − 0.020) + ( 2.33 + 1.28 n σˆ = (0.550 + 0.020) − 2.33 + 1.28 ) 20 ) (0.000976) = 0.5326 20 (0.000976) = 0.5674 Chart control limits for part (b) are slightly narrower than for part (a). Graph > Time Series Plot > Simple Note: Reference lines have been used set to the control limit values. Control Chart of Xbar Values (Ex9-8Xbar) 0.57 UCL = 0.5674 Ex9-8Xbar 0.56 0.55 0.54 LCL = 0.5326 0.53 2 4 6 8 10 12 Ex9-8Samp 14 16 18 20 The process mean falls within the limits defined by 0.90 probability of detecting a 1% fraction nonconforming. 9-12 Chapter 9 Exercise Solutions 9-9. (a) 3-sigma limits: n = 5, δ = 0.001, Zδ = Z 0.001 = 3.090 USL = 40 + 8 = 48, LSL = 40 − 8 = 32 ( UCL = USL − Zδ − 3 ( ) ( n σ = 48 − 3.090 − 3 ) ( n σ = 32 + 3.090 − 3 LCL = LSL+ Zδ − 3 ) 5 (2.0) = 44.503 ) 5 (2.0) = 35.497 Graph > Time Series Plot > Simple Note: Reference lines have been used set to the control limit values. Modified Control Chart of Xbar Values (Ex9-9Xbar) 3-sigma Control Limits 45.0 UCL = 44.5 Ex9-9Xbar 42.5 40 40.0 37.5 LCL = 35.5 35.0 2 4 6 8 10 12 Ex9-9Samp 14 16 18 20 Process is out of control at sample #6. 9-13 Chapter 9 Exercise Solutions 9-9 continued (b) 2-sigma limits: UCL = USL − Zδ − 2 ( ( LCL = LSL+ Zδ − 2 ) ( n σ = 48 − 3.090 − 2 ) ( n σ = 32 + 3.090 − 2 ) 5 (2.0) = 43.609 ) 5 (2.0) = 36.391 Graph > Time Series Plot > Simple Note: Reference lines have been used set to the control limit values. Modified Control Chart of Xbar Values (Ex9-9Xbar) 2-sigma Control Limits 45 44 UCL = 43.61 43 Ex9-9Xbar 42 41 40 40 39 38 37 LCL = 36.39 36 2 4 6 8 10 12 Ex9-9Samp 14 16 18 20 With 3-sigma limits, sample #6 exceeds the UCL, while with 2-sigma limits both samples #6 and #10 exceed the UCL. 9-14 Chapter 9 Exercise Solutions 9-9 continued (c) γ = 0.05, Zγ = Z 0.05 = 1.645 1 − β = 0.95, Z β = Z 0.05 = 1.645 ( LCL = LSL + ( Z UCL = USL − Zγ + zβ γ + zβ 5 ) (2.0) = 43.239 ) ( n ) σ = 32 + (1.645 + 1.645 5 ) (2.0) = 36.761 n σ = 48 − 1.645 + 1.645 Graph > Time Series Plot > Simple Note: Reference lines have been used set to the control limit values. Acceptance Control Chart of Xbar Values (Ex9-9Xbar) 45 44 UCL = 43.24 43 Ex9-9Xbar 42 41 40 40 39 38 37 LCL = 36.76 36 2 4 6 8 10 12 Ex9-9Samp 14 16 18 20 Sample #18 also signals an out-of-control condition. 9-15 Chapter 9 Exercise Solutions 9-10. Design an acceptance control chart. Accept in-control fraction nonconforming = 0.1% → δ = 0.001, Zδ = Z 0.001 = 3.090 with probability 1 − α = 0.95 → α = 0.05, Zα = Z 0.05 = 1.645 Reject at fraction nonconforming = 2% → γ = 0.02, Zγ = Z 0.02 = 2.054 with probability 1 − β = 0.90 → β = 0.10, Z β = Z 0.10 = 1.282 2 ⎛ Z + Z β ⎞ ⎛ 1.645 + 1.282 ⎞ 2 n=⎜ α = = 7.98 ≈ 8 ⎜ Z − Z ⎟⎟ ⎜⎝ 3.090 − 2.054 ⎟⎠ δ γ ⎝ ⎠ ( LCL = LSL + ( Z UCL = USL − Zγ + Z β γ + Zβ 8 ) σ = USL − 2.507σ ) ( n ) σ = LSL + ( 2.054 + 1.282 8 ) σ = LSL + 2.507σ n σ = USL − 2.054 + 1.282 9-16 Chapter 9 Exercise Solutions 9-11. µ = 0, σ = 1.0, n = 5, δ = 0.00135, Zδ = Z0.00135 = 3.00 For 3-sigma limits, Zα = 3 ( UCL = USL − zδ − zα ) ( n σ = USL − 3.000 − 3 ) 5 (1.0) = USL − 1.658 ⎛ USL − 1.658 − µ0 ⎞ ⎛ UCL − µ0 ⎞ Pr{Accept} = Pr{x < UCL} = Φ ⎜ = Φ ⎜ ⎟ = Φ (∆ − 1.658) 5 ⎟ ⎜ ⎟ σ n 1.0 5 ⎝ ⎠ ⎝ ⎠ where ∆ = USL − µ0 ( ( For 2-sigma limits, Zα = 2 ⇒ Pr{Accept} = Φ (∆ − 2.106) 5 ) ) ⎛ USL − µ0 ⎞ p = Pr{x > USL} = 1 − Pr{x ≤ USL} = 1 − Φ ⎜ ⎟ = 1 − Φ (∆ ) σ ⎝ ⎠ Excel : Workbook Chap09.xls : Worksheet Ex9-11 DELTA=USL-mu0 3.50 3.25 3.00 2.50 2.25 2.00 1.75 1.50 1.00 0.50 0.25 0.00 CumNorm(DELTA) 0.9998 0.9994 0.9987 0.9938 0.9878 0.9772 0.9599 0.9332 0.8413 0.6915 0.5987 0.5000 p 0.0002 0.0006 0.0013 0.0062 0.0122 0.0228 0.0401 0.0668 0.1587 0.3085 0.4013 0.5000 Pr(Accept@3) 1.0000 0.9998 0.9987 0.9701 0.9072 0.7778 0.5815 0.3619 0.0706 0.0048 0.0008 0.0001 Pr(Accept@2) 0.9991 0.9947 0.9772 0.8108 0.6263 0.4063 0.2130 0.0877 0.0067 0.0002 0.0000 0.0000 Operating Curves Pr{Acceptance} 1.0000 0.8000 0.6000 0.4000 0.2000 0.0000 0.0000 0.1000 0.2000 0.3000 0.4000 0.5000 Fraction Defective, p Pr(Accept@3) Pr(Accept@2) 9-17 Chapter 9 Exercise Solutions 9-12. Design a modified control chart. n = 8, USL = 8.01, LSL = 7.99, S = 0.001 δ = 0.00135, Zδ = Z0.00135 = 3.000 For 3-sigma control limits, Zα = 3 ( UCL = USL − Zδ − Zα ( LCL = LSL+ Zδ − Zα ) ( n σ = 8.01 − 3.000 − 3 ) ( n σ = 7.99 + 3.000 − 3 ) 8 (0.001) = 8.008 ) 8 (0.001) = 7.992 9-13. Design a modified control chart. n = 4, USL = 70, LSL = 30, S = 4 δ = 0.01, Zδ = 2.326 1 − α = 0.995, α = 0.005, Zα = 2.576 ( LCL = LSL + ( Z UCL = USL − Zδ − Zα δ − Zα 4 ) (4) = 65.848 ) ( n ) σ = (50 − 20) + ( 2.326 − 2.576 4 ) (4) = 34.152 n σ = (50 + 20) − 2.326 − 2.576 9-14. Design a modified control chart. n = 4, USL = 820, LSL = 780, S = 4 δ = 0.01, Zδ = 2.326 1 − α = 0.90, α = 0.10, Zα = 1.282 ( LCL = LSL + ( Z UCL = USL − Zδ − Zα δ − Zα 4 ) (4) = 813.26 ) ( n ) σ = (800 − 20) + ( 2.326 − 1.282 4 ) (4) = 786.74 n σ = (800 + 20) − 2.326 − 1.282 9-18 Chapter 9 Exercise Solutions 9-15. n = 4, R = 8.236, x = 620.00 (a) σˆ x = R d 2 = 8.236 2.059 = 4.000 (b) pˆ = Pr{x < LSL} + Pr{x > USL} = Pr{x < 595} + [1 − Pr{x ≤ 625}] ⎛ 595 − 620 ⎞ ⎡ ⎛ 625 − 620 ⎞ ⎤ = Φ⎜ ⎟ + ⎢1 − Φ ⎜ ⎟⎥ ⎝ 4.000 ⎠ ⎣ ⎝ 4.000 ⎠ ⎦ = 0.0000 + [1 − 0.8944] = 0.1056 (c) δ = 0.005, Zδ = Z 0.005 = 2.576 α = 0.01, Zα = Z 0.01 = 2.326 ( LCL = LSL + ( Z UCL = USL − Zδ − Zα δ − Zα 4 ) 4 = 619.35 ) ( n ) σ = 595 + ( 2.576 − 2.326 4 ) 4 = 600.65 n σ = 625 − 2.576 − 2.326 9-19 48 … The decaying sine wave of the ACFs combined with a spike at lag 1 for the PACFs suggests an autoregressive process of order 1.29 -0. Note: In the textbook.373245 0.658253 0.110802 -0.0 2 4 6 8 10 Lag 12 14 16 18 Partial Autocorrelation Function: Ex9-16mole Lag 1 2 3 4 5 PACF 0.2 0.105969 0.039599 T 5. 9-20 .70 -0.2 -0.072562 -0.29 … Stat > Time Series > Partial Autocorrelation rtial Autocorrelation Function for Molecular Weight Measurements (Ex9-16mo (with 5% significance limits for the partial autocorrelations) 1.055640 T 5.42 -0.2 -0.4 0.Chapter 9 Exercise Solutions 9-16.6 -0.2 0.30 0.033132 -0.0 Partial Autocorrelation 0.8 Autocorrelation 0.84 48.6 0. the 5th column.4 -0.96 -0.16 49.0 -0.4 0.23 LBQ 33.74 49.70 2.6 -0.37 1.92 0.6 0. (a) Stat > Time Series > Autocorrelation Autocorrelation Function for Molecular Weight Measurements (Ex9-16mole) (with 5% significance limits for the autocorrelations) 1.8 -1.81 44.4 -0.0 0.8 0.658253 -0.220536 0. the 5th row should be “2000” not “2006”.0 -0.8 -1.0 2 4 6 8 10 Lag 12 14 16 18 Autocorrelation Function: Ex9-16mole Lag 1 2 3 4 5 ACF 0. AR(1). 13. 14. 37 The process is out of control on the x chart. 9. 34. 13. Failed at points: 8. 53 5. 13. Failed at points: 12. 15. 8. Failed at points: 7. 8 points in a row more than 1 standard deviation from center line (above and below CL).9 2000 6 1975 6 2 1950 2 5 2 1 1 1 1 7 1 1 LCL=1953. 32. 37 8. 35. 12. 33. 36. UCL = 2049. Failed at points: 7. Failed at points: 6. 11. 7. 12. One point more than 3. 6 points in a row all increasing or all decreasing. 9-21 . with big swings and very few observations actually near the mean.97 1. 16. 15 3. 8. 69 2. LCL = 1953 σˆ = MR d 2 = 17. 31. 10. Failed at points: 12. 11.Chapter 9 Exercise Solutions 9-16 continued (b) x chart: CL = 2001. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). 14.93 Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Molecular Weight Measurements (Ex9-16mole) 1 1 2050 UCL=2048.128 = 15. 32. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL).1 1 1 14 21 28 35 42 49 Observation 56 63 70 Test Results for I Chart of Ex9-16mole TEST Test TEST Test TEST Test TEST Test TEST Test TEST Test 1. 35. 40. 70 6. 14. violating many runs tests.00 standard deviations from center line. 9 points in a row on same side of center line. 13. 36. 12. 14.7 8 6 6 6 6 6 2025 Individual Value 3 _ X=2000. 15. There are no other violations of special cause tests.21 7.6979 0.7 0 -25 -50 LCL=-59.1 0.843 2 35687.3 0.4 -75 1 7 14 21 28 35 42 49 Observation 56 63 70 Test Results for I Chart of Ex9-16res TEST 1.19 Constant 605.550 900.756 3 32083.364 256.693 613.942 1 41717.400 1200.4 0.998 6 30897.693 4 30929.196 2. Test Failed at points: 16 Observation 16 signals out of control above the upper limit.0010 Final Estimates of Parameters Type Coef SE Coef T AR 1 0.698 605.Chapter 9 Exercise Solutions 9-16 continued (c) Stat > Time Series > ARIMA ARIMA Model: Ex9-16mole Estimates at each iteration Iteration SSE Parameters 0 50173.7 0.197 5 30898.494 8 30897.9 0.6 0.00 standard deviations from center line.1 0.0 50 Individual Value 25 _ X=-0.000 0. One point more than 3.1 0.196 Relative change in each estimate less than 0.0852 8.82… P 0.956 7 30897. 9-22 .250 1500.02 Mean 2003.100 1800.000 Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Residuals from Molecular Weight Model (Ex9-16res) 1 UCL=58.0 0.698 605.697 606.675 650. 9 Cumulative Sum 50 0 0 -50 LCL=-97. 9-23 .5. k = 0. Let µ0 = 0. h = 5. k = 0.Chapter 9 Exercise Solutions 9-17. The residuals are in control. Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Residuals from Molecular Weight Model (Ex9-16res) mu0 = 0.9 -100 1 7 14 21 28 35 42 Sample 49 56 63 70 No observations exceed the control limit. δ = 1 sigma.5. h = 5 100 UCL=97. L = 2.42 10 EWMA 5 _ _ X=-0.71 0 -5 -10 -2.1.Chapter 9 Exercise Solutions 9-18.7 (approximately the same as a CUSUM with k = 0.83 -15 1 7 14 21 28 35 42 Sample 49 56 63 70 Process is in control.7 +2.1 and L = 2.5 and h = 5). Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Residuals from Molecular Weight Model (Ex9-16res) lambda = 0. Let λ = 0.7SL=11. 9-24 .7SL=-12. 654 CL UCL 2000.947 2044.806 1968.046 1946.304 2017.244 No 1898.731 1984.0762 = 0.533 2031.930 1988.806 1968.1181 0.65 Constant -0.xls : Worksheet Ex9-19 t xt 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 2048 2025 2017 1995 1983 1943 1940 1947 1972 1983 1935 1948 1966 1954 1970 2039 zt 2000.0762 0.415 2026.Chapter 9 Exercise Solutions 9-19.212 No 1890.046 1946.057 1982.859 2029.816 1986.128 = 15.1) (= EWMA = IMA(1.000 2100.780 No 1899.376 2002.282 1964.574 1954.920 No 1948.582 1947.467 1946.842 2033.384 1995.930 λ = 1 – MA1 = 1 – 0.502 1970.467 1946.502 1970.282 1964.217 2074.000 2000.014 1938.1)).000 2050. To find the optimal λ.677 No 1969. CL UCL LCL xt Observation 6 exceeds the upper control limit compared to one out-of-control signal at observation 16 on the Individuals control chart.749 2092.722 1996.128 1940.004 No 1921.014 1938.128 1940.582 1947.93 Excel : Workbook Chap09.09 … P 0.9238 σˆ = MR d 2 = 17.644 LCL OOC? 1953.211 2. fit an ARIMA (0.145 No 1996. No.326 No 1892. Molecular Weight EWMA Moving Center-Line Control Chart for Molecular Weight 2150.000 1900.574 1954.393 -0.731 1984.000 1850. Stat > Time Series > ARIMA ARIMA Model: Ex9-16mole … Final Estimates of Parameters Type Coef SE Coef T MA 1 0.722 1996.000 1800.269 1994. 9-25 .281 2065.057 1982.608 2016.479 2017.524 2044.842 2048.97 1.848 1993.1.929 No 1936.000 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 Obs.613 No 1978.479 2017.000 1950.415 2026.665 No 1898.772 No 1907.947 2044.521 0.700 No 1922.255 No 1934.000 1750.084 2012.040 above UCL … Xt.480 No 1916. 03 150.2 -0.37 3.6 -0.004498 -0. AR(1).0 2 4 6 8 10 12 14 16 18 20 22 24 Lag Autocorrelation Function: Ex9-20conc Lag 1 2 3 4 5 ACF 0.6 0.0 2 4 6 8 10 12 14 16 18 20 22 24 Lag Partial Autocorrelation Function: Ex9-20conc Lag 1 2 3 4 5 PACF 0.2 0.04 -0.05 2.86 144.36 99.0 0.0 -0.8 -1.77 -0.4 0.177336 -0.4 -0.46 4.12 … Stat > Time Series > Partial Autocorrelation Partial Autocorrelation Function for Concentration Readings (Ex9-20conc) (with 5% significance limits for the partial autocorrelations) 1.6 -0.4 -0.58 … The decaying sine wave of the ACFs combined with a spike at lag 1 for the PACFs suggests an autoregressive process of order 1.95 -1.8 0.10 1.6 0.746174 0.8 Autocorrelation 0.390108 0.0 -0.095134 -0.635375 0.8 -1.2 -0.4 0.746174 0.0 Partial Autocorrelation 0. 9-26 .238198 T 7.520417 0.38 127.23 LBQ 57.2 0.Chapter 9 Exercise Solutions 9-20 (a) Stat > Time Series > Autocorrelation Autocorrelation Function for Concentration Readings (Ex9-20conc) (with 5% significance limits for the autocorrelations) 1.46 1.158358 T 7. 95. 29. 95 TEST 2.128 = 3. 37. 44. 95 TEST 6. 40. 9 points in a row on same side of center line. 42. 66. 38. 40. 12. 17. 65. 99 TEST 8. 22. 21. 44 The process is out of control on the x chart. 39. 93. 30. 20. 14. 28. One point more than 3. Test Failed at points: 10. with big swings and very few observations actually near the mean. 8 points in a row more than 1 standard deviation from center line (above and below CL). 37. Test Failed at points: 15. 15. 43. 89. 42. 43. 89. 96. 41. 88. 88.01 11 1 11 30 40 50 60 Observation 70 80 90 100 Test Results for I Chart of Ex9-20conc TEST 1. 13. 68. 9-27 . 68. 43.Chapter 9 Exercise Solutions 9-20 continued (b) σˆ = MR d 2 = 3. 97.34 1 20 _ X=200. 43. 94. 21. 94. 88. 99. Test Failed at points: 11. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). 34. 22. 40. 36. 36.68 6 200 2 195 6 62 5 190 1 1 1 10 6 2 22 2 2 2 2 2 2 2 55 2 6 66 LCL=190. 36. 66. 34. 41. 38. 73. 39. 72. 100 TEST 5. 37. 89. 69. 38. 69.00 standard deviations from center line. violating many runs tests. 23. Test Failed at points: 15. 41. 86. 42. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). 71.64 1. 18. 21. Test Failed at points: 8. 93. 19. 12. 87. 44. 42. 98. 86. 41. 16. 39. 94.227 Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Concentration Readings (Ex9-20conc) 215 1 1 1 1 210 5 Individual Value 5 6 205 1 1 22 52 2 11 UCL=209. 29. 10. 122 1. 9-28 . however this is not unlikely for a dataset of 100 observations. 14 points in a row alternating up and down.1734 0. Consider the process to be in control.4155 120.Chapter 9 Exercise Solutions 9-20 continued (c) Stat > Time Series > ARIMA ARIMA Model: Ex9-20conc … Final Estimates of Parameters Type Coef SE Coef T AR 1 0.7493 0.000 Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Residuals from Concentration Model (Ex9-20res) 15 UCL=13.0669 11. Test Failed at points: 29 Observation 29 signals out of control for test 4.000 0.62 Individual Value 10 5 4 _ X=-0.76 Mean 200.73 -15 1 10 20 30 40 50 60 Observation 70 80 90 100 Test Results for I Chart of Ex9-20res TEST 4.05 0 -5 -10 LCL=-13.657 … P 0.20 Constant 50. 4 0.6 0.0 -0.8 -1.4 -0.6 0.2 -0.0 2 4 6 8 10 12 14 Lag 16 18 20 22 24 Stat > Time Series > Partial Autocorrelation tial Autocorrelation Function for Residuals from Concentration Model (Ex9-20r (with 5% significance limits for the partial autocorrelations) 1.8 Autocorrelation 0.2 0.0 0.2 -0.6 -0.2 0.6 -0.8 0.4 0.0 2 4 6 8 10 12 14 Lag 16 18 20 22 24 9-29 .0 -0.4 -0.8 -1.0 Partial Autocorrelation 0.Chapter 9 Exercise Solutions 9-20 continued (d) Stat > Time Series > Autocorrelation Autocorrelation Function for Residuals from Concentration Model (Ex9-20res) (with 5% significance limits for the autocorrelations) 1. 05075 4.Chapter 9 Exercise Solutions 9-20 (d) continued Stat > Basic Statistics > Normality Test Probability Plot of Residuals from Concentration Model (Ex9-20res) Normal 99.343 80 70 60 50 40 30 20 10 5 1 0.9 Mean StDev N AD P-Value 99 95 Percent 90 -0. PACF and normal probability plot indicates that the residuals are normal and uncorrelated.133 100 0.1 -15 -10 -5 0 Ex9-20res 5 10 Visual examination of the ACF.407 0. 9-30 . 5. k = 0.79 Cumulative Sum 20 10 0 0 -10 -20 LCL=-22. h = 5 UCL=22. k = 0. Let µ0 = 0.79 1 10 20 30 40 50 60 Sample 70 80 90 100 No observations exceed the control limit. δ = 1 sigma.Chapter 9 Exercise Solutions 9-21. Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Residuals from Concentration Model (Ex9-20res) mu0 = 0. and the AR(1) model for concentration should be a good fit. h = 5.5. The residuals are in control. 9-31 . 7SL=2. 9-32 .7SL=-2.1.7 (approximately the same as a CUSUM with k = 0. Let λ = 0.1 and L = 2.773 2 EWMA 1 _ _ X=-0. Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Residuals from Concentration Model (Ex9-20res) lambda = 0.5 and h = 5). L = 2.051 0 -1 -2 -2. The residuals are in control.874 -3 1 10 20 30 40 50 60 Sample 70 80 90 100 No observations exceed the control limit.Chapter 9 Exercise Solutions 9-22.7 3 +2. xls : Worksheet Ex9-23 lamda = 0.2945 0.139 193. Concentration EWMA Moving Center-Line Chart for Concentration 230 220 210 200 190 180 170 160 150 1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97 Obs. 56. To find the optimal λ.010 202.691 212.243 201.825 202. fit an ARIMA (0.366 201.813 198.418 200.706 sigma^ = t xt 0 1 2 3 4 5 6 7 8 9 10 204 202 201 202 197 201 198 188 195 189 zt 200.02 Constant -0.544 190.882 λ = 1 – MA1 = 1 – 0.820 203.432 3.099 209.562 191.494 208.Chapter 9 Exercise Solutions 9-23.458 184.7055 σˆ = MR d 2 = 3. 9-33 .341 200.924 211.132 188. No. 90).685 192.1) (= EWMA = IMA(1.2945 = 0.0452 0.0975 3.3034 -0.239 198.003 0.558 188.506 211.813 198.863 UCL = LCL = OOC? 209.144 192.010 202.863 190.139 193.737 190.329 193. Stat > Time Series > ARIMA ARIMA Model: Ex9-20conc … Final Estimates of Parameters Type Coef SE Coef T MA 1 0.23 CL 200.979 below LCL 181.047 211.660 191.128 = 3.1.64 1.239 198.182 0 0 0 0 0 0 0 0 0 … Xt.920 208.1)). xt CL UCL = LCL = The control chart of concentration data signals out of control at three observations (8.418 200.660 191.825 202.15 … P 0.366 201.227 Excel : Workbook Chap09.243 201. 43 0.4 0.8 -1.592580 0.143236 0.2 0.2 -0.8 -1.36 1.6 -0.78 -1.Chapter 9 Exercise Solutions 9-24.94 170.66 4.489422 0.4 -0.13… Slow decay of ACFs with sinusoidal wave indicates autoregressive process.737994 0.2 0.373763 T 8.865899 -0.078040 -0.0 -0. PACF graph suggest order 1.865899 0. 9-34 . (a) Stat > Time Series > Autocorrelation Autocorrelation Function for Temperature Measurements (Ex9-24temp) (with 5% significance limits for the autocorrelations) 1.66 -0.0 0.047106 -0.2 -0.0 -0.0 2 4 6 8 10 12 14 16 18 20 22 24 Lag Partial Autocorrelation Function: Ex9-24temp Lag 1 2 3 4 5 PACF 0.47 -1.25 133.67 3.13 2.6 0.8 Autocorrelation 0.31 211.71 LBQ 77.0 2 4 6 8 10 12 14 16 18 20 22 24 Lag Autocorrelation Function: Ex9-24temp Lag 1 2 3 4 5 ACF 0.4 -0.31… Stat > Time Series > Partial Autocorrelation Partial Autocorrelation Function for Temperature Measurements (Ex9-24temp) (with 5% significance limits for the partial autocorrelations) 1.4 0.0 Partial Autocorrelation 0.6 0.112785 T 8.6 -0.86 196.8 0. 9-35 . 36. 25. 22. 17. 25. 20. Test Failed at points: 2. 23. 27. 18. 16. … TEST 3. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). 21. 21. 33. Test Failed at points: 65. Test Failed at points: 20. 3. 16. 24. 23. 22. One point more than 3. 23. 18. violating many of the tests for special causes. … TEST 2. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). 9 points in a row on same side of center line. 23. 71 TEST 5. 21. 21.81 22 22 2 2 500 490 5 8 2 1 _ X=506. 26. Test Failed at points: 4. 22. … TEST 8. 23. 21. 4. 17. Test Failed at points: 1. 3. 24. 32. 5.23 5 11 1 1 1 1 1 1 1 470 1 1 10 20 30 40 50 60 Observation 70 80 90 1 100 Test Results for I Chart of Ex9-24temp TEST 1.Chapter 9 Exercise Solutions 9-24 continued (b) Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Temperaure Measurements (Ex9-24temp) 540 1 1 11 530 11 1 1 Individual Value 520 52 2 510 480 2 2 2 8 1 1 11 1 1 1 6 11 1 UCL=521. 22. … Process is out of control. 24. 37. 24. 8 points in a row more than 1 standard deviation from center line (above and below CL). 27. 39. 19. 18. 34. 6. 19.52 6 6 5 6 66 5 52 2 2 2 1 1 1 11 1 5 11 1 LCL=491.00 standard deviations from center line. 20. Test Failed at points: 17. 28. 26. 2. … TEST 6. 6 points in a row all increasing or all decreasing. 20. 18. The temperature measurements appear to wander over time. 24. 22. 19. 38. 19. 80 1 1 10 20 30 40 50 60 Observation 70 80 90 100 Test Results for I Chart of Ex9-24res TEST 1. One point more than 3.67 Constant 52.000 0. The residuals do not exhibit cycles in the original temperature readings.00 standard deviations from center line. and points are distributed between the control limits.22 0 -10 -20 LCL=-21. 9-36 .Chapter 9 Exercise Solutions 9-24 continued (c) Stat > Time Series > ARIMA ARIMA Model: Ex9-24temp … Final Estimates of Parameters Type Coef SE Coef T AR 1 0.0480 18.23 20 5 Individual Value 10 _ X=0.7263 72.12 Mean 503.985 … P 0.3794 0. Test Failed at points: 94 TEST 5. and observation 71 fails Test 5.727 6.8960 0. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). Test Failed at points: 71 Observation 94 signals out of control above the upper limit.000 Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Residuals from Temperature Model (Ex9-24res) UCL=22. The chemical process is in control. h = 5 40 UCL=36. 9-37 .Chapter 9 Exercise Solutions 9-25. This is the same conclusion as applying an Individuals control chart to the model residuals.69 30 Cumulative Sum 20 10 0 0 -10 -20 -30 LCL=-36. MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Residuals from Temperature Model (Ex9-24res) k = 0.69 -40 1 10 20 30 40 50 60 Sample 70 80 90 100 No observations exceed the control limits.5. The residuals are in control. indicating the process is in control. 7SL=-4.5 -2.0 1 10 20 30 40 50 60 Sample 70 80 90 100 No observations exceed the control limits.22 0. MTB > Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of Residuals from Temperature Model (Ex9-24res) lambda = 0.7 5.0 -2.1.Chapter 9 Exercise Solutions 9-26.76 EWMA 2.5 _ _ X=0. This is the same conclusion as applying the Individuals and CUSUM control charts to the model residuals. 9-38 .33 -5. indicating the process is in control.0 +2. L = 2.7SL=4. The residuals are in control. 037 CL 5. 94). 69) and the lower limit (1.940 467. fit an ARIMA (0. similar to the two out-of-control signals on the Individuals control chart (71. 9-39 .465 LCL 521.0794 = 0.560 524.921 sigma^ = zt 506.173 0 0 0 0 0 0 0 0 0 EWMA Moving Center-Line Chart for Temperature Xt.972 506.227 below LCL 476.558 498.363 503.xls : Worksheet Ex9-27 lambda = t xt 0 1 2 3 4 5 6 7 8 9 10 491 482 490 495 499 499 507 503 510 509 0.722 509.465 509.972 506.232 482. To find the optimal λ.1019 0.0711 0.917 λ = 1 – MA1 = 1 – 0.75 1.647 498.850 513.520 492.812 489.070 487.10 … P 0. 94).137 479.098 UCL 506.355 483.679 491.128 = 5.1) (= EWMA = IMA(1.267 509.520 474.265 483.0794 0.940 514.0975 (from a Moving Range chart with CL = 5.438 0.9206 σˆ = MR d 2 = 5.265 521.75) Excel : Workbook Chap09. Stat > Time Series > ARIMA ARIMA Model: Ex9-24temp … Final Estimates of Parameters Type Coef SE Coef T MA 1 0.363 503.520 492.558 498.812 489.655 518.974 494.105 504.429 494.6784 -0.758 OOC? 491.267 509.1. 58.429 494.Chapter 9 Exercise Solutions 9-27.813 507.1)). xt CL UCL LCL A few observations exceed the upper limit (46.525 498.647 498.232 482. Temperature 570 550 530 510 490 470 450 1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97 Sample No.78 Constant -0. adjacent observations will tend to be similar. (a) When the data are positively autocorrelated. Since it is difficult to determine whether a process generating autocorrelated data – or really any process – is in control. (c) If assignable causes are present.Chapter 9 Exercise Solutions 9-28. There is nothing in the derivation of the expected value of S2 = σ2 that depends on an assumption of independence. (b) S2 is still an unbiased estimator of σ2 when the data are positively autocorrelated. This would tend to produce an estimate of the process standard deviation that is too small. 9-40 . therefore making the moving ranges smaller. it is not good practice to estimate σ2 from S2. it is generally a bad practice to use S2 to estimate σ2. 071963 T 4.54 LBQ 25.49.4 0.29 41.049610 -0.8 Autocorrelation 0.94 -0.22 -0.17 -2.6 -0. 9-41 .16 25.2 0.78 41.2 -0. There is a serious problem with autocorrelation in viscosity readings.85 r1 = 0.494137 -0.41 -2.264612 -0. (a) Stat > Time Series > Autocorrelation Autocorrelation Function for Viscosity Readings (Ex9-29Vis) (with 5% significance limits for the autocorrelations) 1.8 -1.283150 -0.0 2 4 6 8 10 12 14 Lag 16 18 20 22 24 Autocorrelation Function: Ex9-29Vis Lag 1 2 3 4 5 … ACF 0.0 -0.0 0.4 -0.6 0.41 32.Chapter 9 Exercise Solutions 9-29. indicating a strong positive correlation at lag 1. Test Failed at points: 22 TEST 8. The viscosity measurements appear to wander over time.Chapter 9 Exercise Solutions 9-29 continued (b) Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Viscosity (Ex9-29Vis) 40 UCL=37.57 25 6 20 6 55 LCL=20.11 5 Individual Value 35 6 6 7 30 _ X=28.03 1 1 1 1 10 20 30 40 50 60 Observation 1 70 80 90 100 Test Results for I Chart of Ex9-29Vis TEST 1. 38. 60. 86. 92 TEST 5. Test Failed at points: 2. Test Failed at points: 64 Process is out of control. 8 points in a row more than 1 standard deviation from center line (above and below CL). 9-42 . violating many of the tests for special causes. 59.00 standard deviations from center line. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). 75 TEST 7. One point more than 3. 58. Test Failed at points: 38. 63. 64. 86 TEST 6. Test Failed at points: 40. 15 points within 1 standard deviation of center line (above and below CL). 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). 569 MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM CUSUM Chart of Viscosity (Ex9-29Vis) target = 28. k = 0. 9-43 .5.569.24 10 0 0 -10 LCL=-14.Chapter 9 Exercise Solutions 9-29 continued (c) Let target = µ0 = 28. h = 5 20 Cumulative Sum UCL=14.24 -20 1 10 20 30 40 50 60 Sample 70 80 90 100 Several observations are out of control on both the lower and upper sides. 569 29 28 27 -2.7SL=26.759 EWMA 30 _ _ X=28.Chapter 9 Exercise Solutions 9-29 continued (d) MTB > Stat > Control Charts > Time-Weighted Charts > CUSUM EWMA Chart of Ex9-29Vis lambda = 0. L = 2.7 32 31 +2.7SL=30. 9-44 .15.380 26 25 1 10 20 30 40 50 60 Sample 70 80 90 100 The process is not in control. There are wide swings in the plot points and several are beyond the control limits. 000 10.898 33.57 Constant 0.0231 0.464 18.665 26.xls : Worksheet Ex9-29 lambda = l Xi 0 1 2 3 4 5 6 7 8 9 10 … 29.750 30.022 38.05 P 0.1579) = 1.441 42.328 2.482 26.1579 0.873 39.030 32.367 20.909 29.756 24.962 λ = 1 – MA1 = 1 – (– 0.000 30.298 32.128 = 2. 37.1)).482 26.298 32.356 25.980 25.243 0 0 0 0 0 0 0 0 0 EWMA Moving Center-Line Chart for Viscosity Xt.330 19.527 26. Stat > Time Series > ARIMA ARIMA Model: Ex9-29Vis … Final Estimates of Parameters Type Coef SE Coef T MA 1 -0.1) (= EWMA = IMA(1.1579 σˆ = MR d 2 = 3.680 33.464 18. 55.21 1.922 below LCL 9.550 Zi 28.788 22.479 29. Viscosity 50.000 1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97 Obs.330 31.500 26.785 37.120 0.85 LCL OOC? 19.452 37.785 31. fit an ARIMA (0.144 1.527 26.007 27. 9-45 .909 29.841 41.75) Excel : Workbook Chap09.000 31. 85).Chapter 9 Exercise Solutions 9-29 continued (e) To find the optimal λ.122 17.937 20.940 18.000 0. No.8457 (from a Moving Range chart with CL = 5. Xi CL UCL LCL A few observations exceed the upper limit (87) and the lower limit (2.158 sigma^ = CL UCL 28.025 35.330 31.560 27.4839 0.1.207 35.000 40.760 29.984 18.069 35.1007 -1.000 20.898 33.665 26.479 29. 9-46 . Test Failed at points: 18. with a good distribution of points between the control limits and no patterns.3278 62.000 0.Chapter 9 Exercise Solutions 9-29 continued (f) Stat > Time Series > ARIMA ARIMA Model: Ex9-29Vis … Final Estimates of Parameters Type Coef SE Coef T AR 1 0. Otherwise the process appears to be in control.000 0.54 Mean 28.68 -10 1 10 20 30 40 50 60 Observation 70 80 90 100 Test Results for I Chart of Ex9-29res TEST 7.000 Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Residuals from Viscosity AR(2) Model (Ex9-29res) 10 UCL=9.5017 0.4349 0.7193 0.6514 0.60 Individual Value 5 7 7 7 _ X=-0. 21.04 0 7 7 -5 LCL=-9. 20.79 AR 2 -0. 19.4581 … P 0. 22 The model residuals signal a potential issue with viscosity around observation 20.0922 -4.72 Constant 20. 15 points within 1 standard deviation of center line (above and below CL).0923 7. 50. α = 0.0027 ⎛ ( µ + k σ n ) − ( µ0 + 2σ ) ⎞ ⎛ ( µ0 − k σ n ) − ( µ0 + 2σ ) ⎞ β = Φ ⎜⎜ 0 ⎟⎟ − Φ ⎜⎜ ⎟⎟ σ n σ n ⎝ ⎠ ⎝ ⎠ ( ) ( = Φ 3 − 2 5 − Φ −3 − 2 5 ) = Φ (−1.10/unit.50/sample. h = 1. λ = 0.Chapter 9 Exercise Solutions 9-30. D = 2hr (a) Excel : workbook Chap09. 1 − β = 0. a3 = $2.79/hr (b) n = 3. k = 3.0705 − 0. hopt = 1.0 a1 = $0.05hr/sample. a'3 = $5.210.00. a4 = $100/hr g = 0.895 E(L) = $3.231. α = 0. a2 = $0.472) = 0.4992 τ≅ − 2 12 α e−λh α ≅ = 0.027. δ = 2.472) − Φ (−7. kopt = 2.6098/hr 9-47 .01/hr or 1/λ = 100hr.0705 h λ h2 = 0.xls : worksheet Ex9-30a n = 5.0000 = 0.27 −λh λh 1− e ( ) E(L) = $3. 12/hr (b) n = 5.0705 − 0. a'3 = $50. hopt = 1.Chapter 9 Exercise Solutions 9-31. h = 0. D = 2hr (a) Excel : workbook Chap09. a4 = $100/hr g = 0. 1 − β = 0.00207.472) = 0.98/hr (c) n = 5. a2 = $0.4992 τ≅ − 2 12 2 12 α e−λh α 0. α = 0. α = 0. h = 1.368.5.01(12 ) = − = 0.52 ) = − = 0.5 0.080.01392/hr 9-48 .01/hr or 1/λ = 100hr.0000 = 0.918 E(L) = $4. λ = 0.0 a1 = $0. β = 0. k = 3.5) 1− e ( ) E(L) = $4.50/sample.0027 ⎛ ( µ + k σ n ) − ( µ + δσ ) ⎞ ⎛ ( µ − k σ n ) − ( µ + δσ ) ⎞ 0 0 ⎟−Φ⎜ 0 ⎟ β = Φ⎜ 0 ⎜ ⎟ ⎜ ⎟ σ n σ n ⎝ ⎠ ⎝ ⎠ ( ) ( ) = Φ ( 3 − 2 5 ) − Φ ( −3 − 2 5 ) = Φ k − δ n − Φ −k − δ n = Φ (−1. a3 = $25. α = 0.0705 h λ h 2 0.472) − Φ (−7.10/unit.0705 h λ h 2 1 0. δ = 2. kopt = 3. k = 3.0027 ≅ = = 0.05hr/sample.01(0.0027 α e−λh α ≅ = = 0.xls : worksheet Ex9-31 n = 5.01(1) 1− e ( ) E(L) = $4.0027.27 −λh λ h 0.2498 τ≅ − 2 12 2 12 0.01(0.54 −λh λ h 0. 8218083 E(L) = $10.52 ) = − = 0. 1 − β = 0.01(0.775 h λ h 2 0. β = 0.5 0. α = 0. hopt = 2. α = 0.05 hr/sample D = 1 hr (a) n = 5.240.xls : worksheet Ex9-32 D0 = 2hr.01/hr or 1/λ = 100hr δ = 2. α = 0.39762/hr 9-49 .17/hr (b) n = 10.0705 E(L) = $13.0027.0 a1 = $2/sample a2 = $0.0027 β = Φ k − δ n − Φ −k − δ n ( ) ( ) = Φ ( 3 − 1 5 ) − Φ ( −3 − 1 5 ) = Φ (−1.54 −λh λ h 0.025091.472) = 0. ∆ = $25 n = 5.472) − Φ (−7. D1 = 2hr V0 = $500. k = 3.xls : worksheet Ex9-33 λ = 0.2498 τ≅ − 2 12 2 12 0. Excel : workbook Chap09. h = 0.5. k = 3.16/hr 9-33. Excel : workbook Chap09. h = 1.Chapter 9 Exercise Solutions 9-32.489018.50/unit a'3 = $75 a3 = $50 a4 = $200/hr g = 0.5) 1− e ( ) E(L) = $16.0000 = 0.01(0. kopt = 2.0027 α e−λh α ≅ = = 0.775 − 0. 0015 1 2 3 4 5 6 7 8 9 10 11 12 Ex9-34Sample n = 5.0035 − 3(0.00028 5 ) = 1.00304 LCL = CL − 3σ x ( ) = 1.00228 9-50 . LCL = 0 x chart initial settings: CL = LSL + 3σ = 1.00266 − 3 ( 0.00064.00234 UCL = CL + 3σ x = 1.0035 1.00266 (maximum permissible average) UCL = CL + 3σ x = 1. It is good practice visually examine data in order to understand the type of tool wear occurring. Graph > Time Series Plot > With Groups Time Series Plot of Ex9-34Xb USL = 1.0020 LSL = 1.00272 LCL = CL − 3σ x ( ) = 1.00028 CL = R = 0.0025 1. The plot below shows that the tool has been reset to approximately the same level as initially and the rate of tool wear is approximately the same after reset.00196 x chart at tool reset: CL = USL − 3σ = 1.00064 2.Chapter 9 Exercise Solutions 9-34.0035 Ex9-34Reset A fter Before Ex9-34Xb 1.00135.00028 5 = 1.114(0.00028) = 1.326 = 0.00234 + 3 0. R = 0.0015 1.00028) = 1. UCL = D4 R = 2.00266 + 3 0.0030 1. σˆ = R d 2 = 0.0015 + 3(0.00064) = 0.00064.00234 − 3 ( 0.00028 5 = 1.00028 5 ) = 1. 5408 13. 10-1. Let α = 0.4254 10.001.00 0.7042 0.00 T^2 40.0817 2. 10-1 .00 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Sample Number Process is out of control at samples 7 and 14.mn − m − p +1 mn − m − p + 1 2(50 + 1)(25 − 1) = F0.1850 0 14.5211 0.0282 0.1850 0 14.948) = 14. p .0676 1.00 20.1850 0 14.6338 14.1690 0.1850 0 14.0423 0.1268 0.1850 0 14.Chapter 10 Exercise Solutions Note: MINITAB’s Tsquared functionality does not use summary statistics.1850 0 14.6901 0. Phase 2 T 2 control charts with m = 50 preliminary samples.1850 0 14.1690 0.186 LCL = 0 Excel : workbook Chap10.1199 50(25) − 50 − 2 + 1 = ( 2448 1199 ) (6.001.1850 0 14.9127 22.8169 0. p = 2 characteristics.1268 3.xls : worksheet Ex10-1 Sample No.00 50.0451 1.1268 3. so many of these exercises have been solved in Excel. xbar1 xbar2 1 58 32 2 60 33 3 50 27 4 54 31 5 63 38 6 53 30 7 42 20 8 55 31 9 46 25 3 2 5 3 -5 -3 -1 1 8 8 -2 0 -13 -10 0 1 -9 -5 0.00 30. p (m + 1)(n − 1) UCL = Fα .1850 0 diff1 diff2 matrix calc t2 = n * calc UCL = LCL = OOC? In control In control In control In control In control In control Above UCL In control In control … T^2 Control Chart for Quality Characteristics 60.2. n = 25 sample size.00 10. 0397 10.Chapter 10 Exercise Solutions 10-2.2 0.2 1.00 T^2 15.4 3 2.6 3 2.8084 1.6 1.4249 17.2372 2.0593 10.4 -0.2 0.5 0.4249 17.8 diff1 diff2 diff3 0.3966 1.1216 0.8525 matrix calc t2 = n * calc UCL = LCL = OOC? 17.00 10.3 2.3.5 7 3.mn − m − p +1 mn − m − p + 1 3(30 + 1)(10 − 1) = F0.3 -0.1 4.8 4.2 -0.8 6 4 4.2706 0. n = 10 sample size.4 4 2.00 0.1 3.4 0.5 -0.5 5 3 3.6 2.3 0.4249 17.4249 17.3 3.1 0.3990 0.4249 17.4249 17.1189 1.2 3.001.xls : worksheet Ex10-2 Sample No.8800 0.2 10 2 2. p = 3 characteristics.1 3 2.6 -0.0852 0.5 0.1880 1.1271 11. p (m + 1)(n − 1) UCL = Fα .1887 0.4249 17.425 LCL = 0 Excel : workbook Chap10.4 0.3 2.7 9 2.2 0 1 1.1399 1.0528 0.9 3.9 -1 0.7 4 3 13 4. p .7 0.3122 3.00 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Sample Number Process is out of control at sample 13.4249 17. Let α = 0.1 -0.1 0.6844 0.2 3.7 3.8 3 2.8692 8.2 0.0684 0.5 -0.5932 0.5 -0.00 5.9 15 3.579) ⎝ 268 ⎠ = 17.2 0.4249 17.7 0.8 11 3.8 4 2.8 0.0793 20.2 14 3.2 0.0808 0.2 0 -0.4249 17. xbar1 xbar2 xbar3 1 3.1 0 0.5741 2.2 3 8 3 3.4249 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 In control In control In control In control In control In control In control In control In control In control In control In control Above UCL In control In control T^2 Control Chart for Quality Characteristics 25.3 0 -0. Phase 2 T 2 control limits with m = 30 preliminary samples.001.5279 0.6 -1 -0.9 3 12 3.4249 17.7 3 2 3.00 20.3719 0.7927 1.2 0.4249 17.4249 17.8 0.268 30(10) − 30 − 3 + 1 ⎛ 837 ⎞ =⎜ ⎟ (5.7 0.1 1.4249 17.6922 0.6 3.2 -0.4 0. 10-2 .6574 6.4 -0.1 0. 0000 T^2 40.1268 3.001.8150 0 13.8150 0 diff1 diff2 matrix calc t2 = n * calc 3 58 32 In control In control In control In control In control In control Above UCL In control In control In control In control In control In control Above UCL In control Phase II T^2 Control Chart 60.8150 0 13.0704 0.0000 0.5408 13.0451 1.7042 0.001.2 Excel : workbook Chap10.4254 10.0000 50.9127 22.0282 0.816 control limit: UCL = χα2 . Phase 2 T 2 control limits with p = 2 characteristics.8150 0 13.Chapter 10 Exercise Solutions 10-3. Same results as for parameters estimated from samples.0000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Sample Number Process is out of control at samples 7 and 14.5211 0.0000 20.0000 30.8150 0 13.0676 1.1690 0.8150 0 13.8150 0 13.1268 0.6901 0.6901 0.8150 0 13. Let α = 0.6338 0.1690 0.0676 1.8150 0 13.8169 0. p = χ 0. Since population parameters are known.8150 0 13.8150 0 13.8150 0 13.xls : worksheet Ex10-3 Sample No.8150 0 13.0000 10. 10-3 .0761 1.9014 2.0423 0.2676 6.1127 52.8150 0 13.2535 6.2028 5. xbar1 xbar2 1 2 UCL = LCL = OOC? 4 5 6 7 8 9 10 11 12 13 14 15 60 33 50 27 54 31 63 38 53 30 42 20 55 31 46 25 50 29 49 27 57 30 58 33 75 45 55 27 3 2 5 3 -5 -3 -1 1 8 8 -2 0 -13 -10 0 1 -9 -5 -5 -1 -6 -3 2 0 3 3 20 15 0 -3 0.1268 3.8169 0.6901 0.3380 13.0817 2. the chi-square formula will be used for the upper 2 = 13.8150 0 13. Same as results for parameters estimated from samples.3 0.2660 16.1 3.001.2660 16.0593 10.2 0.2 10 2 2.8800 0.001. the chi-square formula will be used for the upper 2 = 16. xbar1 xbar2 xbar3 1 3.3 2.4 3 2.3990 0.1 -0.5932 0. Since population parameters are known.5 7 3.1880 1.5 0.2660 16.2660 16.2 0 -0.2660 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 In control In control In control In control In control In control In control In control In control In control In control In control Above UCL In control In control Phase II T^2 Control Chart 25.9 15 3.2 3.2 0.2660 16.2 14 3. 10-4 .1887 0.2 0.6922 0.8 0.1216 0.0684 0.2 3.5 -0.xls : worksheet Ex10-4 Sample No.9 -1 0.3 2.8 4 2.6 -0.1 4.00 5.1271 11.1189 1.5 5 3 3. Let α = 0.7 0.0528 0.8 0.0793 20.4 0.6 2.2660 16.2660 16.9 3.8 6 4 4.7 3.00 T^2 15.1 0.2 0 1 1.6574 6.1 0 0.3 Excel : workbook Chap10.00 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Sample Number Process is out of control at sample 13.4 -0.5279 0.7 9 2.00 20.3719 0.4 0.2660 16.6 1.00 0. p = χ 0.2660 16.1 0.5 0.8525 matrix calc t2 = n * calc UCL = LCL = OOC? 16.8084 1.6844 0.2660 16.1 1.2 1.2372 2.5 -0.2 0.2660 16.2706 0.0852 0.6 3 2.Chapter 10 Exercise Solutions 10-4.7 0.8 11 3.8 diff1 diff2 diff3 0.4 4 2.8 3 2.2660 16.1399 1.0808 0.7927 1.3 -0.2 -0.9 3 12 3.6 3.4 0.1 3 2.2 -0.8 4.8692 8.5741 2.2660 16.1 0.7 0.6 -1 -0.3 0 -0.00 10.2 0.2660 16. Phase 2 T 2 control limits with p = 3 characteristics.2 0.5 -0.3966 1.2 3 8 3 3.7 4 3 13 4.266 control limit: UCL = χα2 .7 3 2 3.3122 3.3 3.0397 10.4 -0. 531) ⎝ 55 ⎠ = 23.7331 18.531 3.xls : worksheet Ex10-5 m 30 40 50 60 70 80 90 100 num denom 372 55 492 75 612 95 732 115 852 135 972 155 1092 175 1212 195 F 3. n = 3 sample size. p = 6. n = 3.107 3. p = χ 0.7325 18. 1.548 The Phase II UCL is almost 30% larger than the chi-square limit.7332 18.8820 22. α = 0.882 LCL = 0 (b) 2 chi-square limit: UCL = χα2 .107 3.294 3.01(18.209 UCL 23.7337 18. Excel : workbook Chap10. Samples size.407 3. p = 6 characteristics.3184 20.223 3.1095 19.263 3.107 18.005.005. Find "m" such that exact Phase II limit is within 1% of chi-square limit.107 3.005 (a) Phase II limits: p (m + 1)(n − 1) UCL = Fα . 10-5 .107 3.5042 20.6.3527 21.733. α = 0.9447 717 718 719 720 721 722 8616 8628 8640 8652 8664 8676 3.548) = 18.6 = 18.7324 … 1429 1431 1433 1435 1437 1439 720 preliminary samples must be taken to ensure that the exact Phase II limit is within 1% of the chi-square limit.Chapter 10 Exercise Solutions 10-5.7328 18.mn − m − p +1 mn − m − p + 1 6(30 + 1)(3 − 1) = F0.240 3. m = 30 preliminary samples.55 30(3) − 30 − 6 + 1 ⎛ 372 ⎞ =⎜ ⎟ (3.5920 20.107 3.005. (c) Quality characteristics.9650 20. p .338 3. 294) ⎝ 115 ⎠ = 21.209 3.005.155 3. 1.105 3.4119 19.xls : worksheet Ex10-6 m 30 40 50 60 70 80 90 100 num 744 984 1224 1464 1704 1944 2184 2424 denom 115 155 195 235 275 315 355 395 F 3.7330 18.733. n = 5 sample size.8641 19.7376 18.3087 20.005 (a) Phase II UCL: p (m + 1)(n − 1) UCL = Fα .005. α = 0.mn − m − p +1 mn − m − p + 1 6(30 + 1)(5 − 1) = F0.6 = 18. n = 5.105 3.294 3.3232 390 400 410 411 412 9384 9624 9864 9888 9912 1555 1595 1635 1639 1643 3.5237 19.7424 18.1422 19.5692 20. 10-6 . m = 30 preliminary samples.240 3. Find "m" such that exact Phase II limit is within 1% of chi-square limit.149 UCL 21. (c) Quality characteristics. p = 6 characteristics.548) = 18.7324 18.005.6.106 3. Samples size.Chapter 10 Exercise Solutions 10-6.6685 19. p = 6. α = 0.01(18.548 The Phase II UCL is almost 15% larger than the chi-square limit.164 3. p . p = χ 0.174 3.115 30(5) − 30 − 6 + 1 ⎛ 744 ⎞ =⎜ ⎟ (3.105 3.7318 … 411 preliminary samples must be taken to ensure that the exact Phase II limit is within 1% of the chi-square limit.309 (b) 2 chi-square UCL: UCL = χα2 . Excel : workbook Chap10.105 18.189 3. n = 3.530 25.4398 25.005.326 (b) 2 chi-square UCL: UCL = χα2 .440. Find "m" such that exact Phase II limit is within 1% of chi-square limit.188) = 25. 10-7 .530 2.3259 34.530 2.188 The Phase II UCL is more than 55% larger than the chi-square limit.101) ⎝ 41 ⎠ = 39.41 25(3) − 25 − 10 + 1 ⎛ 520 ⎞ =⎜ ⎟ (3. p = 10.742 2.4399 25.629 UCL 39.7953 30.01(25. Samples size.677 2.530 2.10.530 2.4394 … 988 preliminary samples must be taken to ensure that the exact Phase II limit is within 1% of the chi-square limit.101 2. p .10 = 25.005 (a) Phase II UCL: p (m + 1)(n − 1) UCL = Fα .0154 27. α = 0.4405 25.xls : worksheet Ex10-7 m 25 35 45 55 65 75 85 95 105 num 520 720 920 1120 1320 1520 1720 1920 2120 denom 41 61 81 101 121 141 161 181 201 F 3.4940 28. m = 25 preliminary samples.641 2.704 2.657 2. α = 0.1991 31.Chapter 10 Exercise Solutions 10-7. p = 10 characteristics.4401 25.7246 986 987 988 989 990 19740 19760 19780 19800 19820 1963 1965 1967 1969 1971 2.3808 28. (c) Quality characteristics. Excel : workbook Chap10.005.4024 29. n = 3 sample size.897 2. p = χ 0.005. 1.8549 28.mn − m − p +1 mn − m − p + 1 10(25 + 1)(3 − 1) = F0.799 2. 4405 25.4141 3040 291 2. n = 5 sample size.188) = 25.529 25.4399 25.578 26.594 27.529 2.4408 25. α = 0.8495 2640 251 2.005.10.585 26.188 The Phase II UCL is more than 25% larger than the chi-square limit.572 26.005. p .xls : worksheet Ex10-8 m 25 35 45 55 65 75 85 95 105 num denom F UCL 1040 91 2.648 28. α = 0.689 29.440.005.767 31.01(25. p = 10 characteristics.91 25(5) − 25 − 10 + 1 ⎛ 1040 ⎞ =⎜ ⎟ (2. (c) Quality characteristics.8651 3840 371 2.10 = 25.529 2.529 2. n = 5.5595 1840 171 2.529 2. p = χ 0. Find "m" such that exact Phase II limit is within 1% of chi-square limit. Samples size.6812 4240 411 2.767) ⎝ 91 ⎠ = 31. Excel : workbook Chap10.4413 25.1011 3440 331 2.Chapter 10 Exercise Solutions 10-8. 10-8 .4394 544 preliminary samples must be taken to ensure that the exact Phase II limit is within 1% of the chi-square limit.4967 2240 211 2.6251 1440 131 2.529 2. 1.625 (b) 2 chi-square UCL: UCL = χα2 .606 27.623 27.mn − m − p +1 mn − m − p + 1 10(25 + 1)(5 − 1) = F0.4419 25. m = 25 preliminary samples. p = 10.005 (a) Phase II UCL: p (m + 1)(n − 1) UCL = Fα .5335 540 541 542 543 544 545 21640 21680 21720 21760 21800 21840 2151 2155 2159 2163 2167 2171 2. 41 25(3) − 25 − 10 + 1 ⎛ 520 ⎞ =⎜ ⎟ (2. p = χ 0.7 0.788) ⎝ 41 ⎠ = 35.xls : worksheet Ex10-10 (a) ⎡ 1 0.7 0.01. n = 3 sample size. p .7 ⎥ ⎥ Σ=⎢ ⎢0.7 ⎤ ⎢0.41 25(3) − 25 − 10 + 1 ⎛ 480 ⎞ =⎜ ⎟ (2.7 0.Chapter 10 Exercise Solutions 10-9.01. Excel : workbook Chap10.7 0.mn − m − p +1 mn − m − p + 1 10(25 − 1)(3 − 1) = F0.277 10-9 .7 0.10.7 0.01. p = 10 quality characteristics. p .638 Phase II UCL: p (m + 1)(n − 1) UCL = Fα . Phase I UCL: p (m − 1)(n − 1) UCL = Fα .4 = 13.360 10-10.7 1 ⎦ (b) 2 UCL = χα2 .7 1 0.788) ⎝ 41 ⎠ = 32.10. Assume α = 0.7 ⎥ ⎢ ⎥ ⎣0.01.7 1 0.mn − m − p +1 mn − m − p + 1 10(25 + 1)(3 − 1) = F0. m = 25 preliminary samples. an out-of-control signal is generated.280) > (UCL = 13.7 1 ⎥ ⎜ ⎢3. since the standardized observations are equal (that is.7 1 0.7 1 ⎦ ⎜⎝ ⎣3.806 Yes. (e) Since (T 2 = 28. (d) −1 T(1)2 = n ( y (1) .5⎤ ⎡0 ⎤ ⎞′ ⎡ 1 0.5 0 0.01.7 1 0.7 0.5⎥ ⎢ 0⎥ ⎟ ⎥ ⎜⎢ ⎥ ⎢ ⎥⎟ 0.7 0. d 2 = 6.689 Investigate variables 1 and 3.7 ⎥⎥ ⎜ ⎢⎢3.7 ⎥ ⎜ ⎢3.806 − 15.µ ) ⎛ ⎡3.313 2 2 2 2 T(1) = T(2) = T(3) = T(4) = 15.7 ⎜⎜ ⎢ ⎥ ⎢ ⎥ ⎟⎟ ⎢ ⎝ ⎣3.Chapter 10 Exercise Solutions 10-10 continued (c) T 2 = n ( y . all variables had the same shift).5⎤ ⎡ 0⎤ ⎞ ⎜ ⎟ ⎜ ⎟ = 1⎜ ⎢⎢3. Since T 2 = 15.7 ⎤ −1 ⎛ ⎡3.313 di = T 2 − T(2i ) d1 = d 2 = d3 = d 4 = 15.01.800 2 T(4) = 25.1 .7 0. an out-of-control signal is generated.300 T(3)2 = 14.5⎦ ⎣ 0 ⎦ ⎠ ⎣ 0. Second.5⎦ ⎣ 0⎦ ⎟⎠ ) > ( UCL = 13.5⎥⎥ ⎢⎢ 0⎥⎥ ⎟ − 0. d 4 = 2.µ (1) )′ Σ (1) ( y (1) . since all di are smaller than χ 0.5⎤ ⎡ 0⎤ ⎞ ⎜ ⎟ 1 0. d3 = 13.01.277) .5⎥ ⎢ 0⎥ ⎟ ⎦ ⎝⎣ ⎦ ⎣ ⎦⎠ ⎝⎣ ⎦ ⎣ ⎦⎠ ⎣ = 15.806 ( 0.µ (1) ) ⎛ ⎡3.585 2 T(2) = 21.635 2 No.635 T(1)2 = 15. First.1 = 6.µ )′ Σ −1 ( y .277 ) .7 = 15.5⎥⎥ − ⎢⎢ 0⎥⎥ ⎟ ⎜ ⎢3.493 2 χ 0.313 = 0.979.1 = 6. (f) 2 χ 0.7 0.7 = 1⎜ ⎢ ⎥ − ⎢ ⎥ ⎟ ⎢ ⎜ ⎢3.7 0.5⎤ ⎡ 0 ⎤ ⎞′ ⎡ 1 ⎜⎢ ⎥ ⎢ ⎥⎟ ⎢ 3.5⎥ ⎢0 ⎥ ⎟ ⎢ 0. d1 = 12. 10-10 .7 ⎥⎥ ⎜ ⎢⎢3.590. no variable is identified as a relatively large contributor.7 0. this information does not assist in identifying which a process variable shifted.694.5⎥⎥ − ⎢⎢0 ⎥⎥ ⎟ ⎢⎢ 0.7 ⎤ −1 ⎛ ⎡3.5⎥ ⎢ 0 ⎥ ⎟ ⎢ 0.479. d3 = 2. ( ) (d) 2 χ 0. p = χ 0. d 2 = 8.538 T(3)2 = 4.841 T(1)2 = 11.000. This is confirmed since none of the di’s exceeds the UCL.043 2 T(2) = 2. d3 = 6.815) .3 = 7.538 2 T(2) = 5.094 Since an out-of-control signal was not generated in (e).154 Yes.154 Variables 2 and 3 should be investigated. d1 = 1. Since T 2 = 11.444.xls : worksheet Ex10-11 (a) ⎡ 1 0.Chapter 10 Exercise Solutions 10-11. d1 = 0.538) > (UCL = 7.05. an out-of-control signal is generated.1 = 3.05. an out-of-control signal is not generated.8 0.841 T(1)2 = 5.000. Excel : workbook Chap10.154 > ( UCL = 7.8⎥⎥ ⎢⎣ 0.1 = 3.111.8⎤ Σ = ⎢⎢ 0. 10-11 .8 1 0.815 (c) T 2 = 11.376 T(3)2 = 5. (e) Since (T 2 = 6.000. d 2 = 1. it is not necessary to calculate the diagnostic quantities. (f) 2 χ 0.05.815 ) .8 1 ⎥⎦ (b) 2 UCL = χα2 .8 0.778. 003 0.440 −0.104] .561 0.641 V'V 133.950 ⎥ .012 7.339 0.003⎤ .016 ⎤ x′ = [15.395 ⎤ x′ = [15.003 −0.526 ⎥⎦ ⎢⎣ 43.xls : worksheet Ex10-14 xbar xbar1 xbar2 10.780 80.Chapter 10 Exercise Solutions 10-12.553 −0.720 ⎤ ⎢ ⎥ V′V = ⎢ −0.256 0.000 ⎥⎦ 10-14.392 1.256 43.001 0.282 3. Excel : workbook Chap10.xls : worksheet Ex10-13 m = 40 ⎡ 4.150 S2 2.305 3.001 ⎦ 10-13.950 587.125] .101 −0.256 0. S 2 = ⎢⎢ −0.016 5.720 0. Excel : workbook Chap10.071 ⎦ ⎣ −0.553 −0.207 S1 3.395 −0.003 0.561⎤ ⎡121.104 88. S1 = ⎢ ⎥ ⎣ −0.014 27.xls : worksheet Ex10-12 m = 40 ⎡ 4.016 0.305 5. Excel : workbook Chap10.599 ⎥⎦ ⎡ 1.307 1.256 ⎤ ⎡ 1.071 0.001 −0. S1 = ⎢⎢ −0. S2 = ⎢ V′V = ⎢ ⎥ ⎥ ⎣ −0.440 −0.158 10-12 .001 ⎦ ⎡121.740 67.339 0.014 ⎥⎥ ⎢⎣ 5.101 −0.012 ⎥⎥ ⎢⎣ −0.607 21.016 0.392 1.740 80. 73.75 0.9231 -0.75 0.17 6.xls : worksheet Ex10-15 p= mu' = Sigma = y' = 4 0 0 0 0 1 0. This gives an ARL1 between 7.2 lambda = 12.9231 -0.9231 3.308 0.75 0.Chapter 10 Exercise Solutions 10-15.9231 -0.0769 -0.308 0.75 0. select (lambda.9231 3.9231 -0.0769 -0. H) pair that closely minimizes ARL1 1 1.53 ARL1 = Select λ = 0.9231 -0.87 UCL = H = 12.308 0.9231 -0.1 with an UCL = H = 12.5 delta = 0.231 delta = 1.75 1 0.109 ARL0 = 200 Sigma-1 = 3.75 1 1 1 1 1 y' Sigma-1 = 0.75 0.308 y' Sigma-1 y = 1.0769 -0.75 1 0.0769 1 1 1 1 From Table 10-3.75 0.1 0.75 0. 10-13 .73 13. Excel : workbook Chap10.9231 y= -0.75 0.22 and 12.75 0.9231 -0.9231 -0.9231 3.17. 270 0. Excel : workbook Chap10.432 -2.270 0.65 Select λ = 0.270 0.432 -2.568 1 1 1 1 From Table 10-4.568 -2.040 ARL0 = 500 Sigma-1 = y= 7.60 7.432 7.xls : worksheet Ex10-16 p= mu' = Sigma = y' = 4 0 0 0 0 1 0. H) pair delta = 1 1.03 ARLmin = 14.432 7.26.5 lambda = 0.9 1 0.432 -2.9 1 1 1 1 1 y' Sigma-1 = 0.9 0.26 16.9 1 0. 10-14 .9 0.432 7.105 with an UCL = H = 15.432 -2.568 -2.568 -2.Chapter 10 Exercise Solutions 10-16.432 -2.9 0. This gives an ARLmin near 14.432 -2.18 UCL = H = 15. select (lambda.9 0.9 0.9 0.432 -2.081 delta = 1.9 0.105 0.60.270 y' Sigma-1 y = 1.432 -2.432 -2.9 0.9 0. 65 9. Excel : workbook Chap10.64 9.65 10.2 0.556 0.2 0.49 5.08 UCL = H = 10. select (lambda.xls : worksheet Ex10-17 p= mu' = Sigma = y' = 2 0 0 1 0.2 with an UCL = H = 9.7778 -2.Chapter 10 Exercise Solutions 10-17.3 lambda = 8.49 and 10.20. H) pair that closely minimizes ARL1 1 1 1.8 1 1 1 y' Sigma-1 = 0.8 0.20 5.1 0.2222 -2.48 ARL1 = Select λ = 0.054 ARL0 = 200 Sigma-1 = 2.5 delta = 0. This gives an ARL1 between 5.65.111 delta = 1.15 10.5 1.7778 y= 1 1 From Table 10-3.2222 2.556 y' Sigma-1 y = 1. 10-15 . 103. 9.25 90 6 85 6 6 5 5 5 5 LCL=82. 9. 37.Chapter 10 Exercise Solutions 10-18. Test Failed at points: 8. 39. 8 points in a row more than 1 standard deviation from center line (above and below CL). 40 TEST 8. 38. 12. 10. and 10 should be 100. (a) Note: In the textbook Table 10-5. the y2 values for Observations 8. 36. 11. 11. 39. and 107. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). Test Failed at points: 40 10-16 . 35.12 5 6 95 _ X=91. 10 TEST 5. 37. 10. Test Failed at points: 9. 40 TEST 6. Test Failed at points: 5. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of y2 Output Variable (Tab10-5y2) 110 1 Individual Value 105 1 5 100 UCL=100. One point more than 3.00 standard deviations from center line.38 80 4 8 12 16 20 24 Observation 28 32 36 40 Test Results for I Chart of Tab10-5y2 TEST 1. Chapter 10 Exercise Solutions 10-18 continued (b) Stat > Regression > Regression Regression Analysis: Tab10-5y2 versus Tab10-5x1. Tab10-5x2.666 Tab10-5x1 .215 Tab10-5x9 Stat > Control Charts > Variables Charts for Individuals > Individuals I Chart of Regression Model Residuals (Ex10-18Res) 7. .435 Tab10-5x3 + 0. Test Failed at points: 21 Plot points on the residuals control chart are spread between the control limits and do not exhibit the downward trend of the response y2 control chart.3..5 _ X=-0. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). Test Failed at points: 19.9 Tab10-5x8 .192 Tab10-5x4 . 25 TEST 6.2 Tab10-5x5 + 0.0 5 LCL=-6.11.00 0.57 5 5 Individual Value 5. 10-17 .00 standard deviations from center line. One point more than 3.0..57 4 8 12 16 20 24 Observation 28 32 36 40 Test Results for I Chart of Ex10-18Res TEST 1. 18 TEST 5.5 1 1 UCL=6.0 -2.0 2.1 Tab10-5x7 + 10.5 -5. The regression equation is Tab10-5y2 = 215 . 2 out of 3 points more than 2 standard deviations from center line (on one side of CL).6 Tab10-5x2 + 0.73 Tab10-5x6 + 6. 21. Test Failed at points: 7. 6 -0.8 -1.6 -0.2 0.0 1 2 3 4 5 6 7 8 9 10 Lag The decaying sine wave of ACFs for Response y2 suggests an autoregressive process. while the ACF for the residuals suggests a random process.4 -0.0 0.6 0.8 -1.2 -0.2 0.8 Autocorrelation 0.0 -0.0 1 2 3 4 5 6 7 8 9 10 Lag Autocorrelation Function for Regression Model Residuals (Ex10-18Res) (with 5% significance limits for the autocorrelations) 1.4 -0.0 -0.0 0.6 0.8 Autocorrelation 0. 10-18 .2 -0.4 0.4 0.Chapter 10 Exercise Solutions 10-18 continued (c) Stat > Time Series > Autocorrelation Autocorrelation Function for y2 Output Variable (Tab10-5y2) (with 5% significance limits for the autocorrelations) 1. 6387 X X X 5 67.2 3..7 8..83550 X X X X X 6 79.3 3.9 2.6 0. Tab10-5x2.1 75.9 60.0 52.4 5..6149 X X X X X X 7 67.9 3.5 0.7 48.8 62.5 71.1 24.8 60. 10-19 .5 4.9 1.1 3. Response is Tab10-5y2 T T T T T T T T a a a a a a a a b b b b b b b b 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 .9 1. This solution uses MINITAB’s “Best Subsets” functionality to identify the best-fitting model with as few variables as possible.8 69..9751 X 2 50.8 61.6147 X X X 4 64.3 76.0 4.84292 X X X X X X … Best Subsets Regression: Tab10-5y2 versus Tab10-5x1.2 3. a regression model of input variables x1. x8.5208 X X X X 5 65.8 0.6816 X 1 35.1799 X 3 67.4 60.5 1.5 5 5 5 5 5 5 5 Mallows x x x x x x x x Vars R-Sq R-Sq(adj) C-p S 1 2 3 4 5 6 7 8 1 43.0 3.9 60.3087 1 31.0 4.8 64. Stat > Regression > Best Subsets Best Subsets Regression: Tab10-5y1 versus Tab10-5x1.. Different approaches can be used to identify insignificant variables and reduce the number of variables in a regression model.1 13.6 60.7 18.6 52.5 5 5 5 5 5 5 5 Mallows x x x x x x x x Vars R-Sq R-Sq(adj) C-p S 1 2 3 4 5 6 7 8 1 36.7 3.8 56.93966 X X X X 6 79.5684 X X X X X 7 67.3 0.4 3.0 0.2 60.4378 X 2 62.8 76.1 5.5660 X X X X X 6 67.0273 X X 4 72...7288 X X X 3 59.0760 X 2 55..6921 2 55.8 6.. and x9 maximize adjusted R2 (minimize S) and minimize Mallow’s C-p. .1 41.0 19. Tab10-5x2.. .Chapter 10 Exercise Solutions 10-19.1665 X X 3 61.1 0..3 29.1 52.4 1.9 4.6 58.4 24.95522 X X X 5 79.5 76.9 2.95201 X X X 4 72.1 4.0 0.0 6...4 4.5 64.5 36.3 1.8 34.9 76.1 12.1 3. x3.0171 X X 3 66.2 5.9 13.9 1..2 4.7 3.83693 X X X X X 7 80..8160 X X 4 64.7 62.1 68.83020 X X X X 5 73.3 3.8 7.9 6.6526 X X X X 6 67.6 0. Response is Tab10-5y1 T T T T T T T T a a a a a a a a b b b b b b b b 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 .9 13.83914 X X X X X X 7 80.2 4.3 69. x4..6200 X X X X X X … T a b 1 0 5 x 9 X X X X X X X X X X X X X ****** T a b 1 0 5 x 9 X X X X X X X X X X X ****** For output variables y1 and y2.1 34.5 24.4 3. 689 F 26.7 S = 3.Chapter 10 Exercise Solutions 10-19 continued Stat > Regression > Regression Regression Analysis: Tab10-5y1 versus Tab10-5x1.47 Total 39 1303. .1034 3.000 10-20 ...35 P 0.52081 SE Coef 123.1497 0.5% MS 18.51 -3.7604 23.830201 SE Coef 29. Tab10-5x3.175 -235.70 3.056 0.198 0.09146 2.64 Tab10-5x8 + 115 Tab10-5x9 Predictor Constant Tab10-5x1 Tab10-5x3 Tab10-5x4 Tab10-5x8 Tab10-5x9 Coef 818.434 Total 39 114.001 0..23 P 0.000 0.02438 0.40 P 0.098 0.7% Analysis of Variance Source DF SS Regression 5 882.03530 0.6367 114.6 0.236 Tab10-5x9 Predictor Constant Tab10-5x1 Tab10-5x3 Tab10-5x4 Tab10-5x8 Tab10-5x9 Coef 244.0.3441 0.0.1758 11.43080 -0.40 F 14.41 12.5% Analysis of Variance Source DF SS Regression 5 90.47 -2.990 Residual Error 34 23.14 0.000 Regression Analysis: Tab10-5y2 versus Tab10-5x1.3 R-Sq = 67.4540 0.03 Residual Error 34 421.001 0.80 0.424 T 28.001 0.6329 0.10 5. Tab10-5x3.124 Tab10-5x3 .03 1.08113 0.65 R-Sq = 79.176 Tab10-5x4 + 11.12396 -0.75 3.47 4.0.9% MS 176.454 Tab10-5x3 + 0.81 S = 0.005 0. The regression equation is Tab10-5y2 = 244 . .4 -0.000 0.85 P 0.633 Tab10-5x1 + 0.000 R-Sq(adj) = 76.431 Tab10-5x1 .2 Tab10-5x8 .025 R-Sq(adj) = 62..0915 Tab10-5x4 + 2.84 3.98 -1.001 0.50 T 1.225 100.075 0. The regression equation is Tab10-5y1 = 819 + 0.31 -3. One point more than 3.105 1 _ X=-0. 9 points in a row on same side of center line. Failed at points: 11 10-21 .00 standard deviations from center line.105 4 8 12 16 20 24 O bser vation 28 32 36 40 1 M oving Range 3 U C L=2. 9 points in a row on same side of center line. 11 Test Results for MR Chart of Ex10-19Res1 TEST Test TEST Test 1.791 1 0 LC L=0 2 4 8 12 16 20 24 O bser vation 28 32 36 40 Test Results for I Chart of Ex10-19Res1 TEST Test TEST Test 1. One point more than 3. Failed at points: 10. Failed at points: 26 2.000 0 2 -1 2 -2 LC L=-2.00 standard deviations from center line.Chapter 10 Exercise Solutions 10-19 continued Stat > Control Charts > Variables Charts for Individuals > Individuals I-MR Chart of y1 Regression Model Residuals (Ex10-19Res1) 1 Individual V alue 2 U C L=2. Failed at points: 25 2.586 2 __ M R=0. 2 out of 3 points more than 2 standard deviations from center line (on one side of CL). Test Failed at points: 7.Chapter 10 Exercise Solutions 10-19 continued I-MR Chart of y2 Regression Model Residuals (Ex10-19Res2) Individual Value 8 1 1 5 U C L=6.00 0 -4 5 LC L=-6. there is not a significant difference between control charts for residuals from either the full regression model (Figure 10-10. One point more than 3. For response y2. Test Failed at points: 19. Test Failed at points: 26 For response y1. 4 out of 5 points more than 1 standard deviation from center line (on one side of CL). no out-of-control observations) and the subset regression model (observation 25 is OOC).52 5 4 _ X=-0. 21 Test Results for MR Chart of Ex10-19Res2 TEST 1. 18 TEST 5.02 6 4 __ M R=2. 25 TEST 6. there is not a significant difference between control charts for residuals from either the full regression model (Exercise 10-18.45 2 0 LC L=0 4 8 12 16 20 24 O bser vation 28 32 36 40 Test Results for I Chart of Ex10-19Res2 TEST 1.00 standard deviations from center line. observations 7 and 18 are OOC) and the subset regression model (observations 7 and 18 are OOC). Test Failed at points: 7. 21.00 standard deviations from center line.52 -8 4 8 12 16 20 24 O bser vation 28 32 36 40 1 M oving Range 8 U C L=8. 10-22 . One point more than 3. 347 -1.5 -1. One point beyond control limits.435 4 8 12 16 20 24 Sample 28 32 36 40 EWMA Chart of y2 Regression Model Residuals (Ex10-19Res2) 2.7SL=0.1 and L = 2. Stat > Control Charts > Time-Weighted Charts > EWMA EWMA Chart of y1 Regression Model Residuals (Ex10-19Res1) 0.4 -2.347 EWMA 1.5 _ _ X=-0.1 _ _ X=-0.0 -0.2 -0. A potential advantage to using the EWMA control chart for residuals from a regression model is the quicker detection of small shifts in the process.435 0.0 -0.000 0.000 0. Use λ = 0.Chapter 10 Exercise Solutions 10-20.2 0. Test Failed at points: 21.5 +2.1 -0.7SL=1.7SL=-1.0 -2.7SL=-0.0 0.0 1. 22 The EWMA control chart for residuals from the response y1 subset model has no out-ofcontrol signals.4 0.5 4 8 12 16 20 24 Sample 28 32 36 40 Test Results for EWMA Chart of Ex10-19Res2 TEST. However the chart for y2 residuals still indicates a problem beginning near observation 20.5 +2.3 -0.3 EWMA 0. 10-23 .7. 02496 1.29428 0. Ex10-21X4 Eigenanalysis of the Correlation Matrix Eigenvalue 2.31445 -0.685 -0.13529 -1. Principal Component Analysis: Ex10-21X1.985 1. (a) Stat > Multivariate > Principal Components Note: To work in standardized variables in MINITAB.23100 0.6088 0.65608 2.70241 -0.257 0.32286 0.20787 0.12378 -1. Ex10-21X3.444 PC2 -0.12420 0.015 Cumulative 0.36528 0.42176 -0.104 Principal Component Scores Ex10-21z1 0.27825 -2.33164 0.83902 3.0118 0.11243 0.23823 -0.50471 -0.08767 1.58815 1.3181 1.286 0.387 PC3 0.17976 -1.14662 Ex10-21z2 -0.29514 1.580 0.11787 -1.49153 0.20488 0.21893 -1.99498 0.03929 0.09129 -0.67046 -0.29168 0.64094 1.31493 0.30494 -0.86838 10-24 .64480 -0.60867 -0.Chapter 10 Exercise Solutions 10-21.01299 -1.718 -0.47259 -0.46958 0.083 0.10423 -0.41131 -2. Ex10-21X2.832 0.534 -0.594 0.794 0. select Storage and enter columns for Scores.334 -0.82157 0.152 0.99211 -1.0613 Proportion 0.000 Variable Ex10-21X1 Ex10-21X2 Ex10-21X3 Ex10-21X4 PC1 0.56081 -0.14246 -0.19166 -0.17849 Ex10-21z3 0.60340 0.253 0.87917 -2.20327 -0.330 0.94470 -1.19734 0.061 0.21950 2.19592 0.36089 0. select Correlation Matrix.607 0. Note: To obtain principal component scores.62545 -0.801 PC4 0.580 0.94763 0. 5 -2 0 2 -0.9 (c) Note: Principal component scores for new observations were calculated in Excel. Ex10-21z2all.5 0.xls : worksheet Ex10-21. Graph > Matrix Plot > Matrix of Plots with Groups Matrix Plot of Ex10-21z1all. Ex10-21z3 Principal Component Scores -2 0 2 2 0 Ex10-21z1 -2 2 0 Ex10-21z2 -2 1.0 -0.0 -0.0 Ex10-21Obs New Original 4 Ex1 0 -2 1 z1 all 0 -4 2.7 Ex10-21z3 -0.5 2. Ex10-21z2.Chapter 10 Exercise Solutions 10-21 continued (b) Graph > Matrix Plot > Simple Matrix of Plots Matrix Plot of Ex10-21z1. See Excel : workbook Chap10.5 Ex1 0 -2 1 z2 all -3.0 2 Ex1 0 -2 1 z3 all 0 -2 -4 0 4 -2 0 2 Although a few new points are within area defined by the original points. Ex10-21z3all Principal Component Scores -3.9 0.7 1. 10-25 . the majority of new observations are clearly different from the original observations. 074 -0. select Correlation Matrix.529 0.471 0.419 0.115 PC4 -0.232 0.022 -0.028 0.513 -0.3292 1.230 0. Ex10-22z3 Principal Component Scores -3 0 3 3 0 Ex10-22z1 -3 3 Ex10-22z2 0 -3 2 0 Ex10-22z3 -2 -3 0 3 -2 0 2 10-26 .525 -0.575 0. Ex10-22z2.105 0.166 -0.323 0.436 -0.349 0. Ex10-22x2.123 0.448 PC2 0.277 0.193 -0. Note: To obtain principal component scores. (a) Stat > Multivariate > Principal Components Note: To work in standardized variables in MINITAB.975 PC7 0.048 0.256 0.0520 Proportion 0.2542 0.1973 0.244 -0.090 -0. (c) Graph > Matrix Plot > Simple Matrix of Plots Matrix Plot of Ex10-22z1.380 0.3121 0.267 0.261 -0.120 0.000 PC9 0.175 0.163 -0.035 0.727 0.332 -0.204 -0.912 PC5 0.6129 0.068 0.368 0.Chapter 10 Exercise Solutions 10-22.000 0.032 -0.068 0.0730 1.406 0.018 0.508 0.003 1.297 0.021 0.632 0.425 -0.200 0.022 0.322 0.006 0.135 0.247 0. select Storage and enter columns for Scores.357 0.997 PC8 0. Ex10-22x9 Eigenanalysis of the Correlation Matrix Eigenvalue 3.947 PC6 -0.133 (b) 72.262 0.372 0.089 0.579 -0.874 0.465 0.240 0.579 0.050 0.844 Variable Ex10-22x1 Ex10-22x2 Ex10-22x3 Ex10-22x4 Ex10-22x5 Ex10-22x6 Ex10-22x7 Ex10-22x8 Ex10-22x9 PC1 -0.127 -0.7% of the variability is explained by the first 3 principal components. Principal Component Analysis: Ex10-22x1.089 0.156 -0.662 -0.651 -0.262 0.035 0.022 0.188 0.407 0. ….014 0.108 0.723 0.1407 2.0287 0.124 -0. Ex10-22x3.152 -0.431 -0.162 0.349 0.409 -0.202 -0.602 -0.145 -0.148 0.199 0.238 PC3 -0.467 -0.099 -0.117 Cumulative 0. 10-27 . indicating that the process is not in control. Graph > Matrix Plot > Matrix of Plots with Groups Matrix Plot of Ex10-22z1all.xls : worksheet Ex10-22.Chapter 10 Exercise Solutions 10-22 continued (d) Note: Principal component scores for new observations were calculated in Excel. See Excel : workbook Chap10. Ex10-22z2all. Ex10-22z3all All Principal Component Scores -4 0 4 3 Ex10-22Obs First Last 0 Ex1 0 -2 2 z1 all -3 4 0 Ex1 0 -2 2 z2 all -4 2 0 Ex1 0 -2 2 z3 all -2 -3 0 3 -2 0 2 Several points lie outside the area defined by the first 30 observations. yt : observation zt : EWMA (a) zt = λ yt + (1 − λ ) zt −1 zt = λ yt + zt −1 − λ zt −1 zt − zt −1 = λ yt + zt −1 − zt −1 − λ zt −1 zt − zt −1 = λ yt − λ zt −1 zt − zt −1 = λ ( yt − zt −1 ) (b) zt −1 − zt − 2 = λ et −1 (as a result of part (a)) zt −1 − zt − 2 + (et − et −1 ) = λ et −1 + (et − et −1 ) zt −1 + et − zt − 2 − et −1 = et − (1 − λ )et −1 yt − yt −1 = et − (1 − λ )et −1 11-11 .Chapter 11 Exercise Solutions 11-1. 800 10.625 19.xls : worksheet Ex 11-2 0 0.625 7.969 5.688 4.975 -13.0 L = +10 10 4.715 -5.0 12.000 0.975 -13. 11-12 .375 -22.000 0. Excel : workbook Chap11.854 0.0 -10.975 -1.0 0 2.375 -18.000 0.000 -13.923 10.975 -13.975 -13.0 0.975 -10.0 40 -8.690 Bounded Adjustment Chart for Ex 11-2 50 -12.375 -18.560 6.625 4.359 45 46 47 48 49 50 22 -9 3 12 3 12 9 -31 12 9 -9 9 8.975 -10.0 -9.0 -18.625 10.Chapter 11 Exercise Solutions 11-2.000 25.76. The chart with λ = 0.0 -9.000 0.318 -7.0 -50 1 6 11 16 21 26 31 36 41 46 Obs Orig_out Adj_out_t EWMA_t Adj_Obs_t+1 Chart with λ = 0.700 1.975 … SS = Average = 21468 17.2 gives SS = 9780 and average deviation from target = 1.000 -3.984 0.000 0.000 11.179 -6.000 0.982 -8.3 exhibits less variability and is closer to target on average.0 0.025 -22.0 -40 10.0 -20 6.000 5.24 6526.375 -18.8 T= lambda = L= g= Obs Orig_out Orig_Nt Adj_out_t EWMA_t |EWMA_t|>L? Adj_Obs_t+1 Cum_Adj 1 2 3 4 5 6 7 8 9 10 0 16 24 29 34 24 31 26 38 29 0 16 8 5 5 -10 7 -5 12 -9 16 24 20.0 L = -10 -10 Adjustment Scale -4.766 9.0 20 -2.375 -18.375 0.993 no no no no no no 0.000 0.127 -6.000 0.625 12.0 -9.975 -13.975 -0.3 10 0.0 30 -6.0 -9.0 -30 8.134 no yes no yes no no no no yes 0.975 -1. 5 -11. but is further from target on average than for the chart with λ = 0.800 12.061 -7.440 3.037 -4.288 no no no no no Adj_Obs_t+1 Cum_Adj 0 -12 0 -11 0 0 0 0 0 0 -12 -12 -23 -23 -23 -23 -23 -23 0 0 0 0 0 -11.264 7. 11-13 .Chapter 11 Exercise Solutions 11-3.813 -3.5 -11.5 -8.4 10 0.958 7.5 0.24 5610.3.175 no yes no yes no no no no no 46 47 48 49 50 -9 3 12 3 12 -31 12 9 -9 9 -20. Excel : workbook Chap11.xls : worksheet Ex 11-3 Target yt = lambda = L= g= 0 0.25 0.400 3.5 -11.91 Bounded Adjustment Chart for Ex 11-3 80 -15 70 60 -10 50 40 30 L = +10 10 0 0 L = -10 -10 -20 Adjustment Scale -5 20 5 -30 -40 -50 10 -60 -70 15 -80 1 6 11 16 21 26 31 36 41 46 Obs Orig_out Adj_out_t EWMA_t Adj_Obs_t+1 The chart with λ = 0.5 -11.440 6.5 … SS = Average = 21468 17.5 -6.022 -5.8 Obs Orig_out Orig_Nt Adj_out_t EWMA_t |EWMA_t|>L? 1 2 3 4 5 6 7 8 9 10 0 16 24 29 34 24 31 26 38 29 0 16 8 5 5 -10 7 -5 12 -9 16 24 17 22 1 8 3 15 6 6.400 13.880 0.5 0.4 exhibits less variability.5 -8. 7 -6.9 -6.0 10.3 -2.2 -1.7 -0.3 -13. Excel : workbook Chap11.7 45 46 47 48 49 50 22 -9 3 12 3 12 9 -31 12 9 -9 9 11.7 -27.3 -4.3 -27.0 -20.0 -5.3 -7.7 -4.0 5.5 -7.2 -1.6 -22.8 -3.5 -4.0 -9.3 -22.4 1.0 20.8 Obs Orig_out Adj_out_t Orig_Nt Adj_Obs_t+1 Cum_Adj 1 2 3 4 5 6 7 8 9 10 0 16 24 29 34 24 31 26 38 29 0 16 8 5 5 -10 7 -5 12 -9 16.7 5.9 0.2 0.9 5.0 20.0 -5.6 1.0 -14.4 -4.9 5.7 Integral Control for Ex 11-4 50 -12 -10 40 -8 30 -6 20 -2 0 0 2 -10 Adjustment Scale -4 10 4 -20 6 -30 8 -40 10 -50 12 1 6 11 16 Orig_out 21 26 Adj_out_t 31 36 41 46 Adj_Obs_t+1 The chart with process adjustment after every observation exhibits approximately the same variability and deviation from target as the chart with λ = 0.Chapter 11 Exercise Solutions 11-4.4.0 20.0 -19.0 -1.8 3. 11-14 .0 -5.6 -0.3 1.9 -23.3 -2.7 -8.1 14.24 5495.0 … SS = Average = 21468 17.xls : worksheet Ex 11-4 T= lambda = g= 0 0. 000 0.xls : worksheet Ex 11-5 t Yt 1 2 3 4 5 6 7 8 9 10 … / m => 0 16 24 29 34 24 31 26 38 29 Var_m = Var_m/Var_1 = 1 2 3 4 5 6 7 8 9 16 8 5 5 -10 7 -5 12 -9 24 13 10 -5 -3 2 7 3 29 18 0 2 -8 14 -2 34 8 7 -3 4 5 24 15 2 9 -5 31 10 14 0 26 22 5 38 13 29 147.60 151.029 1.195 1.500 1.72 147.000 1.43 201.000 0 5 10 15 20 25 Vm/V1 11-15 .370 0.104 1.02 136.943 Variogram for Ex 11-5 2. Excel : workbook Chap11.39 162.70 1.002 1.000 m 1.217 0.47 179.500 2.Chapter 11 Exercise Solutions 11-5.500 0.11 175.929 1.53 138. 149321 0.164698 0.4 -0.01 2.Chapter 11 Exercise Solutions 11-5 continued MTB : Chap11.33 0.14 32.440855 0.26 LBQ 10.59 54.0 0.58 1.345327 0.334961 0.6 0.05 0.44 67.2 0.0 -0.066173 T 3.325056 0.03 55.86 63.85 1.8 Autocorrelation 0.012158 0.31 16.228540 0.440819 0.68 1.389094 0.54 1.80 41.28 0.55 48.0 1 2 3 4 5 6 7 Lag 8 9 10 11 12 13 Autocorrelation Function: Yt Lag 1 2 3 4 5 6 7 8 9 10 11 12 13 ACF 0.45 1.6 -0.59 0. The slow decline in the sample ACF also indicates the data are correlated and potentially nonstationary.4 0.71 62.41 63.8 -1.75 Variogram appears to be increasing.90 0.316478 0.87 67.2 -0.12 2. so the observations are correlated and there may be some mild indication of nonstationary behavior. 11-16 .299822 0.39 27.mtw : Yt Stat > Time Series > Autocorrelation Function Autocorrelation Function for Data in Table 11-1 (Yt) (with 5% significance limits for the autocorrelations) 1. t 1 2 3 4 5 6 7 8 9 10 Orig_out Orig_Nt Adj_out_t Adj_Obs_t+1 Cum_Adj 215.5 24.2 1.8 -1.7 3.0 0.818.0 185.4 62.8 162.1 4.4 Adjusted 1.0 260.4 0.0 30.0 290.510.3 2.7 4.2 T= lambda = g= Obs.3 -6.5 -7.8 0.0 -10.7 220.4 170.0 187.2 7.0 -5. (a) and (b) Excel : workbook Chap11.0 170.3 467.0 195.8 0.0 200.8 -20.0 1 6 11 16 21 26 31 36 41 46 Obs.2 1.0 195.1 216.5 192.0 -15.0 80.0 5.1 -3.8 -15.0 1.5 3.1 177.0 Adjustment Scale Integral Control for Ex 11-6(a) 20.3 2.0 140.0 15.9 -39.4 4.8 161.5 208.7 180.4 191.9 320.0 129.Chapter 11 Exercise Solutions 11-6.9 12.7 0.3 192.8 … 145.2 0.323.4 1.0 193.2 160.9 200.0 230.5 176.1 13.5 Unadjusted 1.0 -0.0 -20.0 12.6 -13.5 49 50 SS = Average = Variance = 11.3 -4.0 63.xls : worksheet Ex 11-6a 200 0.8 176.0 10.0 110.7 191.871. 11-17 .9 180.7 187. t Orig_out Adj_out_t Adj_Obs_t+1 Significant reduction in variability with use of integral control scheme. 995.0 3.5 176.0 170.0 200.0 1 6 11 16 21 26 31 36 41 46 Obs.4 3.xls : worksheet Ex 11-6c 200 0.0 320.3 -6.0 -30.0 7.6 8.0 66.0 230.0 10.0 -10.9 183.5 24.0 6.888.4 Adjusted 1.6 -13.323.2 and λ = 0.6 -2.9 164.5 216.0 290.7 2.0 195.0 20.8 -0.5 -7.0 SS = Average = Variance = -15.4 176.0 9.9 2.1 3.8 161.6 7.0 10.0 -40.5 192.4 18.1 187.8 0. 11-18 .4 191.0 Orig_out Orig_Nt Adj_out_t Adj_Obs_t+1 Cum_Adj 215. t Orig_out Adj_out_t Adj_Obs_t+1 Variances are similar for both integral adjustment control schemes (λ = 0.0 8.2 T= lambda = g= Obs.0 -20.Chapter 11 Exercise Solutions 11-6 continued (c) Excel : workbook Chap11.7 … 129.0 Adjustment Scale Integral Control for Ex 11-6(c) 40.0 140.0 30.5 5.3 -4.8 -20.4 1.0 260.5 50.0 110.0 162.0 2.8 7.0 80.0 198.7 223.0 195.0 195.0 -0.8 193.3 467.0 189.1 176.0 4.2 180.9 18.8 185.4 175.5 13.7 0.5 -7.0 0.4 1.4).9 208.0 5.871.8 1.1 21. t 1.9 -39.5 Unadjusted 1.5 7.0 30. 0 -20.233 0.516 48.0 -10. t 1 2 3 4 5 6 7 8 9 10 Orig_out Orig_Nt Adj_out_t EWMA_t 215.2 and λ = 0.000 6.900 -5.233 yes no no no no 5.8 0 195. Excel : workbook Chap11.8 -20 196 -0.000 0. 11-19 .516 … 122.0 110.7 216.0 0.696 176.0 1 6 11 16 21 26 31 36 41 46 Obs.0 30.516 48.0 5.2 145.233 6.412 185.0 -15.0 200.5 24.0 0.0 129.233 6.0 290.2 Target yt = lambda = L= g= Obs.1 176.000 0.Chapter 11 Exercise Solutions 11-7.000 0.5 46 47 48 49 50 -7 3.016 -12.733 -1.969 -5.7 320.0 -5.000 0.051 172.2 133.0 230.0 0.699 174.000 0.941 162.0 Adjustment Scale Bounded Adjustment Chart for Ex 11-7 20.9 193.323.8 -15.000 0.633 187.2 12 1.0 140.3 7 11.0 170.300 -2.0 -0.8 SS = Average = Variance = 165.177 176.872 161.516 178.656 |EWMA_t|>L? Adj_Obs_t+1 Cum_Adj no no no no no no yes no no 0.5 191.632.000 0.5 -7 186.600 -14.717 0.4 162.000 -8.265 182.9 176.716 181.253 180.702 -7.0 10.840 191.0 80.870 216.000 0.000 -0.459 -10. t Orig_out Adj_out_t EWMA_t Adj_Obs_t+1 Behavior of the bounded adjustment control scheme is similar to both integral control schemes (λ = 0.716 193.516 48.9 -39.xls : worksheet Ex 11-7 200 0.516 48.0 0.5 1.000 0.000 6.0 15.364 1.3 -4.0 260.000 0.304 467.300 -4.000 48.057 -7.6 -13.733 -3.3 -6 185.000 0.9 126.4).000 0. xls : worksheet Ex 11-8 200 0.9 -39.000 0.7 216.304 467.0 110.256 176.4 15 1.000 0.5 191.2 T= lambda = L= g= Obs.000 0.849 -10.9 176.0 1 6 11 16 21 26 31 36 41 46 Obs.5 24.5 181.658 184.958 203.967 -2. 11-110 .6 -13.323.9 199.0 230.0 0.600 -23.573 216.083 189.000 0.0 16.0 -32.758 58.758 188.0 0.0 -0.0 -48.0 80.900 -8.0 8.692 162.0 -40.3 7 11.1 176. t 1 2 3 4 5 6 7 8 9 10 Orig_out Orig_Nt Adj_out_t EWMA_t 215.0 24.000 0.967 -0.0 129.775 187.000 0.5 -7 192.300 -8.0 320.86 Bounded Adjustment Chart for Ex 11-8 350.0 200.958 191.0 290.000 0.000 0.0 -8.821 |EWMA_t|>L? Adj_Obs_t+1 Cum_Adj no no no no no no yes no no 0.784 170.758 58.000 0. Excel : workbook Chap11.000 0.0 32. t Orig_out Adj_out_t EWMA_t Adj_Obs_t+1 Behavior of both bounded adjustment control schemes are similar to each other and simlar to the integral control schemes.000 0.0 50.000 0.758 58.326 -4.486 176.488 185.680 191.81 -9.513 1.258 1.000 12.8 -15.693 -7.758 58.0 40.467 12.467 no no no no no 0.0 140.3 -4.0 Adjustment Scale -16.0 170.8 -20 196 -1.872 161.9 126.467 12.000 12.Chapter 11 Exercise Solutions 11-8.300 -4.0 30.0 -24.758 … 122.4 162.720 -11.773.0 260.467 0.3 -6 185.5 46 47 48 49 50 SS = Average = Variance = -7 3.0 48.0 0.0 0.8 0 195.000 0.000 1.013 180.2 133.000 58.2 145.000 -14. 0 18 4.1 48.520 44.2 22.xls : worksheet Ex 11-9a T= lambda = g= 50 0.7 49 50 23 26 3.6 -1.6 0.4 65.262 78.0 58.7 64. 90 -5.0 -9.4 223.0 -4.8 -0.3 -1.6 55.0 45.4 -0.7 22.Chapter 11 Exercise Solutions 11-9.7 -0.2 -1.0 10 1 6 11 16 21 26 31 36 41 46 Obs. (a) and (b) Excel : workbook Chap11.0 42 1.0 43.0 10.0 74 -3.0 58 -1.7 0.4 0.0 -1.0 34 2.629 46.2 50.0 11.2 51. t Orig_out Adj_out_t Adj_Obs_t+1 11-111 .0 3.3 -1.0 14.4 54.0 26 3.8 -3.0 -1.6 -1.7 -5.0 66 -2.0 0.0 -0.2 1.0 -13.51 Significant reduction in variability with use of an integral control scheme.0 -11.9 0.6 Obs Orig_out Adj_out_t Orig_Nt Adj_Obs_t+1 Cum_Adj 1 2 3 4 5 6 7 8 9 10 50 58 54 45 56 56 66 55 69 56 8.0 82 -4.9 … SS = Average = Variance = Adjusted 108.32 Unadjusted 109.6 -3.8 -5.0 53.0 Adjustment Scale Integral Control for Ex 11-9 (a) 5.0 50 0.8 -0. 3 -2. 11-112 .3 -5.0 -3.Chapter 11 Exercise Solutions 11-9 continued (c) Excel : workbook Chap11.819 47.520 44.0 54 -4.5 114.4 223.0 46.9 69 14.0 42.2 55 -11.8 27.833 56.0 -0.4 1.4 -0.1 -3.9 -0.0 52.0 66 10.51 3.0 58.0 48.0 55.40 There is a slight reduction in variability with use of λ = 0. as compared to λ = 0.8 0.0 56 0.5 1.0 3.3 -9.4.xls : worksheet Ex 11-9c T= lambda = g= 50 0.0 63.0 -3.5 45 -9.0 50.1 -0.4 … 49 50 SS = Average = Variance = 23 26 109.8 0.2.6 56 11.4 26.4 -1.0 63.5 -2.0 54.8 -8.0 -2.5 53.2 56 -13.3 -6. with a process average slightly closer to the target of 50.6 Obs 1 2 3 4 5 6 7 8 9 10 Orig_out Orig_Nt Adj_out_t Adj_Obs_t+1 Cum_Adj 50 58 8.0 -1.0 -2. Chapter 11 Exercise Solutions Integral Control for Ex 11-9(c) 90 -10.0 18 8.0 1 6 11 16 21 26 31 36 41 46 Obs.0 74 -6.t Orig_out Adj_out_t Adj_Obs_t+1 11-113 .0 26 6.0 50 0.0 10 10.0 34 4.0 42 2.0 58 -2.0 66 -4.0 82 -8. 00 -2.6 Target yt = lambda = L= g= Obs.79 38.0 58 -1.48 31.00 5.731 56 0 56.520 44.080 45 -9 45.268 55 -11 53.00 -2.00 no yes no yes no 0.00 0.19 … 24 18 20 23 26 46 47 48 49 50 8 -6 2 3 3 37.880 56 -13 54.40 0.0 42 1.0 2 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 Obs.48 15.51 SS = Average = Variance = -2. but different variances (bounded adjustment variance is larger).79 43.2 4 1.0 26 3.620 121.4 223.709 -2.0 90 -5.585 66 10 66.842 -4.19 109.00 0.515 -1. Excel : workbook Chap11.0 Adjustment Scale Bounded Adjustment Chart for Ex 11-10 6.00 0.xls : worksheet Ex 11-10 50 0.00 3.0 66 -2.00 0.00 0.48 35.00 0.600 69 14 67.0 0.0 10 5.505 -5. with similar means and sums of squares.0 0.32 0.00 13.362 107.00 0.00 2.Chapter 11 Exercise Solutions 11-10. t 1 2 3 4 5 6 7 8 9 10 Orig_out Orig_Nt Adj_out_t EWMA_t 50 0 58 8 58 1.822 45.00 0.79 17.19 17.00 2.00 0.00 3.00 0.00 1.0 74 -3. t Orig_out Adj_out_t EWMA_t Adj_Obs_t+1 Nearly the same performance as the integral control scheme.00 0. 11-114 .0 34 2.00 1.00 -2.0 82 -4.664 56 11 56.600 54 -4 54.00 -2.904 |EWMA_t|>L? Adj_Obs_t+1 Cum_Adj no no no no no yes no no no 0.0 18 4.0 50 0.00 -2.00 2.79 15.00 0.72 98 -6. The remaining exercises are worked in a similar manner. the initial experimental layout must be created in MINITAB or defined by the user. 12-1. Since the experiment layout was not created in MINITAB. one column for phosphor type. the design must be defined before the results can be analyzed. check “General full factorial”. Enter the data into the MINITAB worksheet using the first three columns: one column for glass type. This is how the Excel file is structured (Chap12. Select the two factors (Glass Type and Phosphor Type).Chapter 12 Exercise Solutions Note: To analyze an experiment in MINITAB. The dialog box should look: 12-1 . therefore some information required by MINITAB is not included. select Stat > DOE > Factorial > Define Custom Factorial Design. then for this exercise. The Excel data sets contain only the data given in the textbook.xls). After entering the data in MINITAB. This experiment is three replicates of a factorial design in two factors—two levels of glass type and three levels of phosphor type—to investigate brightness. Detailed MINITAB instructions are provided for Exercises 12-1 and 12-2 to define and create designs. and only the solutions are provided. and one column for brightness. Chapter 12 Exercise Solutions 12-1 continued Next. and click “OK” twice. MINITAB recognizes these contiguous seven columns as a designed experiment. DO NOT insert or delete columns between columns 1 through 7. Note that MINITAB added four new columns (4 through 7) to the worksheet. inserting or deleting columns will cause the design layout to become corrupt. or blocks. select “Designs”. For this exercise. no information is provided on standard order. 12-2 . The design and data are in the MINITAB worksheet Ex12-1. run order. so leave the selections as below.MTW. point type. Click “OK”. click “OK”.004 Ex12-1Glass*Ex12-1Phosphor 2 133. then click on “Terms”. Glass type (A) and phosphor type (B) significantly affect television tube brightness (P-values are less than 0. since the factor levels in this experiment are categorical. verify that the selected terms are Glass Type.318 Error 12 633.3 466. using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Ex12-1Glass 1 14450. Phosphor Type. Ex12-1Phosphor Factor Ex12-1Glass Ex12-1Phosphor Type fixed fixed Levels 2 3 Values 1.3 52.3 66.44% No indication of significant interaction (P-value is greater than 0.0 14450.3 133. and their interaction. General Linear Model: Ex12-1Bright versus Ex12-1Glass.84 0.79 0.26 0.3 933.10).10).3 633. select “Residuals Plots : Four in one”. and click “OK” twice.000 Ex12-1Phosphor 2 933.26483 R-Sq = 96.0 273.Chapter 12 Exercise Solutions 12-1 continued Select Stat > DOE > Factorial > Analyze Factorial Design.8 Total 17 16150.7 8.7 1.0 14450.0 S = 7. 2 1. The option to plot residuals versus variables is for continuous factor levels. select “Fits” and “Residuals”. do not select this option. 2. Click on “Storage”.08% R-Sq(adj) = 94. Select the response (Brightness). Residual Plots for Ex12-1Bright Normal Probability Plot of the Residuals Residuals Versus the Fitted Values 99 10 Residual Percent 90 50 0 10 -10 1 -10 0 Residual 10 220 Histogram of the Residuals 240 260 280 Fitted Value 300 Residuals Versus the Order of the Data 10 3 Residual Frequency 4 2 0 1 0 -10 -10 -5 0 5 Residual 10 15 2 4 6 8 10 12 14 Observation Order 16 18 12-3 . Click on “Graphs”. 3 Analysis of Variance for Ex12-1Bright. and versus fitted values reveals no problems. select Graph > Individual Value Plot > One Y with Groups. select the “Reference Lines” tab. the residuals should be re-examined.Chapter 12 Exercise Solutions 12-1 continued Visual examination of residuals on the normal probability plot. histogram. To plot residuals versus the two factors. and enter “0” for the Y axis. Select the column with stored residuals (RESI1) as the Graph variable and select one of the factors (Glass Type or Phosphor Type) as the Categorical variable for grouping. Click on “Scale”. Individual Value Plot of RESI1 vs Ex12-1Glass 15 RESI1 10 5 0 0 -5 -10 1 2 Ex12-1Glass Individual Value Plot of RESI1 vs Ex12-1Phosphor 15 RESI1 10 5 0 0 -5 -10 1 2 Ex12-1Phosphor 3 12-4 . then click “OK” twice. The plot of residuals versus observation order is not meaningful since no order was provided with the data. If the model were re-fit with only Glass Type and Phosphor Type. select the graph to make it active then: Editor > Select Item > Individual Symbols and then Editor > Edit Individual Symbols > Jitter and de-select Add jitter to direction. To remove the jitter. Individual Value Plot of RESI1 vs Ex12-1Glass 15 RESI1 10 5 0 0 -5 -10 1 2 Ex12-1Glass Individual Value Plot of RESI1 vs Ex12-1Phosphor 15 RESI1 10 5 0 0 -5 -10 1 2 Ex12-1Phosphor 3 Variability appears to be the same for both glass types. 12-5 .Chapter 12 Exercise Solutions 12-1 continued Note that the plot points are “jittered” about the factor levels. there appears to be more variability in results with phosphor type 2. however. and click “OK” twice. Final selected combination of glass type and phosphor type depends on the desired brightness level. Select “Interaction Plot” and click on “Setup”.Chapter 12 Exercise Solutions 12-1 continued Select Stat > DOE > Factorial > Factorial Plots. 12-6 . Interaction Plot (data means) for Ex12-1Bright 310 Ex12-1Glass 1 2 300 290 Mean 280 270 260 250 240 230 220 1 2 Ex12-1Phosphor 3 The absence of a significant interaction is evident in the parallelism of the two lines. select the response (Brightness) and both factors (Glass Type and Phosphor Type). 0 280.000 Ex12-1Phosphor 2 933.000 273. The DOE functionality was selected to illustrate the approach that will be used for most of the remaining exercises.08% R-Sq(adj) = 94.318 Error 12 633.0 12-7 .0 S = 7.79 0.004 Interaction 2 133.0 264. select Stat > ANOVA > Two-Way.3 66.26 0.7 1. Two-way ANOVA: Ex12-1Bright versus Ex12-1Glass.3 52.667 235. To obtain results which match the output in the textbook’s Table 12.3 466.44% Ex12-1Glass 1 2 Mean 291. and complete the dialog box as below.0 273. Ex12-1Phosphor Source DF SS MS F P Ex12-1Glass 1 14450.0 14450.5.333 256.265 R-Sq = 96.000 Ex12-1Phosphor 1 2 3 Individual 95% CIs For Mean Based on Pooled StDev -----+---------+---------+---------+---(--*-) (--*-) -----+---------+---------+---------+---240 260 280 300 Mean 260.7 8.Chapter 12 Exercise Solutions 12-1 continued Alternate Solution: This exercise may also be solved using MINITAB’s ANOVA functionality instead of its DOE functionality.667 Individual 95% CIs For Mean Based on Pooled StDev -------+---------+---------+---------+-(-------*-------) (-------*-------) (-------*-------) -------+---------+---------+---------+-256.84 0.0 272.8 Total 17 16150. 008 0. (a) To analyze the experiment.96 P 0. Since the standard order (Run) is provided.81 Metal Hardness*Cutting Angle -24. a reduced model in Metal Hardness.25 -5.199 R-Sq(adj) = 72.000 0. Both approaches would achieve the same result.12 12. BC. then enter the data. This solution uses the first approach. Another approach would be to create a worksheet containing the data.009 ** 0.74 Metal Hardness 84. and change the Number of factors to “3”.13 12. and click “OK”. Metal Hardness.36% P 0.662 0. Select “Factors”. highlight full factorial. The worksheet is in run order.09 1.98 Cutting Speed*Metal Hardness* -34.25 -12. Select “Terms” and verify that all terms (A. to change to standard order (and ease data entry) select Stat > DOE > Display Design and choose standard order.75 35. Cutting Angle Estimated Effects and Coefficients for Life (coded units) Term Effect Coef SE Coef T Constant 413. C. then define a customer factorial design. and Cutting Speed*Cutting Angle is more appropriate. a full factorial model is not necessary. select Stat > DOE > Factorial > Analyze Factorial Design.12 12.37 12. one approach to solving this exercise is to create a 23 factorial design in MINITAB. AB.62 12. Select Stat > DOE > Factorial > Create Factorial Design. Factorial Fit: Life versus Cutting Speed.89 Cutting Speed*Metal Hardness -11. change number of replicates to “2”.41 -0. Cutting Angle.75 -17.014 0.13 12.199 Based on ANOVA results.41 3. Select “Designs”.40 Cutting Angle 71. Based on P-values less than 0.483 0.62 12.81 8.40 Cutting Angle S = 49.357 0.41 33. AC.25 42.45 Cutting Speed*Cutting Angle -119. Leave the design type as a 2-level factorial with default generators.001 ** 0. leave factor types as “Numeric” and factor levels as -1 and +1.41 -0.41 -4. and click “OK” twice.56% Analysis of Variance for Life (coded units) Source DF Seq SS Adj SS Adj MS Main Effects 3 50317 50317 16772 2-Way Interactions 3 59741 59741 19914 3-Way Interactions 1 4830 4830 4830 Residual Error 8 19700 19700 2462 Pure Error 8 19700 19700 2463 Total 15 134588 … F 6.25 -59. B.88 12.30 Cutting Speed 18.41 -1. Cutting Speed will also be retained to maintain a hierarchical model. enter the factor names.020 ** 0. The design and data are in the MINITAB worksheet Ex12-2.Chapter 12 Exercise Solutions 12-2.MTW.41 0. 12-8 .6236 R-Sq = 85.41 2.25 9.10. ABC) are included. 0 -1 1 -1 Cutting Speed Cutting Angle -1 Longest tool life is at A-.25 -59.000 0. for an average predicted life of 552.5 351.47 -4.12 12.015 0.Chapter 12 Exercise Solutions 12-2(a) continued Factorial Fit: Life versus Cutting Speed.74 22.78 S = 49.88 Cutting Speed*Cutting Angle -119. Cube Plot (data means) for Life Exercise 12-2(b) 552.13 12. Cutting Angle Estimated Effects and Coefficients for Life (coded units) Term Effect Coef SE Coef T Constant 413.04 0.001 R-Sq(adj) = 72.47 2. we see that the low level of cutting speed and the high level of cutting angle gives good results regardless of metal hardness.75 35.001 1.38 Cutting Angle 71.25 42.5 1 266.5 391.0 512.25 9.425 (b) The combination that maximizes tool life is easily seen from a cube plot.65% P 0. 12-9 .0 380.13 12.47 3.62 12.008 0.8988 R-Sq = 79.88 12. Select Stat > DOE > Factorial > Factorial Plots.0 1 Metal Hardness 446. (c) From examination of the cube plot.5 405.85 P 0. Choose and set-up a “Cube Plot”.25% Analysis of Variance for Life (coded units) Source DF Seq SS Adj SS Adj MS Main Effects 3 50317 50317 16772 2-Way Interactions 1 56882 56882 56882 Residual Error 11 27389 27389 2490 Lack of Fit 3 7689 7689 2563 Pure Error 8 19700 19700 2463 Total 15 134588 F 6.006 0.5.47 0.73 Metal Hardness 84.12 Cutting Speed 18.47 33. Metal Hardness.480 0. B+ and C+. To find the residuals. B. select Stat > DOE > Factorial > Analyze Factorial Design. Residuals versus fitted values plot shows that the equal variance assumption across the prediction range is reasonable. Select “Terms” and verify that all terms for the reduced model (A. select “Storage” and choose “Residuals”. and for residuals plots choose “Normal plot” and “Residuals versus fits”. Normal Probability Plot of the Residuals (response is Life) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -100 -50 0 Residual 50 100 Residuals Versus the Fitted Values (response is Life) 50 Residual 25 0 -25 -50 -75 250 300 350 400 Fitted Value 450 500 550 Normal probability plot of residuals indicates that the normality assumption is reasonable. Select “Graphs”. 12-10 . C.Chapter 12 Exercise Solutions 12-3. To save residuals to the worksheet. AC) are included. Since there are two replicates of the experiment.062 -4.50 462. select “Terms” and verify that all terms are selected.177 0. sweetener*temperature.015 0.000 0.10 Carbonation*Temperature 1.16 7.9504 0.188 -0.688 33.Chapter 12 Exercise Solutions 12-4.531 0.781 0.125 101. sweetener*syrup to water. Select Stat > DOE > Factorial > Analyze Factorial Design.9504 1.594 0.53 47.91 Total 31 1967.15 3-Way Interactions 4 405.656 0.9504 192.50 462. Syrup to Water.9504 -0.187 -2.094 0.9504 2.10 and select terms with P-value less than 0.36 Sweetener*Temperature -2.844 0.49 Temperature Sweetener*Syrup to Water* 2.938 -0.629 * * * * * 0. Estimated Effects and Coefficients for Total Score (coded units) Term Effect Coef SE Coef T Constant 182. temperature.969 0.10.64 Residual Error 16 462. as well as lower-order terms included in the significant terms (main effects: syrup to water.9504 -0.281 0. type of sweetener is dominant.9504 -4.77 Syrup to Water -1.688 2.313 -0.001 0.563 -0.219 0. 12-11 .025 0.9504 -0.73 Sweetener*Syrup to Water*Temperature 4.438 1.500 28.469 0.379 0.31 Sweetener -9.37 2-Way Interactions 6 199.9504 0. carbonation.629 0.9504 2.031 0.062 2.500 0.938 -0. and then enter the data.91 Pure Error 16 462.188 -1.89 Sweetener*Syrup to Water*Carbonation -5. syrup to water*temperature).771 0..13 405.500 28.688 -1.218 P 0.9504 -1.9504 2.9504 -0.687 0.000 0.688 0.388 0.41 Temperature 3.28 Carbonation*Temperature Analysis of Variance for Total Score (coded units) Source DF Seq SS Adj SS Adj MS F Main Effects 4 852.531 47.923 0. along with interactions involving both sweetener and the ratio of syrup to water.344 0.69 Carbonation -2.50 4-Way Interactions 1 47.15 Syrup to Water*Carbonation -0.048 0.14 Sweetener*Carbonation 0.07 Sweetener*Syrup to Water 4.469 0.344 0.094 0.47 Sweetener*Carbonation*Temperature -0.MTW.722 0.267 0. the reduced model will contain the significant terms (sweetener.47 P 0.938 1. sweetener*syrup to water*carbonation.055 0. Use an α = 0.28 3.9504 -1. Create a 24 factorial design in MINITAB.49 Syrup to Water*Carbonation* -0.9504 -0.. .28 1.218 From magnitude of effects.30 Syrup to Water*Temperature -0. The design and data are in the MINITAB worksheet Ex12-4. two-factor interactions: sweetener*carbonation.031 0.53 1.63 852. Factorial Fit: Total Score versus Sweetener. syrup to water*carbonation.9504 -2. To preserve model hierarchy.625 213.69 199. sweetener*syrup to water*temperature).344 0. 09 0.38 84.9244 -1.764 0.531 0.54 Analysis of Variance for Total Score (coded units) Source DF Seq SS Adj SS Adj MS F Main Effects 4 852.29 3-Way Interactions 2 391.656 0. Syrup to Water.046 0.47 P 0.80 2-Way Interactions 5 176.344 0.344 0.005 0.9244 2.9244 -0.062 -4.031 0.88 546. Estimated Effects and Coefficients for Total Score (coded units) Term Effect Coef SE Coef T Constant 182.50 28.06 391.001 0.9244 2.251 0.71 Carbonation -2.306 0. .9244 197.188 -1.30 Syrup to Water*Temperature -0.18 Syrup to Water*Carbonation -0.34 Lack of Fit 4 84.9244 -0.88 27.91 176.020 P 0.000 0.094 0.9244 -1.90 Syrup to Water -1.91 Total 31 1967.15 Residual Error 20 546.563 -0.16 7.63 852.969 0.344 0.38 1.000 0.585 12-12 .Chapter 12 Exercise Solutions 12-4 continued Factorial Fit: Total Score versus Sweetener.81 Sweetener*Syrup to Water*Temperature 4.91 35.162 0.73 Sweetener -9.688 -1.9244 0.781 0.714 0.13 Sweetener*Syrup to Water 4.281 0..920 0.9244 2.188 -0.688 2.9244 -4.9244 -2.688 0.37 Sweetener*Temperature -2.50 462.486 0.63 213.10 Sweetener*Syrup to Water*Carbonation -5.06 195.53 7.011 0.094 0.062 2.594 0.9244 -0..313 -0.20 Sweetener*Carbonation 0.188 -2.38 21.73 Pure Error 16 462.938 1.45 Temperature 3.040 0. 5 1. and then select the variables.Chapter 12 Exercise Solutions 12-5. as well as a slight indication of an outlier.5 1. select Stat > DOE > Factorial > Analyze Factorial Design.0 Syrup to Water Residuals Versus Carbonation Residuals Versus Temperature (response is Total Score) (response is Total Score) 0. Select “Terms” and verify that all terms for the reduced model are included.0 8 Residual 8 Residual 0 0 -8 0 -8 -1.0 0.0 -0.5 0. choose “Normal plot” of residuals and “Residuals versus variables”. 12-13 .5 1.5 1. Select “Graphs”. Normal Probability Plot of the Residuals (response is Total Score) 99 Percent 90 50 10 1 -10 -5 0 Residual 5 10 Residuals Versus Sweetener Residuals Versus Syrup to Water (response is Total Score) (response is Total Score) 8 Residual Residual 8 0 -8 -8 -1.5 0.0 Carbonation 0. This is not serious enough to warrant concern.5 0.0 -1.0 Sweetener 0. To find the residuals.5 0.0 -1.0 Temperature There appears to be a slight indication of inequality of variance for sweetener and syrup ratio.0 -0.0 -0.0 -0. 000 0. .13 Sweetener*Syrup to Water 4. AB.281 0.062 2.9244 -0.251 0.73 Sweetener -9.656 0.531 0.9244 2.046 0. D.9244 -4.313 -0. if the confidence interval includes zero.9244 -1.688 0.20 Sweetener*Carbonation 0.688 2.486 0. Factorial Fit: Total Score versus Sweetener.37 Sweetener*Temperature -2.45 Temperature 3.10 Sweetener*Syrup to Water*Carbonation -5.344 0.54 P 0.9244 197.9244 2.162 0. Estimated Effects and Coefficients for Total Score (coded units) Term Effect Coef SE Coef T Constant 182.764 0.71 Carbonation -2.9244 -0.30 Syrup to Water*Temperature -0.18 Syrup to Water*Carbonation -0.011 0.344 0..9244 2.188 -0.781 0. the factor is not significant. and ABD appear to be significant. Select Stat > DOE > Factorial > Analyze Factorial Design.9244 -2.040 0..094 0.344 0.714 0.000 0.594 0. ABC.094 0.969 0.9244 -0.90 Syrup to Water -1. and a value greater than approximately |2| would be considered significant.563 -0. From examination of the above table. factors A. 12-14 .920 0.188 -2.031 0.938 1. Also.81 Sweetener*Syrup to Water*Temperature 4.062 -4. Syrup to Water.9244 0.9244 -1.688 -1.Chapter 12 Exercise Solutions 12-6.188 -1. Select “Terms” and verify that all terms for the reduced model are selected.020 The ratio of the coefficient estimate to the standard error is distributed as t statistic. 000 0. Since there is only one replicate of the experiment. and set alpha to 0. select “Terms” and verify that all terms are selected.625 Sweetener -10. choose the normal effects plot.Chapter 12 Exercise Solutions 12-7.125 Syrup to Water*Carbonation* -2.750 Sweetener*Syrup to Water*Temperature 4.500 -3..250 Temperature Sweetener*Syrup to Water* 3.750 -0.125 Carbonation 0.500 Sweetener*Temperature -6.250 -3.500 -1.MTW. Then select “Graphs”.375 Temperature 5.250 0.875 Carbonation*Temperature … Normal Probability Plot of the Effects (response is Total Score. and then enter the data. Create a 24 factorial design in MINITAB. Estimated Effects and Coefficients for Total Score (coded units) Term Effect Coef Constant 183.750 1. .875 Syrup to Water*Temperature -3.125 Syrup to Water*Carbonation -1. Select Stat > DOE > Factorial > Analyze Factorial Design.750 0. Syrup to Water..500 Carbonation*Temperature 1.000 Sweetener*Carbonation 1.250 -0.10) 99 Effect Type Not Significant Significant 95 90 F actor A B C D Percent 80 70 60 50 40 N ame S w eetener S y rup to Water C arbonation Temperature 30 20 10 5 1 A -10 -5 0 Effect 5 10 Lenth's PSE = 4.10 Factorial Fit: Total Score versus Sweetener.000 0.000 2. The design and data are in the MINITAB worksheet Ex12-7.750 Sweetener*Syrup to Water 4.250 2.125 Sweetener*Carbonation*Temperature 0.000 -1.500 Sweetener*Syrup to Water*Carbonation -7.500 2.5 12-15 .250 Syrup to Water -0. Alpha = .500 -5. 93 Residual Error 14 778.865 -2.5 190. as well as in the predicted values.63 Total 15 1219.014 S = 7. This is not serious enough to warrant concern.75 778.000 Sweetener -10.75 778. Re-fit and analyze the reduced model.500 -5.625 1. Factorial Fit: Total Score versus Sweetener Estimated Effects and Coefficients for Total Score (coded units) Term Effect Coef SE Coef T P Constant 183. only factor A (sweetener) is significant.0 187.250 1.45822 R-Sq = 36.0 FittedValue Residuals Versus Sweetener (responseisTotal Score) Residual 10 0 -10 -1.00 441.59% Analysis of Variance for Total Score (coded units) Source DF Seq SS Adj SS Adj MS F Main Effects 1 441.865 98.5 185.0 -0.0 Sweetener 0.750 55.63 Pure Error 14 778.00 7.014 Normal Probability Plot of the Residuals (responseisTotal Score) 99 Percent 90 50 10 1 -20 -10 0 Residual 10 20 Residuals Versus the FittedValues (responseisTotal Score) Residual 10 0 -10 180.000 441.5 1.15% R-Sq(adj) = 31.Chapter 12 Exercise Solutions 12-7 continued From visual examination of the normal probability plot of effects.48 0.5 0.82 0.75 P 0.750 55.0 182.0 There appears to be a slight indication of inequality of variance for sweetener. 12-16 . A 25 design in two blocks will lose the ABCDE interaction to blocks. 12-9. Day 1 a b c abc Day 2 d abd acd bcd (1) ab ac ad bc bd cd abcd Treatment combinations within a day should be run in random order.Chapter 12 Exercise Solutions 12-8. The ABCD interaction is confounded with blocks. or days. Block 1 (1) ae ab be ac ce bc abce ad de bd abde cd acde abcd bcde Block 2 a e b abe c ace abc bce d ade abd bde acd cde bcd abcde 12-17 . Alpha = . and then enter the data.1062 Normal Probability Plot of the Effects (response is Color.7700 Solv/React 1.10.2725 -0. Then select “Graphs”.1462 Cat/React*React Purity 0.1625 0.4275 0. Cat/React.7025 -0. select “Terms” and verify that all main effects and interaction effects are selected.1200 0. (a) Create a 25-1 factorial design in MINITAB.1363 React Purity 4. Estimated Effects and Coefficients for Color (coded units) Term Effect Coef Constant 2.1500 0.4650 -0.5450 2.2300 -0.2138 Cat/React*Temp 0.3513 Solv/React*Cat/React 1.. Since there is only one replicate of the experiment.8375 -0.10) 99 Effect Ty pe Not Significant Significant D 95 90 F actor A B C D E Percent 80 70 60 50 40 N ame S olv /React C at/React Temp React P urity React pH 30 20 10 5 1 -2 -1 0 1 2 Effect 3 4 5 Lenth's PSE = 0.Chapter 12 Exercise Solutions 12-10.4187 Temp*React pH -0. and set alpha to 0.9125 -0.0600 Cat/React*React pH 0. The design and data are in the MINITAB worksheet Ex12-10.6150 Solv/React*React pH 0. Factorial Fit: Color versus Solv/React.2125 0.4350 0.0812 Temp*React Purity -0.4562 Solv/React*React Purity -1.1825 React Purity*React pH 0.8475 12-18 .MTW.2725 React pH -0. . Select Stat > DOE > Factorial > Analyze Factorial Design.7325 Temp -0.5750 Solv/React*Temp -0.7175 Cat/React -1.2925 0. choose the normal effects plot..3650 -0. 63 82.63 82.545 2.68 0.65876 R-Sq = 68. only factor D (reactant purity) is significant.Chapter 12 Exercise Solutions 12-10 (a) continued From visual examination of the normal probability plot of effects.770 0.628 30.272 0.20% R-Sq(adj) = 65.93% Analysis of Variance for Color (coded units) Source DF Seq SS Adj SS Adj MS F Main Effects 1 82.52 38.0 -0. Factorial Fit: Color versus React Purity Estimated Effects and Coefficients for Color (coded units) Term Effect Coef SE Coef T P Constant 2.15 P 0.52 2.0 React Purity 0.0 Residual plots indicate that there may be problems with both the normality and constant variance assumptions.03 Residual Error 14 38. Re-fit and analyze the reduced model.5 1.52 2.52 38.751 Total 15 121.000 (b) Nor mal P r obability P lot of the Residuals (response is Color) 99 Percent 90 50 10 1 -4 -3 -2 -1 0 Residual 1 2 3 4 Residuals V er sus the Fitted V alues (response is Color) Residual 2 0 -2 0 1 2 3 4 5 Fitted Value Residuals Ver sus React P ur ity (response is Color) Residual 2 0 -2 -1.751 Pure Error 14 38.4147 5. 12-19 .48 0.5 0.000 S = 1.4147 6.000 React Purity 4. Looking at the original normal probability plot of effects and effect estimates.005 5. the 2nd and 3rd largest effects in absolute magnitude are A (solvent/reactant) and B (catalyst/reactant). A cube plot in these factors shows how the design can be collapsed into a replicated 23 design. so this design collapses to a onefactor experiment. the lowest at low reactant purity. Cube Plot (data means) for Color 3.795 1 Cat/React 6. D (reactant purity). or simply a 2-sample t-test.715 1.865 -2.865 -1 1 -1 React Purity -1 Solv/React 12-20 . The highest color scores are at high reactant purity.Chapter 12 Exercise Solutions 12-10 continued (c) There is only one significant factor.425 1 0.875 4.385 1. 762 B:Matl1 -5. The design and data are in the MINITAB worksheet Ex12-11.638 … Alias Structure I + A:Temp*C:Vol*E:Matl2 + B:Matl1*D:Time*E:Matl2 + A:Temp*B:Matl1*C:Vol*D:Time A:Temp + C:Vol*E:Matl2 + B:Matl1*C:Vol*D:Time + A:Temp*B:Matl1*D:Time*E:Matl2 B:Matl1 + D:Time*E:Matl2 + A:Temp*C:Vol*D:Time + A:Temp*B:Matl1*C:Vol*E:Matl2 C:Vol + A:Temp*E:Matl2 + A:Temp*B:Matl1*D:Time + B:Matl1*C:Vol*D:Time*E:Matl2 D:Time + B:Matl1*E:Matl2 + A:Temp*B:Matl1*C:Vol + A:Temp*C:Vol*D:Time*E:Matl2 E:Matl2 + A:Temp*C:Vol + B:Matl1*D:Time + A:Temp*B:Matl1*C:Vol*D:Time*E:Matl2 A:Temp*B:Matl1 + C:Vol*D:Time + A:Temp*D:Time*E:Matl2 + B:Matl1*C:Vol*E:Matl2 A:Temp*D:Time + B:Matl1*C:Vol + A:Temp*B:Matl1*E:Matl2 + C:Vol*D:Time*E:Matl2 From the Alias Structure shown in the Session Window. Factorial Fit: yield versus A:Temp. E:Matl2 Estimated Effects and Coefficients for yield (coded units) Term Effect Coef Constant 19. Since there is only one replicate of the experiment.337 E:Matl2 2. D:Time. B:Matl1.138 A:Temp*B:Matl1 1.275 1.913 A:Temp*D:Time -1.Chapter 12 Exercise Solutions 12-11. 12-21 . Enter the factor levels and yield data into a MINITAB worksheet.587 C:Vol 2. (a) and (b) Select Stat > DOE > Factorial > Analyze Factorial Design.275 1.138 D:Time -0. The aliases are: A*I = A*ACE = A*BDE = A*ABCD ⇒ A = CE = ABDE = BCD B*I = B*ACE = B*BDE = B*ABCD ⇒ B = ABCE = DE = ACD C*I = C*ACE = C*BDE = C*ABCD ⇒ C = AE = BCDE = ABD … AB*I = AB*ACE = AB*BDE = AB*ABCD ⇒ AB = BCE = ADE = CD The remaining aliases are calculated in a similar fashion. then define the experiment using Stat > DOE > Factorial > Define Custom Factorial Design. select “Terms” and verify that all main effects and twofactor interaction effects are selected.275 -0. C:Vol.525 -0.MTW.238 A:Temp -1.825 0.175 -2.675 -0. the complete defining relation is: I = ACE = BDE = ABCD. 1) = ¼ (–6.825 This are the same effect estimates provided in the MINITAB output above.9 -16.913 4. The other main effects and interaction effects are calculated in the same way.8 + 18. (d) Select Stat > DOE > Factorial > Analyze Factorial Design. E:Matl2 … … Analysis of Variance for yield (coded units) Source DF Seq SS Adj SS Adj MS Main Effects 5 79.2 15.913 9.826 79.2 23.5 + 16. Since there is only one replicate of the experiment.5 – 16.2 – 23. and set alpha to 0.9 – 16.739 … F * * P * * 12-22 .3) = 1.5 16.9 16. Factorial Fit: yield versus A:Temp. B:Matl1.2 +23. C:Vol.525 [AB] = AB + BCE + ADE + CD = ¼ (+23. Then select “Graphs”.8 – 23. choose the normal effects plot.8 + 18.1 [A] = A + CE + BCD + ABDE = ¼ (–23.10.826 15.8 23.1) = ¼ (7.4 – 16.Chapter 12 Exercise Solutions 12-11 continued (c) A B -1 1 1 -1 -1 1 -1 1 C -1 1 -1 1 -1 -1 1 1 D -1 -1 -1 1 1 1 -1 1 E -1 -1 1 -1 1 -1 1 1 1 -1 -1 -1 -1 1 1 1 yield 23.8 18.8 + 23.2 + 15.4 16. D:Time. select “Terms” and verify that all main effects and twofactor interaction effects are selected.2 +15.4 – 16.1) = –1.965 2-Way Interactions 2 9.956 Residual Error 0 * * * Total 7 89. 8682 -2.5 5.Chapter 12 Exercise Solutions 12-11 (d) continued Normal Probability Plot of the Effects (response is yield.030 Pure Error 6 36.175 -2.5 -5. Select Stat > DOE > Factorial > Analyze Factorial Design.000 B:Matl1 -5. and pool the remaining terms to estimate error.16 0.18 36.74 … 12-23 .5 0.025 Residual Error 6 36.0 Effect 2. Then select “Graphs”. Alpha = . Select “Terms” and select “B”.10) 99 Effect Ty pe Not Significant Significant 95 90 F actor A B C D E Percent 80 70 60 50 40 N ame A :Temp B:M atl1 C :V ol D :Time E :M atl2 30 20 B 10 5 1 -7.56 53.0 -2.18 6. and select the “Normal plot” and “Residuals versus fits” residual plots.030 Total 7 89.561 8.0 Lenth's PSE = 2.025 … Analysis of Variance for yield (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 1 53. Factorial Fit: yield versus B:Matl1 Estimated Effects and Coefficients for yield (coded units) Term Effect Coef SE Coef T P Constant 19.98 0.7375 Although none of the effects is significant at 0.18 6.10.587 0. main effect B (amount of material 1) is more than twice as large as the 2nd largest effect (absolute values) and falls far from a line passing through the remaining points.18 36. Re-fit a reduced model containing only the B main effect.8682 22.56 53.238 0.88 0. 0 Residual plots indicate a potential outlier.0 -2.5 0. If no issues can be identified.0 Residual 2. it may be necessary to make additional experimental runs 12-24 .5 5. The run should be investigated for any issues which occurred while running the experiment.Chapter 12 Exercise Solutions 12-11 continued (e) Residuals Versus the Fitted Values Exercise 12-11 (e) (response is yield) 2 1 Residual 0 -1 -2 -3 -4 -5 16 17 18 19 Fitted Value 20 21 22 Normal Probability Plot of the Residuals Exercise 12-11(e) (response is yield) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -5. 00 -15.953 0.953 2.79 0. Re-fit and analyze a reduced model containing A.5 9.765 A*D 5.50 1.50 3. choose the normal effects plot. and BC. D Estimated Effects and Coefficients for Mole Wt (coded units) Term Effect Coef SE Coef T P Constant 837.953 -0.32 0.953 0. C. Then select “Graphs”.25 3.50 -3.953 0.386 … Analysis of Variance for Mole Wt (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 4 9850 9850 2462. B.50 3.765 C*D 7.25 3. BC) are significant.95 0. AB. (a) Select Stat > DOE > Factorial > Analyze Factorial Design.953 -0.10) 99 95 AB 90 Percent 80 70 60 50 40 30 F actor A B C D N ame A B C D BC 20 C 10 A 5 1 Effect Ty pe Not Significant Significant -5 -4 -3 -2 -1 0 Standardized Effect 1 2 3 The main effects A and C and two two-factor interactions with B (AB. and then enter the data.63 0.00 3.013 * D -7. The main effect B must be kept in the model to maintain hierarchy.75 3.00 3. B.85 0.953 -4. Factorial Fit: Mole Wt versus A.953 -2.25 3.67 0.50 3.95 0.74 0.00 3.00 5.00 -10.386 A*B 22.953 -3.75 3.00 2. Create a 24 factorial design in MINITAB.953 1.53 0.10.32 0. Since this is a single replicate of the experiment.MTW.262 C -30.005 * B 10. C.000 A -37.151 Residual Error 5 1250 1250 250. and set alpha to 0.26 0.87 0.953 211.014 2-Way Interactions 6 4000 4000 666.75 3.50 -18. 12-25 .7 2.Chapter 12 Exercise Solutions 12-12.036 * A*C -2.50 -1. Alpha = .50 11.053 * B*D 2.0 Total 15 15100 Normal Probability Plot of the Standardized Effects (response is Mole Wt. The design and data are in the MINITAB worksheet Ex12-12.555 B*C -20. select “Terms” and verify that all main effects and twofactor interaction effects are selected.85 0. 015 * … Analysis of Variance for Mole Wt (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 3 9625. Select “Terms” and select “A.80 0.0 185.50 -18.400 246.004 Residual Error 10 1850.25 3. and BC.000 2-Way Interactions 2 3625.Chapter 12 Exercise Solutions 12-12 continued (b) Select Stat > DOE > Factorial > Analyze Factorial Design.63 0.000 A -37.0 250. B.0 1600.00 -10. A.41 0.400 3. Factorial Fit: Mole Wt versus A.5 9.94 0.00 -15.0 Lack of Fit 2 250.008 * B*C -20. 12-26 . C.400 -2.0 1812.75 3.50 3.30 0.50 11.0 125.3 17.001 * A*B 22.0 … The same terms remain significant. B.00 3.400 1. AB.51 0.0 0.31 0.0 3625.0 1850.0 Total 15 15100.0 9625.00 5.559 Pure Error 8 1600. Then select “Graphs”.00 3.0 200. C. BC”.47 0. C Estimated Effects and Coefficients for Mole Wt (coded units) Term Effect Coef SE Coef T P Constant 837.34 0.00 3. and select the “Normal plot” and “Residuals versus fits” residual plots.172 C -30.000 * B 10.400 -5.0 3208.400 -4. AB. however neither plot reveals a major problem with the normality and constant variance assumptions. 12-27 .Chapter 12 Exercise Solutions 12-12 continued (c) Normal Probability Plot of the Residuals (response is Mole Wt) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -30 -20 -10 0 Residual 10 20 30 Residuals Versus the Fitted Values (response is Mole Wt) 20 Residual 10 0 -10 -20 -30 790 800 810 820 830 840 Fitted Value 850 860 870 880 A “modest” outlier appears on both plots. MTW. Select “Terms” and verify that all main effects and two-factor interactions are selected. This will ensure that if both lack of fit and curvature are not significant. The design and data are in the MINITAB worksheet Ex12-13. DO NOT include the center points in the model (uncheck the default selection). lack of fit. See the dialog box below. (a) Select Stat > DOE > Factorial > Analyze Factorial Design. all three (curvature. To summarize MINITAB’s functionality. 12-28 . Also. curvature is always tested against pure error and lack of fit (if available). and pure error) should be included in the residual mean square. and then enter the data. regardless of whether center points are included in the model. The inclusion/exclusion of center points in the model affects the total residual error used to test significance of effects. Assuming that lack of fit and curvature tests are not significant. the main and interaction effects are tested for significance against the correct residual error (lack of fit + curvature + pure error).Chapter 12 Exercise Solutions 12-13. Create a 24 factorial design with four center points in MINITAB. pure error.50 -18.00 9.25 0. the three-factor and four-factor interactions).527 0.05 0. the first test to consider is the “lack of fit” test.75 9.60 0.0 Total 19 26920 … (b) The test for curvature is significant (P-value = 0.527 -1.004 * Lack of Fit 5 1250 1250 250. (In MINITAB. D Estimated Effects and Coefficients for Mole Wt (coded units) Term Effect Coef SE Coef T P Constant 848.150 D -7.13 0.75 9.00 9.521 99.0 16.39 0.00 -10.2 Curvature 1 8820 8820 8820. and lack of fit (if available) and use this as the basis for testing for significant effects.527 -1.7 0.527 -0. no further statistical analysis should be performed because the model is inadequate.50 1.527 -1.97 0.75 9. This approach to sequential experimentation is presented in Chapter 13. which is a test of significance for terms not included in the model (in this exercise.57 0. this is accomplished by not including center points in the model.70 0. Although one could pick a “winning combination” from the experimental runs. the model is not correctly specified. which is a test of significance for the pure quadratic terms.234 2-Way Interactions 6 4000 4000 666.527 0. If lack of fit is significant.50 9.004).25 9.) Factorial Fit: Mole Wt versus A.5 1.527 1.26 0. If tests for both lack of fit and curvature are not significant.321 B*D 2.527 -0.00 -15.268 A*C -2.527 0.703 A*B 22. the next test to consider is the “curvature” test. If this test is significant.915 Pure Error 3 3000 3000 1000.00 8.799 B*C -20.703 … Analysis of Variance for Mole Wt (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 4 9850 9850 2462.00 5.25 9. C.52 0. then it is reasonable to pool the curvature.50 -3.081 B 10.13 0.18 0. and some terms need to be added to the model.612 C -30.50 -1.46 0.00 2.527 0.00 9.000 A -37. If lack of fit is not significant.39 0.25 9.898 A*D 5. a better strategy is to add runs that would enable estimation of the quadratic effects.0 0.898 C*D 7.52 0. B.50 11.50 3.Chapter 12 Exercise Solutions 12-13 (a) continued When looking at results in the ANOVA table. 12-29 .822 Residual Error 9 13070 13070 1452. The source of interest for any combined main and 2-factor interaction effect would be in question. A lower resolution design would have some 2-factor interactions and main effects aliased together.Chapter 12 Exercise Solutions 12-14. so this is a resolution IV design. Main effects are clear of 2-factor interactions. a 28IV− 4 design has a complete defining relation of: I = BCDE = ACDF = ABCG = ABDH = ABEF = ADEG = ACEH = BDFG = BCFH = CDGH = CEFG = DEFH = AFGH = ABCDEFGH The runs would be: Run A 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 + 11 – 12 + 13 – 14 + 15 – 16 + B – – + + – – + + – – + + – – + + C – – – – + + + + – – – – + + + + D – – – – – – – – + + + + + + + + E=BCD – – + + + + – – + + – – – – + + F=ACD – + – + + – + – + – + – – + – + G=ABC – + + – + – – + – + + – + – – + H=ABD – + + – – + + – + – – + + – – + A=BCDE=CDF=BCG=BDH=BEF=DEG=CEH=ABDFG=ACDGH=ABCFH=ACEFG=ADEFH=FGH=BCDEFGH B=CDE=ACDF=ACG=ADH=AEF=ABDEG=ABCEH=DFG=CFH=BCDGH=BCEFG=BDEFH=ABFGH=ACDEFGH C=BDE=ADF=ABG=ABDH=ABCEF=ACDEF=AEH=BCDFG=BFH=DGH=EFG=CDEFH=ACFGH=ABDEFGH D=BCE=ACF=ABCG=ABH=ABDEF=AEG=ACDEH=BFG=BCDFH=CGH=CDEFG=EFH=ADFGH=ABCEFGH E=BCD=ACDEF=ABCEG=ABDEH=ABF=ADG=ACH=BDEFG=BCEFH=CDEGH=CFG=DFH=AEFGH=ABCDFGH F=BCDEF=ACD=ABCFG=ABDFH=ABE=ADEFG=ACEFH=BDE=BCH=CDFGH=CEG=DEH=AGH=ABCDEGH G=BCDEG=ACDFG=ABC=ABDGH=ABEFG=ADE=ACEGH=BDF=BFGH=CDH=CEF=DEFGH=AFH=ABCDEFH H=BCDEH=ACDFH=ABCGH=ABD=ABEFH=ADEGH= ACE=BDFGH=BCF=CDG=CEFGH=DEF=AFG=ABCDEFG AB=ACDE=BCDF=CG=DH=EF=BDEG=BCEH=ADFG=ACFH=ABCDFH=ABCEFG=ABDEFH=BFGH=CDEFGH AC=ABDE=DF=BG=BCDH=BCEF=CDEG=EH=ABCDFG=ABFH=ADGH=AEFG=ACDEFH=CFGH=BDEFGH etc. From Table 12-23 in the textbook. Since significant 2-factor interactions often occur in practice. and at least some 2-factor interactions are aliased with each other. 12-30 . this problem is of concern. 000 -1.433 0. Enter the factor levels and resist data into a MINITAB worksheet.500 -0.600 40.6 4.000 * D -2.700 23.000 Residual Error 4 18.338 Pure Error 3 13.000 * A*D -2.3 4388.6 5.33 Total 11 28184.759 C 80.7621 31.250 0. including a column indicating whether a run is a center point run (1 = not center point.260 … Analysis of Variance for Resist (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 4 17555.7621 -1.10) 99 Effect Ty pe Not Significant Significant 95 C 90 Percent 80 F actor A B C D AC 70 A 60 50 40 N ame A B C D 30 20 10 5 1 0 10 20 30 40 Standardized Effect 50 60 12-31 .200 0.400 -1. choose the normal effects plot.65 Curvature 1 5. B.6 5.7621 0.550 0. C. (a) Select Stat > DOE > Factorial > Analyze Factorial Design.0 4. D Estimated Effects and Coefficients for Resist (coded units) Term Effect Coef SE Coef T P Constant 60.300 0.61 1.MTW.7621 -0. Factorial Fit: Resist versus A.57 0.000 * B -0.850 0.29 0. Alpha = . DO NOT include the center points in the model (uncheck the default selection).190 A*B 1.7621 52.6 18.33 0.51 0. Then select “Graphs”.13 0.Chapter 12 Exercise Solutions 12-15. The design and data are in the MINITAB worksheet Ex12-15. 0 = center point).6223 97.000 A 47.3 17555.76 0.1 3536.12 0.88 0.0 13.30 0. Also.000 2-Way Interactions 3 10610.0 Normal Probability Plot of the Standardized Effects (response is Resist. Select “Terms” and verify that all main effects and two-factor interactions are selected.1 10610.7621 47.400 0.800 36.000 0. and set alpha to 0.83 944. Then define the experiment using Stat > DOE > Factorial > Define Custom Factorial Design.510 A*C 72.31 0.7621 -1.10.70 761.72 0.100 0. the main effects A and C and their twofactor interaction (AC) are significant.0 5.3 8771.000 2-Way Interactions 1 10599.4 5.8007 50.8007 29.79 0.1 Total 11 28184.327 Pure Error 7 35.6 1.4 35.44 0. C.6 1710.43 0.000 A 47.8007 45.43 0. so continue with analysis.000 * C 80.7 10599. 12-32 . (b) Factorial Fit: Resist versus A.6 5.7 10599.70 23. Select Stat > DOE > Factorial > Analyze Factorial Design.000 * A*C 72.60 40.7 2066.11 0.0 Curvature is not significant (P-value = 0. Then select “Graphs”. Re-fit and analyze a reduced model containing A. Select “Terms” and select “A.Chapter 12 Exercise Solutions 12-15 continued Examining the normal probability plot of effects.30 0. C Estimated Effects and Coefficients for Resist (coded units) Term Effect Coef SE Coef T P Constant 60.0 41. (c) Residuals Versus the Fitted Values (response is Resist) 3 2 Residual 1 0 -1 -2 -3 0 20 40 60 80 100 Fitted Value 120 140 160 180 A funnel pattern at the low value and an overall lack of consistent width suggest a problem with equal variance across the prediction range.000 * … Analysis of Variance for Resist (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 2 17543.80 36.6 5.89 0.1 Curvature 1 5. and select the “Normal plot” and “Residuals versus fits” residual plots.327).33 0.3 17543.000 Residual Error 8 41.40 0. C.46 0. AC”.6537 92.85 0. and AC. 0 Residual 2.5 5.Chapter 12 Exercise Solutions 12-15 continued (d) Normal Probability Plot of the Residuals (response is Resist) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -5.0 The normal probability plot of residuals is satisfactory. The concern with variance in the predicted resistivity indicates that a data transformation may be needed.5 0.0 -2. 12-33 . 5 1.65 65 .Chapter 13 Exercise Solutions Note: To analyze an experiment in MINITAB.1 ≤ x2 ≤ 1 x2 6 = = 0.0 − 1 ≤ x1 ≤ 1.0 -0. The Excel data sets contain only the data given in the textbook.85 > 85 Ex13-1x2 0. the initial experimental layout must be created in MINITAB or defined by the user.5 -1.6 13-1 .0 Ex13-1y < 60 60 .5 0.70 70 .75 75 .0 Ex13-1x1 0.0 (b) yˆ = 75 + 10 x1 + 6 x2 -0. 13-1.6 x1 10 ∆x1 = 1 ∆x2 = 0. Ex13-1x1 y = 75 + 10x1 + 6x2 1.80 80 . therefore some information required by MINITAB is not included.0 -1. The MINITAB instructions provided for the factorial designs in Chapter 12 are similar to those for response surface designs in this Chapter. (a) Graph > Contour Plot Contour Plot of Ex13-1y vs Ex13-1x2.5 0. yˆ = 50 + 2 x1 − 15 x2 + 3 x3 − 1 ≤ xi ≤ +1. The design is not rotatable.3 select x2 with largest absolute coefficient.13 βˆ ∆x −15 1.20 −15 1.0 2 ∆x3 = 2 βˆ 3 βˆ2 ∆x2 = 3 = −0.0 2 βˆ1 ∆x1 = = = −0. and set ∆x2 = 1.5. βˆ2 = −15. 13-2 . (a) This design is a CCD with k = 2 and α = 1.0 13-3. 2. i = 1.Chapter 13 Exercise Solutions 13-2. 0 823.000 Square 2 1647.577 7. Estimated Regression Coefficients for y Term Coef SE Coef T P Constant 160. Response Surface Regression: y versus x1.314 0.3 2.2 0.5 15.875 … Analysis of Variance for y Source DF Seq SS Adj SS Adj MS F P Regression 5 30583. Then define the experiment using Stat > DOE > Response Surface > Define Custom Response Surface Design.769 0.590 0.18 0.991 Pure Error 3 486.082 x1*x2 -1.088 0.3 2.000 Linear 2 28934. 0 = center point).2 28934.704 -18. Select “Terms” and verify that all main effects.85 0.441 4.461 -3.618 4.013 Interaction 1 2.5 5.423 7.6 Lack-of-Fit 3 15.4 6116. two-factor interactions.4 30583.5 9.555 35.MTW.0 162.0 Total 11 31084.8682 x1 -58. x2 The analysis was done using coded units.688 10.000 x1 -87.0 1647.000 x2 3.285 -0.9 … Estimated Regression Coefficients for y using data in uncoded units Term Coef Constant 160.704 0.0 486.09 0.5 83.273 0.7500 13-3 .2 14467.2941 x2 2.8546 x2*x2 6.461 2.Chapter 13 Exercise Solutions 13-3 continued (b) Enter the factor levels and response data into a MINITAB worksheet.03 0.1 173. and quadratic terms are selected.3 0.9231 x1*x2 -0.017 x2*x2 15.471 x1*x1 -24. including a column indicating whether a run is a center point run (1 = not center point.4118 x1*x1 -10. The design and data are in the MINITAB worksheet Ex13-3. Select Stat > DOE > Response Surface > Analyze Response Surface Design.868 4.03 0.875 Residual Error 6 501.164 0.7 73.5 501. The level for temperature could be established based on other considerations. (d) Temp = 50 x1 + 750 = 50(+1. such as cost.Chapter 13 Exercise Solutions 13-3 continued (c) Stat > DOE > Response Surface > Contour/Surface Plots Contour Plot of y vs x2. x1 Surface Plot of y vs x2.0 240 180 y 120 60 1 0 -1 -1.5 x1 = 1.49384 x2 = -0.0 -0.217615 y = 49.5.22) + 30 = 26.5 -1 0 x1 1 x1 0 x2 -1 1 From visual examination of the contour and surface plots.7 13-4 . A flag is planted at one option on the contour plot above.5 40 50 75 100 125 150 175 200 y < > 40 50 75 100 125 150 175 200 225 225 -1.5 1. it appears that minimum purity can be achieved by setting x1 (time) = +1.5 and letting x2 (temperature) range from −1.471).6101 0. x1 1. The range for x2 agrees with the ANOVA results indicating that it is statistically insignificant (P-value = 0.50) + 750 = 825 Time = 15 x2 + 30 = 15(−0.0 x2 0.5 to + 1. 64 64 .60 60 .012 at x1 ≈ +0.885)] (−2. Ex13-4x1 y = 69.4) = 0.68 68 .1x1^2 .1 − 2.3(0.0 + 1.6 (b) ∂ yˆ ∂ (69.1.3 x1 = 0 ∂ x2 x1 = −13.3 x2 = 0 ∂ yˆ = 1.62 62 .1x2 .885 x2 = [−1.6x1 + 1.6 − 2 x1 + 0. Graph > Contour Plot Contour Plot of Ex13-4y vs Ex13-4x2.Chapter 13 Exercise Solutions 13-4.0 + 1. x2 ≈ +0.9 (−15.3x1 x2 ) = ⇒0 ∂ x1 ∂ x1 = 1.569 13-5 .1x2 − 1x12 − 1.66 66 .2 x22 + 0.6 x1 + 1.4 x2 + 0.2x2^2 + 0.70 > 70 Ex13-4x2 1 0 -1 -2 -2 -1 0 Ex13-4x1 1 2 yˆ max = 70.9.3x1x2 2 Ex13-4y < 56 56 .1 − 0.58 58 .7) = 0. 0550 13-6 . Select Stat > DOE > Response Surface > Create Response Surface Design.00781 108. then enter the data.03706 -3. Leave the design type as a 2-factor. central composite design.13355 0. and enter a custom alpha value of exactly 1.35 0.1249 0.01210 0.56 0.000 x1*x1 -0.012 x2*x2 -0. The design is rotatable.MTW.01210 0.03424 8. (b) Since the standard order is provided.03228 0.0790 0. select Stat > DOE > Response Surface > Analyze Response Surface Design. highlight the design with five center points (13 runs). The worksheet is in run order.4071 x1*x1 -0.16128 0.13355 0. Estimated Regression Coefficients for y Term Coef SE Coef T P Constant 13.291 … Analysis of Variance for y Source DF Seq SS Adj SS Adj MS F P Regression 5 2.06499 0.22628 … Estimated Regression Coefficients for y using data in uncoded units Term Coef Constant 13.7273 x1 0. B2.03706 -2.06678 7.070 x1*x2 0. Response Surface Regression: y versus x1.142 0.2980 x2 -0.000 x2 -0. A2. to change to standard order (and ease data entry) select Stat > DOE > Display Design and choose standard order. Select “Terms” and verify that a full quadratic model (A.371 0.291 Residual Error 7 0.0790 x1*x2 0.Chapter 13 Exercise Solutions 13-5.01210 1.4 (the rotatable design is α = 1. x2 The analysis was done using coded units.19 0.000 x1 0.4.41421).06499 0.03271 0.04309 318.01090 1.889 0.54 0. The design and data are in the MINITAB worksheet Ex13-5.00928 Lack-of-Fit 3 0.020 Interaction 1 0.703 0.132 0.43226 46.4071 0.04818 1.01563 1. one approach to solving this exercise is to create a two-factor response surface design in MINITAB. To analyze the experiment.03228 0.000 Linear 2 2.2980 0.580 0.0550 0. AB) is selected.01563 2. (a) The design is a CCD with k = 2 and α = 1.03424 -11. Select “Designs”.000 Square 2 0.7273 0.1249 x2*x2 -0.30 0.03271 0.16128 2.00807 Total 12 2.377 Pure Error 4 0. B. 0 13.0 x1 0.0 -0.4 will maximize viscosity.5 1 0 -1 -1.00 14. From the plots and the optimizer. setting x1 in a range from 0 to +1.00 0.5 x2 Surface Plot of y vs x2.25 12. x1 0.75 13.5 -1.0 y < > 12.5 0.5 y 13. or using MINITAB’s Response Optimizer.75 14.25 13.75 14.0 x1 0 x2 -1 1 Stat > DOE > Response Surface > Response Optimizer In Setup.00 12.0 12.00 13.00 12.25 13.5 1.25 14.25 12.50 12.50 13. and Weight = 7. Lower = 10. Target = 20.75 13.00 13. 13-7 .0 12. x1 1.25 14.4 and setting x2 between -1 and -1.50 13.50 12. let Goal = maximize.Chapter 13 Exercise Solutions 13-5 (b) continued Values of x1 and x2 maximizing the Mooney viscosity can be found from visual examination of the contour and surface plots.0 -0. Stat > DOE > Response Surface > Contour/Surface Plots Contour Plot of y vs x2. 4708 x2*x3 43.97 1. and then enter the data.4256 x3 131. it must also be defined using Stat > DOE > Response Surface > Define Custom Response Surface Design.000 x2 109.43 17.064 … Analysis of Variance for y1 Source DF Seq SS Adj SS Adj MS F P Regression 9 1248237 1248237 138693 23.3844 x3*x3 -29.000 Linear 3 1090558 1090558 363519 62.06 31.03 21. The design and data are in the MINITAB worksheet Ex13-6.94 0. then select “Factors” to specify the actual level values.5833 13-8 .00 17.38 31. (a) To analyze the experiment. In order to analyze this experiment using the Response Surface functionality.000 Square 3 14219 14219 4740 0.0283 x1*x3 75. Estimated Regression Coefficients for y1 Term Coef SE Coef T P Constant 327.97 3. Select “Terms” and verify that a full quadratic model is selected.328 0.000 x3 131.25 0.983 0. select Stat > DOE > Response Surface > Analyze Response Surface Design.74 0.94 7.003 x2*x3 43.935 0.94 6. Response Surface Regression: y1 versus x1.481 x3*x3 -29. Select Stat > DOE > Factoriall > Create Factorial Design. x2.099 0.58 21.453 0.01 31. x3 The analysis was done using coded units.866 0.435 0. and select the number of factors as “3”.4656 x1*x1 32.000 x1*x1 32.94 9.001 Residual Error 17 98498 98498 5794 Total 26 1346735 … Estimated Regression Coefficients for y1 using data in uncoded units Term Coef Constant 327.0056 x2*x2 -22.502 Interaction 3 143461 143461 47820 8.005 0.6237 x1 177.363 x1*x2 66.47 21.008 x1*x3 75.97 3.Chapter 13 Exercise Solutions 13-6. one approach to solving this exercise is to create a general full factorial design in MINITAB.MTW. Change the design type to a general full factorial design.47 17.82 0.76 8.720 0.0578 x1*x2 66. Since the runs are listed in a patterned (but not standard) order.08 -0.62 38.08 1.08 -0. The design is a full factorial of three factors at three levels.0011 x2 109.000 x1 177.030 0. Select “Designs” to establish three levels for each factor.317 x2*x2 -22. 33 2.0 3408.691 x2*x3 14.31 0.89 0.202 0.404 0.082 12.3 7319.89 -0.5278 x2 15.817 x2*x2 -1.483 0.319 17.3 21957.116 0.815 0.1978 x2*x2 -1.8896 x1 11.192 10.890 22.65 1.235 0.012 x1*x1 4.3233 x3 29.59 P 0. Estimated Regression Coefficients for y2 Term Coef SE Coef T P Constant 34. Response Surface Regression: y2 versus x1.361 x1*x2 7.99 Residual Error 17 32650.281 … Analysis of Variance for y2 Source DF Seq SS Adj SS Adj MS Regression 9 27170.Chapter 13 Exercise Solutions 13-6 continued (b) To analyze the experiment. x3 The analysis was done using coded units.33 1.074 0.33 1.198 17.938 0.81 0.0 1135.5 1805.2 1920.97 Linear 3 21957.3189 x3*x3 16.65 0.826 0.09 Square 3 1805.7 27170.610 0. Select “Terms” and verify that a full quadratic model is selected.1083 x2*x3 14.280 x2 15.564 0.7192 x1*x3 5.7 3018.60 Total 26 59820. x2.156 x3 29.629 data in uncoded units 13-9 .89 0.31 1.550 x1*x3 5.0825 F 1.942 x3*x3 16.9 … Estimated Regression Coefficients for y2 using Term Coef Constant 34. select Stat > DOE > Response Surface > Analyze Response Surface Design.779 17.528 10.57 3.7794 x1*x2 7.323 10.5 601.113 0.65 0.136 x1 11.108 12.82 Interaction 3 3408.2 32650.030 0.1917 x1*x1 4.719 12. Stat > DOE > Response Surface > Overlaid Contour Plot After selecting the responses.0 13-10 .0 -0.0 -0.0 Overlaid Contour Plot of y1.0 -1. The goal is to minimize y2 (standard deviation) set low = 0 (the minimum of the observed results) and high = 80 (the 3rd quartile of the observed results).0 y1 400 600 y2 0 80 0.5 0.5 -1.0 y1 400 600 y2 0 80 0. set low = 400 and high = 600. Overlaid Contour Plot of y1.0 -0. Since the goal is to hold y1 (resistivity) at 500.0 x1 0. select the first two factors x1 and x2.5 x2 Hold Values x3 -1 0.5 1.5 -1. y2 1.0 x1 0.5 1.0 -1.5 x2 Hold Values x3 0 0.Chapter 13 Exercise Solutions 13-6 continued (c) Both overlaid contour plots and the response optimizer can be used to identify settings to achieve both objectives. y2 1.0 -0. Select “Contours” to establish the low and high contours for both y1 and y2.5 0. 0 -0. Lower = 400. Leave all Weight and Importance values at 1. Target = 0.5 1.5 0.0 y1 400 600 y2 0 80 0.5 x2 Hold Values x3 1 0.4.0. 13-11 . At x1 = 1.0 Stat > DOE > Response Surface > Response Optimizer In Setup.16 and standard deviation is 44. x2 = 0. For y2.0 x1 0. set Goal = Minimize.5 -1. y2 1.0 -1.75. and Upper = 80.3 and x3 = -0. Upper = 600.Chapter 13 Exercise Solutions 13-6 (c) continued Overlaid Contour Plot of y1. the predicted resistivity mean is 495. The graph below represents one possible solution.0 -0. for y1 set Goal = Target. Target = 500. All main effects are clear of 2-factor interactions.Chapter 13 Exercise Solutions 13-7. Enter the factor levels and response data into a MINITAB worksheet. The design and data are in the MINITAB worksheet Ex13-7. D. (a) The defining relation for this half-fraction design is I = ABCD (from examination of the plus and minus signs). Stat > DOE > Factorial > Analyze Factorial Design Factorial Fit: Mean versus A. E … Alias Structure I + A*B*C*D A + B*C*D B + A*C*D C + A*B*D D + A*B*C E + A*B*C*D*E A*B + C*D A*C + B*D A*D + B*C A*E + B*C*D*E B*E + A*C*D*E C*E + A*B*D*E D*E + A*B*C*E 13-12 .MTW. B. C. and then define the experiment using Stat > DOE > Factorial > Define Custom Factorial Design. but some 2-factor interactions are aliased with each other. A+BCD B+ACD C+ABD D+ABC E AB+CD AC+BD AD+BC AE+BCDE BE+ACDE CE+ABDE DE+ABCE ABE+CDE ACE+BDE ADE+BCE This is a resolution IV design. 0496 -0.6256 0.02021 5.02021 377.03 0.02021 0.41 0.0098 0.001 C*E -0.89 0.1210 0.575 A*E 0.0012 0.1638 -0.07 0.01994 2.76 0.0229 -0.000 Residual Error 42 0.45224 23.000 B*E 0.57 0.0329 -0.000 B -0.959 A*C*E 0.0819 0.02021 0.02021 3.0596 -0.99 0.8090 1.2806 0.501 Residual Error 32 0.05 0.99 0.02021 -0.02021 0.Chapter 13 Exercise Solutions 13-7 continued (b) The full model for mean: Stat > DOE > Factorial > Analyze Factorial Design Factorial Fit: Height versus A.02021 -1.0913 0.23 0.124 B*E 0.421 D*E 0.02021 -1.2421 0.000 B -0.000 2-Way Interactions 7 0.36769 18.0912 0.37800 0.000 … Analysis of Variance for Height (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 4 1. D.01994 -5.71 0.2806 0.62707 0.000 A 0.01960 Total 47 2.631 A*D*E -0.01994 382.71 0.02021 1.0819 0.62707 0.1529 0.04726 0.02021 -0.80 0.29 0.554 Pure Error 32 0.0006 0.98 0.000 A 0.01575 0.62707 0.0765 0.000 A*B -0.1194 0.78 0.83846 0.000 D 0. B.000 2-Way Interactions 1 0.6256 0.6271 0.89078 The reduced model for mean: Factorial Fit: Height versus A.1529 0.02021 -4.0298 0.000 C -0.0115 0.11 0.6271 0.0165 0.01994 -4.8012 0.62707 0. E Estimated Effects and Coefficients for Height (coded units) Term Effect Coef SE Coef T P Constant 7.229 D 0.01742 0.8012 0.48 0.05 0.81 0.05400 2.26 0.976 A*D -0.84 0.8908 13-13 .8090 0.51 0.73 0.01908 Lack of Fit 10 0.01994 3.83846 1.0456 0.031 E -0. E Estimated Effects and Coefficients for Height (coded units) Term Effect Coef SE Coef T P Constant 7.1638 -0.01960 Total 47 2. C.2387 -0.0196 0.2387 -0.02021 -0.1210 0.1742 0.027 E -0.0456 0.0296 -0.04726 0.02021 2.0148 0.76 0.0021 0.0396 0.0637 0.0010 0.023 3-Way Interactions 3 0.02021 -5.47 0.469 A*C 0.28060 14.0319 0.0198 0.335 A*B*E 0.0248 0.150 … Analysis of Variance for Height (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 5 1.01960 Pure Error 32 0.01994 6. B.1194 0.1742 0.02021 0.91 0.58 0.0765 0. D.2421 0.37800 0. 02625 0.27 0.12625 -0.07426 0.13403 0.02125 -0.07426 0.06312 0.01625 -4.050625 The reduced model for range: Factorial Fit: Range versus A. E Effects and Coefficients for Range (coded units) Effect Coef 0.01313 A DE 95 0.23 0.01625 -0.01375 -0.000 A 0.03380 0.00375 -0.01313 0.683 C*E -0.13403 0.01562 1 0.06937 Percent Estimated Term Constant A B C D E A*B A*C A*D A*E B*E C*E D*E A*B*E A*C*E A*D*E Effect Ty pe Not Significant Significant F actor A B C D E N ame A B C D E Lenth's PSE = 0.011 2-Way Interactions 1 0.096 E -0.15 -0.008 B -0.03062 0.003 Residual Error 8 0.11375 0. B.01625 3.13875 0. C.005 C 0.21937 0.06812 5 CE -0.06937 0.01625 -3.05 0.10 -0.12625 -0.13625 -0.03625 0.003 3-Way Interactions 1 0.01063 0.02437 -0.06125 0.00687 80 0.07701 0.04875 0.10 0.06125 0.01625 13.01625 1.01688 60 50 0.88 0.06312 99 0.31909 … 13-14 .443 D 0.06812 0.34 0.21937 Normal Probability Plot of the Effects 0.004225 Total 15 0.07701 0.003 A*D*E 0.03380 0.05 0.58 0.50 0. C.00188 30 20 0. D.01625 4.026806 6. E Estimated Effects and Coefficients for Range (coded units) Term Effect Coef SE Coef T P Constant 0.01375 -0.00 0.05688 0.003 … Analysis of Variance for Range (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 5 0.074256 17.00812 B 10 -0.02188 70 -0.02625 0.19 0.01625 0.88 0.04375 0.05688 (response is Range.03375 -0.13875 0.50 0.11375 0.01812 40 -0. B. D.42 0.15 Effect 0.Chapter 13 Exercise Solutions 13-7 continued (c) The full model for range: Factorial Fit: Range versus A. Alpha = .03062 90 A -0.03125 0.077006 18.81 0.01625 0.13625 -0.00687 0.20) -0. 00528 0.06259 0.00234 0.00528 A DE 95 0.07148 -0.504 D 0.01540 0.03536 0.11744 0.00684 -0.Chapter 13 Exercise Solutions 13-7 (c) continued The full model for standard deviation: Factorial Fit: StdDev versus A.34 0.00770 70 60 -0.00778 1 -0.00953 30 -0.007559 -0.007314 0.01093 50 40 0.00329 -0.023369 0. Alpha = .00342 D 80 0.001 0.06 0. B.00877 0.00999 Effect 0.020438 0.007559 4.20) 0.003 B -0.0232179 The reduced model for standard deviation: Factorial Fit: StdDev versus A.70 0.041748 0.047 E -0.06 -0. 13-15 .007559 -4.007314 0.03574 0.03574 5 B -0.041748 0. E Effects and Coefficients for StdDev (coded units) Effect Coef Normal Probability Plot of the Effects 0.03822 Percent Estimated Term Constant A B C D E A*B A*C A*D A*E B*E C*E D*E A*B*E A*C*E A*D*E Effect Ty pe Not Significant Significant F actor A B C D E N ame A B C D E Lenth's PSE = 0. D.007559 2.00 0.00468 -0.0009142 Total 15 0.36 3-Way Interactions 1 0.03822 0.0233690 25.07643 0. C.03574 0. E Estimated Effects and Coefficients for StdDev (coded units) Term Effect Coef SE Coef T P Constant 0.00438 CE 10 -0.007559 15. interactions CE (transfer time × quench oil temperature) and ADE=BCE.00684 -0.11744 (response is StdDev.03129 0.001 A*D*E 0.13 2-Way Interactions 1 0.02 0.0204385 22. Factors C and E are included to keep the models hierarchical.07149 -0. B.03536 0.00165 20 0. along with factors B (heating time) and A (furnace temperature) are significant.001 C 0.01556 0.04 0.14 0.03129 99 -0.06 0.54 0.01997 0.07148 -0.000 A 0.02 0.023369 0.56 Residual Error 8 0. C.020438 0.0083496 9.03574 0.08 0.01906 0.001 … Analysis of Variance for StdDev (coded units) Source DF Seq SS Adj SS Adj MS F Main Effects 5 0.08 -0.663 C*E -0.06259 0.45 0. D.007559 5.07643 0.01057 0.02185 -0.07149 -0.73 0.092869 P 0.00342 0.01768 90 A -0.73 0.001 For both models of variability.007559 -4.004 0.01057 0.007559 0.01768 0.04 -0. 0 -0.5 1.0 1.5 1.5 Residuals Versus C (response is Height) 2 -1.4 Residuals Versus the Order of the Data 1 0 -1 1 2 3 4 5 6 7 8 (response is Range) -1.5 0.5 1.5 Standardized Residual Residuals Versus A Residuals Versus the Fitted Values (response is Height) 2 0 -2 7.0 D 0.5 0.5 -1 Residuals Versus B -1 -1.5 0.0 0.0 0 Residuals Versus D 0 -2 -0.0 -0.0 -0.0 -1.2 Fitted Value 0.75 Fitted Value 8.0 0.0 0.0 -1.5 0.5 0.0 1.5 0.0 0.0 D 0.0 -2 -1.0 0 9 10 11 12 13 14 15 16 0 -0.0 (response is Height) 0 -0.0 1.5 -1.0 2 -1.00 Residuals Versus the Order of the Data 2 0 Standardized Residual Percent Normal Probability Plot of the Residuals 99 90 50 10 1 2 0 -2 -2 1 5 -1.5 1.0 A 0.0 1 0 -1 -1.0 1.0 -1.5 Residuals Versus B Standardized Residual Standardized Residual -2 0.0 E 13-16 .0 B 0.0 -0.0 B 0.5 -1.5 0.0 E Standardized Residual Standardized Residual For range: Residual Plots for Range Histogram of the Residuals 3.5 Standardized Residual 0.0 A Residuals Versus C (response is Range) (response is Range) 0.0 -1.5 2 0 -2 1.0 -0.5 1.5 0.5 0.5 1 0 -1 1.0 C Residuals Versus E (response is Height) (response is Height) Standardized Residual Standardized Residual 1.5 0.0 Standardized Residual 0 2 Standardized Residual 1 0 -1 Standardized Residual -2 Residuals Versus A Residuals Versus the Fitted Values Standardized Residual Standardized Residual Frequency Percent Normal Probability Plot of the Residuals 99 90 50 10 1 0.5 0.5 1.50 7.0 -0.5 1 Observation Order 1 -1.0 0.5 2 1.0 10 15 20 25 30 35 40 45 Observation Order -0.Chapter 13 Exercise Solutions (d) Standardized Residual Standardized Residual For mean height: Residual Plots for Height -2 0 Standardized Residual 2 Frequency Histogram of the Residuals 10 5 0 -2.0 0.0 -0.5 1.0 C Residuals Versus E (response is Range) (response is Range) Standardized Residual Residuals Versus D 1 0 -1 -0.0 0. 0 0.0 0 Residuals Versus D -0.5 0.5 0.0 0.0 0.Chapter 13 Exercise Solutions 13-7 (d) continued Standardized Residual Standardized Residual For standard deviation: Residual Plots for StdDev -2 0 Standardized Residual 2 Histogram of the Residuals 4 2 0 -1.5 0.0 2 -1. Standard Deviation Residuals versus predicted plot and residuals normal probability plot support constant variance and normality assumptions.0 -0. Plots of residuals versus each factor shows that variance is less at low level of factor E. Normal probability plot of residuals support normality assumption.2 Fitted Value Residuals Versus the Order of the Data 2 0 -2 1 2 3 4 5 6 7 8 Standardized Residual Frequency Percent Normal Probability Plot of the Residuals 99 90 50 10 1 2 0 -2 -1.0 -0.0 E Mean Height Plot of residuals versus predicted indicates constant variance assumption is reasonable. Plots of residuals versus each factor indicate that the variance may be different at different levels of factor D.5 -1.0 C 1.0 1.0 B 0. 13-17 . (e) This is not the best 16-run design for five factors.5 2 1.5 Standardized Residual 2.0 Residuals Versus E (response is StdDev) Standardized Residual Standardized Residual 0. then none of the 2-factor interactions will be aliased with each other.5 1.5 -1.0 -0.0 Residuals Versus A Residuals Versus the Fitted Values (response is StdDev) 2 0 -2 0.5 0.5 0.0 A 0.0 (response is StdDev) 0 -0.0 2 0 -2 -1.5 Observation Order Residuals Versus B Standardized Residual Standardized Residual -2 0.5 Residuals Versus C (response is StdDev) 2 -1.5 1.0 D 1. Residuals normal probability plot indicate normality assumption is reasonable Plots of residuals versus each factor indicate that the variance may be different at different levels of factor D.0 1.0 9 10 11 12 1 3 14 15 16 -0.1 0. A resolution V design can be generated with E = ± ABCD.5 -2 (response is StdDev) 0 -2 0. Range Plot of residuals versus predicted shows that variance is approximately constant over range of predicted values. h 14 51 04 X -0 .081B + 0. 55 0.00 -0 . B. 49 59 4 POE( free height): 0. Using equations (13-6) and (13-7) the mean and variance models are: Mean Free Height = 7.12 + 0.077B)2 + 0.12A – 0.Chapter 13 Exercise Solutions 13-8. 99 A ctu a l Fa cto rs C: tra n s tim e = 0 .00 -0.00 O ve rl a y Pl o t X = A: fu rn te m p Y = B: h e a t tim e f ree he igh7. This approach is fully described in the solution to Exercise 13-12.056 at A = –0.63 + 0. The overlaid contour plot below (constructed in Design-Expert) shows one solution with mean Free Height ≅ 7.00 A: fu rn te m p 13-18 .49 and minimum standard deviation of 0.15 PO E(f re e 0.4 1 Y 0.046D Variance of Free Height = σ2E (–0.50 0 . and D could be constructed using the Graph > Contour Plot functionality. and optimal settings could be identified from visual examination of both plots.0 0 -1 . Free Height Variance could be calculated in the MINITAB worksheet. and then contour plots in factors A.5 0 -1.99.50 1.0 0 B: h ea t tim e 0.077B)2 + σ2 Assume (following text) that σ2E = 1 and σˆ 2 = MSE = 0.50 f r ee h eight: 7. Overlay Plot DES IG N-E XP ERT P lo t 1. Factor E is hard to control (a “noise” variable). These contour plots could be compared with a contour plot of Mean Free Height.0 0 D: h o ld ti m e = 0 .00 0.44 and B = 0.02 For the current factor levels.0 0 E : o il te m p = 0 . so Variance of Free Height = (–0.12 + 0.02 . 9 9 A ctu a l Fa cto rs C: tra n s tim e = 0 .02 : Variance of Free Height = (0. Free Height Variance could be calculated in the MINITAB worksheet.0 0 D: h o ld ti m e = 0 . These contour plots could be compared with a contour plot of Mean Free Height.00 -0 .50 f r ee h eight: 7.00 0.5 0 -1.16 Y 0 . Assume σ D2 = σ E2 = 1 .50 and minimum standard deviation of Free Height to be: A = – 0. Factors D and E are noise variables.00 -0.00 O ve rl a y Pl o t X = A: fu rn te m p Y = B: h e a t tim e f ree he igh7 .077B)2 + 0.0 0 B: h ea t tim e 0. and then contour plots in factors A.046)2 + (–0. Overlay Plot DES IG N-E XP ERT P lo t 1. Using equations (13-6) and (13-7).42 and B = 0. 42 POE( free height): 0.081B Variance of Free Height = σ2D (+0. This approach is fully described in the solution to Exercise 13-12.63 + 0.99.50 0 . 55 0. The overlaid contour plot below (constructed in Design-Expert) shows one solution with mean Free Height ≅ 7.Chapter 13 Exercise Solutions 13-9. and optimal settings could be identified from visual examination of both plots. and D could be constructed using the Graph > Contour Plot functionality. the mean and variance are: Mean Free Height = 7.046)2 + σ2E (–0. B.1 52 23 3 X -0.12 + 0.12A – 0.00 A: fu rn te m p 13-19 .50 1.0 0 -1 .12 + 0.077B)2 + σ2 Using σˆ 2 = MSE = 0.4 95 01 PO E(f re e 0h.0 0 E : o il te m p = 0 .02 For the current factor levels. 38 292. 91.04 Pure Error 5 45. 78.175 0.446 x1*x2 8.466 2.359 1.0509 x1*x1 -5. 60.33 45.794 0.689 -6.073 x2*x3 -13.3613 x3 -6.723 23.49 Interaction 3 292.29 277.526 0.7759 x1*x2 2.227 uncoded units 13-20 . 90.689 1. 83. Note: Several y values are incorrectly listed in the textbook.8862 x3*x3 0. These values are used in the Excel and MINITAB data files.000 x1*x1 -14.764 -5.74 405. The correct values are: 66.766 Lack-of-Fit 5 92.425 3.195 2. 87. 80.38 97.33 18.764 -8. Since the runs are listed in a patterned (but not standard) order.66 13. 65.008 0. 85.17 329. 100.8750 x1*x3 -2.33 92.081 3.0578 x2*x2 -7.513 57.08 Residual Error 10 137.000 x3*x3 2.000 x1 9.069 0.053 x1*x3 -7. x3 The analysis was done using coded units.710 -3.005 … Analysis of Variance for y Source DF Seq SS Adj SS Adj MS F Regression 9 2499.29 2499.74 1217.067 Total 19 2636.000 x2*x2 -22.000 x2 2.205 x3 -10. 70.027 0. Stat > DOE > Response Surface > Analyze Response Surface Design Response Surface Regression: y versus x1.305 2. 82.192 0.305 2.805 0.356 0.914 29. 100.699 20. one approach to solving this exercise is to create a general full factorial design in MINITAB.000 0.764 0.33 9.6250 P 0.176 1. 75. Estimated Regression Coefficients for y Term Coef SE Coef T P Constant 87.MTW. 88. x2. 63.689 5.801 1. The design and data are in the MINITAB worksheet Ex13-10.730 0.710 2.95 Square 3 1217.17 989.8279 x2 1. 70.132 3.710 -2. and then enter the data.001 0.Chapter 13 Exercise Solutions 13-10. 80.289 1.95 … Estimated Regression Coefficients for y using data in Term Coef Constant 87.6250 x2*x3 -4.17 Linear 3 989.000 0.3589 x1 5.458 7.000 0. 68.66 137. 834 Pure Error 5 45.33 45.0 0 2.994 1.38 292.257 Residual Plots for y Normal Probability Plot of the Residuals (response is y) 5 90 Residual 50 10 5.329 0.006 1.000 0.5 0 -5.704 -8.048 x1*x3 -7.000 0.425 3.0 -5.34 146.523 2.86 0.44 7.00 101.33 9.587 0.176 1.34 13.07 1209.647 2. x2.523 2.801 1. Estimated Regression Coefficients for y Term Coef SE Coef T P Constant 87.660 1.327 Linear 3 989.000 x1 9.5 -5 1 -5 0 Residual 5 60 Histogram of the Residuals 80 Fitted Value 100 Residuals Versus the Order of the Data Residual Percent Residuals Versus x1 Residuals Versus the Fitted Values 99 Residual Frequency 5 4 2 0.00 16.685 0.004 … Analysis of Variance for y Source DF Seq SS Adj SS Adj MS Regression 8 2490.132 3.263 69.0 -2.371 0.33 P 0.647 -2.5 2.95 … F 23.081 3.61 311.458 Residual Error 11 146.131 0.647 -3.0 -2.17 989.5 -5.303 Lack-of-Fit 6 101.07 604.704 -5.5 Residual Residual 0 x1 0.5 0.61 2490.660 5.723 Square 2 1209.067 x2*x3 -13.534 Interaction 3 292.036 0.17 329.229 0.905 0.000 x2*x2 -22. x3 The analysis was done using coded units.0 -5 0 -4 -2 0 2 Residual 4 2 4 -2 6 8 10 12 14 16 18 20 Observation Order -1 Residuals Versus x2 1 2 1 2 Residuals Versus x3 (response is y) (response is y) 5.38 97.000 x2 2.000 0.000 x1*x1 -14.0 5.40 24.289 1.195 x3 -10.78 45.379 0.Chapter 13 Exercise Solutions 13-10 continued Reduced model: Response Surface Regression: y versus x1.000 x1*x2 8.0 -2 -1 0 x2 1 2 -2 -1 0 x3 13-21 .067 Total 19 2636.660 -6.0 2.0 -2. 0 90. Lower = 60.0 80.807. 13-22 .0 100. x1 Surface Plot of y vs x3.292.0 70.0 80.0 60.0 1 x2 Surface Plot of y vs x2. and x3 = −1. Weight = 1.0 60. Upper = 120.0 70. x1 y < > Hold Values x3 0 40.682.0 100.0 Hold Values x3 0 0 80 y 60 2 40 -1 0 -2 0 x1 -1 0 x1 -2 1 Contour Plot of y vs x3.0 > 1 Hold Values x2 0 70.0 50. x2 = 0.0 Hold Values x2 0 x3 2 x2 105 0 90 y 75 2 60 -1 0 -2 0 x1 -1 0 x1 2 x3 -2 1 Stat > DOE > Response Surface > Response Optimizer Goal = Maximize.0 90. x1 40. Importance = 1 One solution maximizing growth is x1 = 1.0 50.Chapter 13 Exercise Solutions 13-10 continued Stat > DOE > Response Surface > Contour/Surface Plots Contour Plot of y vs x2.0 80. x1 y < 70. Predicted yield is approximately 108 grams.0 80. 011 Square 2 123.700 Total 12 470. Since the runs are listed in a patterned (but not standard) order.58 61.119 Linear 2 48.60 139.463 1.100 19.457 1.085 0.200 2.076 x2*x2 2.383 0.09 2.60 46.60 63.660 -1.017 13-23 . The design and data are in the MINITAB worksheet Ex13-11.58 0.616 0.60 315.209 x1*x2 6.02 48.00 144.80 3.788 Interaction 1 144. one approach to solving this exercise is to create a general full factorial design in MINITAB. x2 The analysis was done using coded units.128 0.348 2.MTW.712 1.186 0.409 x1*x1 3.534 Pure Error 4 14.000 2.388 0.Chapter 13 Exercise Solutions 13-11.274 x2 1.000 x1 -1.86 1.40 22.038 … Analysis of Variance for y Source DF Seq SS Adj SS Adj MS Regression 5 315. Stat > DOE > Response Surface > Analyze Response Surface Design Response Surface Regression: y versus x1.00 F 2.660 0.40 154.555 0.781 1.80 6.058 Lack-of-Fit 3 139.878 0.038 12.000 Residual Error 7 154. and then enter the data.970 1.102 0.80 14.58 123.781 2.00 144.02 24.53 P 0. Estimated Regression Coefficients for y Term Coef SE Coef T P Constant 41. 4109 and pressure = -1.0 (a) Goal = Minimize. x1 y < > 40 45 50 55 60 60 70 0. Importance = 1 Recommended operating conditions are temperature = +1. 13-24 .0785. to achieve predicted filtration time of 36.0 40 45 50 55 0.0 60 y -0. to achieve predicted filtration time of 46.3415 and pressure = -0.0.0 x1 0. Upper = 50.4142. Lower = 42. Weight = 1.7.0 -0. x1 1. Importance = 1 Recommended operating conditions are temperature = +1.5 0. Upper = 55.5 50 40 -1.5 0 x2 -1 1 1.Chapter 13 Exercise Solutions 13-11 continued Contour Plot of y vs x2. Target = 46. Target = 0.5 x2 Surface Plot of y vs x2.0 1 0 -1 x1 -1. (b) Goal = Target. Weight = 10. 001 x2 2.Chapter 13 Exercise Solutions 13-12.22 30.4147 2.448 -3.873 5.455 -4.3333 1.8333 3. z The analysis was done using coded units.8750 x1*z -2.084 x1*z -4.873 1.8013 1.209 0.072 0.945 x1*x2 8.7783 2.50 … Estimated Regression Coefficients for y using data in Term Coef Constant 87.458 5. Estimated Regression Coefficients for y Term Coef SE Coef T P Constant 87.115 uncoded units The coefficients for x1z and x2z (the two interactions involving the noise variable) are significant (P-values ≤ 0. The design and data are in the MINITAB worksheet Ex13-12.1517 x1*x2 2.074 3.10).94 2034.75 Interaction 3 292.945 Lack-of-Fit 3 90.6250 P 0.1250 1.003 x2*x2 -21.33 45.34 Linear 3 789.33 9.22 90.38 292.178 0.3613 z -6.448 -1.38 97.256 z -6.8333 3.021 0.109 x2*z -7.8279 x2 1.56 135.116 0.29 953.001 0.8908 x2*x2 -7.001 0.1317 4. x2. 13-25 .105 13.56 16.804 0.001 0.092 15.29 317.1517 2.6250 x2*z -4.116 0.968 0.000 x1 9.32 Pure Error 5 45.3333 x1 5.116 1.764 18.975 0.7192 z*z 0.681 51.28 263.496 0.75 Residual Error 8 135.067 Total 17 2170.53 Square 3 953.94 226.013 … Analysis of Variance for y Source DF Seq SS Adj SS Adj MS F Regression 9 2034.361 -6.1250 x1*x1 -4.MTW Stat > DOE > Response Surface > Analyze Response Surface Design Response Surface Regression: y versus x1.232 0.2894 1. so there is a robust design problem.222 0.003 x1*x1 -13.361 -4.28 789.000 z*z 0. Estimated Regression Coefficients for y Term Coef SE Coef T P Constant 87.49 0.88 17.002 x1*x1 -13.013 2.0 -5 -4 -2 0 2 Residual 4 2 4 6 8 10 12 14 16 Observation Order -2 18 -1 Residuals Versus x2 1 2 Residuals Versus z (response is y) (response is y) 5.0 0 2.072 Lack-of-Fit 4 90.5 2.912 0.0 -2 -1 0 x2 1 2 -1.0 z 0.Chapter 13 Exercise Solutions 13-12 continued Reduced model: Response Surface Regression: y versus x1. z The analysis was done using coded units.000 0.46 31.0 -2.602 Interaction 3 292.067 Total 17 2170.33 45.208 0.088 x2*z -7.5 -5 -5 0 Residual 5 60 Histogram of the Residuals 80 Fitted Value 100 Residuals Versus the Order of the Data Residual Percent 50 1 0.462 0.172 Residual Plots for y Normal Probability Plot of the Residuals Residual 10 5.008 … Analysis of Variance for y Source DF Seq SS Adj SS Adj MS Regression 8 2034.5 1.373 -4.548 0.31 90.132 3.675 0.092 Square 2 953.38 292.50 … F 16.20 476.5 Residual Residual 0 x1 0.5 0.760 3.361 1.767 1.125 1.64 15.62 6.767 5.458 Residual Error 9 135.370 0.28 789.0 -2.31 22.64 135.28 263.33 9.0 5.5 0.0 5 Residual 3.5 0 -5.415 2.86 254.000 x2 2.227 z -6.0 Frequency (response is y) 5 90 1.000 0.095 0.357 Linear 3 789.20 953.558 0.289 1.38 97.019 -7.0 Residuals Versus x1 Residuals Versus the Fitted Values 99 -2.308 -3.308 -1.760 3.000 0.001 x2*x2 -21.0 2.778 2.578 Pure Error 5 45.0 -0.541 56.066 x1*z -4.5 -5.000 x1*x2 8.0 13-26 .000 x1 9.296 0.86 2034.47 P 0.882 2. x2.019 -4.0 -5.801 1.5 0. 0 1 x2 x2 -1 0 x1 Contour Plot of y vs x2.36 + 5.109756 x2 = -0.0 80.0 60.) Contour Plot of y vs x2.0 90.0 60.63x1 – 4.293782 y = 90.0 80.0 Hold Values z -1 0 x1 = -0.0 1 -1 Contour Plot of y vs x2.0 90.63x2)2 + σˆ 2 = (–6.0 40.0 70.13 – 2.83x1 + 1.2054 0 x1 40.0 50.31 (mean yield of about 90 with a standard deviation between 6 and 8). set z = 0: Mean Yield = 87.0 Hold Values z 1 0 -1 -1 0 x1 1 -1 0 x1 1 Examination of contour plots for Free Height show that heights greater than 90 are achieved with z = –1.83x1 + 1.0 1 x2 50.0 40.0 50.0 70.10 10 . Comparison with the contour plot for variability shows that growth greater than 90 with minimum variability is achieved at approximately x1 = – 0.0 60.0 60.Chapter 13 Exercise Solutions 13-12 continued yPred = 87. x1 y < > -1 -1 Contour Plot of sqrt{Vz(y(x.0 80.z)]} < 6 6 8 8 .63x1 – 4.36x2 – 4.11 and x2 = – 0. There are other combinations that would work.36x2 – 4.0 60.0 80.0 70.0 60.555312 y = 90. x1 1 30. 13-27 . x1 sqrt{Vz(y(x.z)]} vs x2.708407 x2 = -0.0 Hold Values z 0 0 1 1 y < > y < > 30.13 – 2. (Refer to MINITAB worksheet Ex13-12.0 50.0 70.0 50.072 This equation can be added to the worksheet and used in a contour plot with x1 and x2. x1 x2 x1 = -0.63x1 – 4.0198 -1 x1 = 0.63x2)2 + 15.1692 50.36 + 5.69x22 For the variance model.393500 x2 = 0.0 70.308367 y = 90.0 70.0 80.69x22 + (–6. assume σz2 = 1: Variance of Yield = σz2 (–6.MTW.13 – 2.86x12 – 7.0 80.86x12 – 7.63x2)z For the mean yield model.12 > 12 0 40. z ) = ∑ γ i + ∑ ∑ δ ui xu + ∑ ∑ λij ( zi + z j ) + 2∑ θ i zi . and u =1 ∂zi r If h(x.Chapter 13 Exercise Solutions 13-13. and i =1 i =1 i =1 u =1 i< j =2 i =1 ∂zi r ∑ r r k r r ⎡ k r ⎤ V [ y (x. z ) ] = V ∑ ⎢γ i + ∑ δ ui xu + ∑ λij ( zi + z j ) + 2θi zi2 ⎥zi + σ 2 i =1 ⎣ u =1 j >i ⎦ There will be additional terms in the variance expression arising from the last two terms inside the square brackets. z ) = ∑ γ i zi + ∑ ∑ δ ij xi z j + ∑ ∑ λij zi z j + ∑ θi zi2 . z ) ] = σ ∑ γ i + ∑ δ ui xu 2 z r r i =1 k u =1 k ) +σ 2 r 2 r If h(x. z ) ] = V ∑ ⎢γ i + ∑ δ ui xu + ∑ λij ( zi + z j ) ⎥zi + σ 2 i =1 ⎣ u =1 j >i ⎦ There will be additional terms in the variance expression arising from the third term inside the square brackets. then i =1 i =1 j =1 r ∂h(x. r k k ∂h(x. z ) = ∑ γ i zi + ∑ ∑ δ ij xi z j . 13-14. i =1 i =1 j =1 i< j =2 r k r ∂h(x. and i =1 i =1 i i =1 u =1 i< j =2 ∂zi r then ∑ r ⎡ k r ⎤ V [ y (x. z ) = γ i + ∑ δ ui xu . z ) r = ∑ γ + ∑ ∑ δ ui xu + ∑ ∑ λij ( zi + z j ) . z ) = ∑ γ i zi + ∑ ∑ δ ij xi z j + ∑ ∑ λij zi z j . r k r r r i< j =2 i =1 If h(x. then i =1 i =1 j =1 ( V [ y (x. 13-28 . 55528 0.26986 0.60 0.05324 0.95190 0.060 0.003 0.63633 0.030 0.04761 0.37160 0. HYPGEOMDIST.050 0.006 0.Chapter 14 Exercise Solutions Note: Many of the exercises in this chapter are easily solved with spreadsheet application software.090 0.010 0.27060 0.30556 0.xls.16434 0.09066 0.01547 0.99540 0.21807 0.002 0.27943 0.98998 0.86051 0.22339 0.080 0.060 0.12649 0.12989 0.02656 0.90475 0. and graphing functions in Microsoft® Excel were used for these solutions.66924 0.74015 0.00 0.00 Pr{acceptance} 0.20 1. p 0.020 0.120 0.070 0.19556 0.000 0. The BINOMDIST.100 0.73577 0.77831 0.07694 0.020 0.81840 0.02863 Pr{d<=c} 0.009 0.14467 0.005 0.60501 0.04533 0.96353 0.70382 0.20 0.140 p 14-1 .040 0.001 0. Solutions are in the Excel workbook Chap14.95121 0.33721 0.100 f(d=0) 0.92528 0.03379 Type-B OC Curve for n=50.97387 0. 14-1.040 0.98274 0.80 0.00896 0.28895 0.36417 0.12947 0.40 0.09994 0.008 0.40048 0.04428 0.91056 0.06725 0. c=1 1.007 0.24807 0.004 0.00515 f(d=1) 0.19000 0.080 0.99881 0.93910 0.08271 0.20249 0. 00194 0.41978 0.090 0.00003 0.01127 0.26899 0.040 0.36773 0.01687 0.120 0.Chapter 14 Exercise Solutions 14-2.67669 0.001 0.070 0.36120 0.16531 0.020 0.27341 0.05661 0.80 0.93796 0.00592 0.13262 0.27065 0.08118 0.050 0.00 0.04755 0.60 0.09880 0.99888 0.002 0.000 0.90479 0.040 0.22515 0.00030 0.97730 0.60577 0.54782 0.40 0.00208 0.00000 f(d=1) 0.00895 0.100 0.00388 0.00000 f(d=2) 0.030 0.00079 0.003 0.22281 0.006 0.03116 0.99649 0.99225 0.020 0.98590 0. c=2 1.16404 0.004 0.010 0.080 0.30441 0.07572 0.66978 0.20 0.92063 0.00008 0.36973 0.23214 0.100 0.00531 0.060 0.00205 0.34920 0.05347 0.01312 0.007 0.36603 0.09057 0.99985 0.14707 0.03319 0.74048 0.44789 0.01627 0.11826 0.080 0.07029 0. p 0.20 1.01978 0.04144 0.008 0.00476 0.02579 0.96641 0.14498 0.00071 0.40492 0.00000 Type-B OC Curve for n=100.00449 0.005 0.95327 0.00000 Pr{d<=c} 0.00162 0.18486 0.49536 0.00 Pr{acceptance} 0.14419 0.33068 0.060 0.12185 0.81857 0.00024 0.140 p 14-2 .009 0.200 f(d=0) 0. or β ≅ 0.600 0.10 (b) Type-B OC Curve for N=5000.Chapter 14 Exercise Solutions 14-3.9521.800 0.040 0.075) = 0.9521.020 0.140 p Pa (d = 35) = 0.10 (c) Based on values for α and β.080 0.00 0. n=50.000 0. or α ≅ 0.100 0.00 Pr{acceptance} 0.060 0. n=50.007) = 0.05 Pa (d = 375) = 0. 14-3 .10133.200 0.100 0.60 0.120 0. either is appropriate. or β ≅ 0.400 0.000 Pr{acceptance} 0.20 0.10133.05 Pa (p = 0.40 0. the difference between the two curves is small.000 0.000 0.120 0.040 0.140 p Pa (p = 0. c=1 1.080 0. or α ≅ 0.020 0.060 0.80 0. c=1 1. (a) Type-A OC Curve for N=5000. 05 = 0. p2 = 0. p2 = 0.01 = 0.04786 and β = 0.15. p1 = 0.12238. β = 0. resulting in actual α = 0.06. 14-6. select n = 35 and c = 1.99. 1 − α = 1 − 0. 14-5. 1 − α = 1 − 0.Chapter 14 Exercise Solutions 14-4.01.95.95. p1 = 0. p2 = 0. β = 0. the sampling plan is n = 80 and c = 7. 1 − α = 1 − 0. β = 0. p1 = 0.10 From the binomial nomograph.10. 14-4 .05.10 From the binomial nomograph. select a sampling plan of n = 300 and c = 12.02.10 From the binomial nomograph.05 = 0. 0000 0.0132 0.0000 0.0250 0.0050 0.025) = 0.8178 0.0000 0.18221 0.00006 -0.000.05 p 0.00000 1.00000 0.05 by Pa(N = 5.Chapter 14 Exercise Solutions 14-7.99903 0.0200 10 N2 = n1 = pmax = cmax = binomial Pr{reject} Pr{d<=20} 0.99991 0.00000 0.0010 0.0090 0.00967 0.0070 0.00000.7060 0.0700 N1 = n1 = pmax = cmax = binomial Pr{d<=10} 1.01175 0.9997 1.0027 0.00030 0.00001 0. LTPD = 0.00001 -0.00000 0.0001 0.0200 20 Pr{reject} 0.0400 0.11479 0.00000 0.00000 Different sample sizes offer different levels of protection.00027 -0.9667 0.0000 1.000.0001 1.00000 -0.0500 0.99729 0.00046 and Pa(N = 10.02395 0.000) = 0.0000 1.0600 0.99359 0.00000 0.9995 0.00001 0.08151 0.0015 0.00000 0. but pays for the high probability of rejecting acceptable lots like those with p = 0.00046 0.00000 0.03328 0.00046 0.00938 0.99999 0.025.99850 0.11183 0.0000 0.0100 0.0000 0.00000 1.9903 0.0000 0. For N = 5.00000 0.98676 0.99999 0.0003 1.00095 -0.0000 1. Pa(p = 0.0020 0.0000 0.8852 0.0030 0.025) = 0.0000 1.0200 0.00263 -0. 14-5 .000) = 0.00000 1.0010 0.00000 0.294.0000 0.00000 5000 500 0.4409 0. Pa(p = 0. while for N = 10.0080 0.0064 0.182.29404 0.99972 0.0004 0.99994 0. Also.00000 1.99959 0.4170 0.0000 0.0040 0.0060 0.0000 1.0000 difference 0.00600 -0. the consumer is protected from a LTPD = 0.0000 1.0300 0.00000 10000 1000 0.58304 0.55910 0. 3884 0.0800 0.0070 0.53107 0.1308 0.9994 1.9938 0.1702 0.0000 0.03815 0.0123 0.00000 1000 32 0.0050 0.0411 0.0090 0.0276 0.9971 0.01) = 0.0676 0.7961 0.77958 0.00002 0.3000 0.53457 0.2677 0. depending on the lot size.00000 5000 71 0.65622 0.0300 0.0000 0.Chapter 14 Exercise Solutions 14-8.98610 0.0008 0.75536 0.38898 0.8814 0.8531 0.0900 0.4689 0.8946 0.7179 0.9990 0.2446 0.0200 0.02758 0.96946 0.0020 0.9426 0.2959 0.5052 0.0000 This plan offers vastly different protections at various levels of defectives.0184 0.95886 0.97550 0.0000 1.6110 0.00622 0.94552 0.10543 0.0100 0.0700 0. at p = 0.0245 0.2444 0.11858 0.9724 0.9618 0.01) = 0.49484 0.70407 0.05741 0. and Pa(p = 0.0305 0.98767 0.0004 0.97238 0.0000 1.00136 0.01 0 N2 = n1 = pmax = cmax = binomial Pr{reject} Pr{d<=1} 0.01315 0.61157 0.1957 0.01 1 Pr{reject} 0.9868 0.9986 0.0030 0.56992 0.28210 0.07541 0.93236 0.0500 0.0400 0.1439 0.88316 0.0040 0.1168 0.7569 0.4301 0.2000 0.9463 0.0602 0.0080 0.00099 0.99382 0.0545 0.93982 0.3500 N1 = n1 = pmax = cmax = binomial Pr{d<=0} 0.05374 0.73230 0.81033 0.86924 0.14688 0.91107 0.1897 0.4949 for N = 5000.0006 0.24312 0.0139 0.85608 0.82981 0.75558 0.00292 0.0600 0.00063 0.0062 0.7323 for N = 1000. For example.01.4654 0.9246 0.0060 0. Pa(p = 0.20391 0.80432 0. p 0.00000 1.98157 0.0002 0. 14-6 .1000 0.3438 0.0010 0.00000 1.2204 0.0889 0. 010 0.15 0.20 0. c=1 2500 2000 ATI 1500 1000 500 0 0.30 p 14-7 .000 ATI = n + (1 − Pa )( N − n) = 35 + (1 − Pa )(2000 − 35) = 2000 − 1965 Pa Pa p( N − n) N = (1965 2000 ) Pa p AOQ = AOQL = 0.15 0.0234 ATI Curve for n=35.25 p AOQ Curve for n=35.05 0.020 AOQ 0. c = 1.000 0. n = 35. N = 2.10 0.25 0.00 0.20 0.025 0. c=1 0.005 0.015 0.Chapter 14 Exercise Solutions 14-9.05 0.10 0.00 0. 060 0.007 0.090 0.025 0.00142 0.42093 0.0000 3000 (a) OC Curve for n=150.010 0.0077 693 0. n = 150.050 0.03292 0.0022 2834 0.88019 0.98927 0.0061 404 0.01815 0.99646 0.0048 2517 0.97716 0.00523 0.0046 264 0.20 0.0053 328 0.003 0.00009 0.040 0.80948 0.008 0.0009 2948 0.80 0.006 0.070 0.00036 0.27341 0. N = 3000.004 0.045 0.60 0.95991 0.16932 0.001 0.0001 2996 0.015 0.060 0.00002 AOQ ATI 0.91092 0.080 0.0014 2906 0.10098 0.030 0.000 0.93769 0. c = 2 p 0.040 0.0000 2999 0.84615 0.020 0.60884 0.030 0.0087 AOQL 1265 0.0003 2985 0.080 p 14-8 .40 0. c=2 1.0009 151 0.002 0.00 Pr{accept} 0.035 0.0028 181 0.99951 0.00 0.0019 160 0.005 0.0000 3000 0.0034 2712 0.0065 2221 0.05840 0.Chapter 14 Exercise Solutions 14-10.0080 1800 0.0067 491 0.020 0.009 0.100 Pa=Pr{d<=2} 0.0037 215 0.070 0.010 0.050 0.0072 588 0. 0020 0.030 0.0080 0.050 0.010 0.050 0.0000 0.070 0.020 0.010 0.0030 0. c=2 AOQL ≅ 0.000 0.020 0.Chapter 14 Exercise Solutions 14-10 continued (b) AOQ Curve for n=150.060 0.080 p (c) ATI Curve for n=150.0070 AOQ 0.0050 0.0010 0.0060 0.080 p 14-9 .0100 0.040 0.0040 0.060 0.0090 0.030 0.070 0.0087 0. c=2 3500 3000 2500 ATI 2000 1500 1000 500 0 0.040 0.000 0. 09614 0.81080 0.41625 0.77403 0.0600 0.67671 0.2000 Fraction defective. Pa 0.0900 0.0500 0.89236 0.1200 0.0700 0.00109 0.31079 0.99474 0.0010 0.10368 0.09985 0.90745 0.0030 0.1010 0.1030 0.00000 OC Curve for n=50.83946 0.90386 0.22597 0.99952 0.0400 0.18920 0.98618 0.16054 0.0400 0.3000 0.99891 0.99794 0.20 0.1040 0.2000 0.45947 0.00000 1.32329 0.99657 0.98957 0.99985 0.00343 0.0070 0.0080 0.0090 0.54053 0.00 Probabilty of Acceptance. (a) N = 5000.00 0.0300 0.60 0.00048 0.01043 0.1600 0.88827 0.0800 0.Chapter 14 Exercise Solutions 14-11.0200 0.00758 0.80 0.00002 0.90015 0.09255 0.40 0.1400 0.1020 0.0050 0.99242 0.92157 0.0040 0. c=2 1.1800 0.00015 0.99998 0.1050 0.00206 0. c = 2 p Pa=Pr{d<=1} Pr{reject} 0.89632 0.1000 0.68921 0.10764 0.58375 0. n = 50.0060 0.1000 0. p 14-10 .0800 0.00526 0.0200 0.07843 0.0000 0.0020 0.11173 0.0100 0.0600 0.01382 0.99871 0.00129 0. 00080 0.Chapter 14 Exercise Solutions 14-11 continued (b) p = 0. not the sampling plan.07703 0.87842 0.0500 0.0030 0.99996 1.98019 0.88638 ≈ 0. Generally.0070 0.70989 0.0040 0.96075 0.05832 0.76576 0. will be extremely hard on the vendor because the Pa is low even if the lot fraction defective is low. (d) From the nomograph.0700 0.15164 0.90.81131 0.0100 0.01981 0.66761 0.11372 = 0.0020 0.4000 0.0060 0. quality improvement begins with the manufacturing process control. select n = 20.0050 0.3000 0.00000 14-11 .0900 0. yielding Pa = 1 – 0.00004 0.88660 0.98847 0.0800 0.5000 Pa=Pr{d<=0} 0.18869 0.33239 0.0010 0.0090 0.2000 0.54379 0.11340 0.1030 will be rejected about 90% of the time.0300 0. The OC curve for this zero-defects plan is much steeper.94168 0.84836 0.23424 0.99920 0. (c) A zero-defects sampling plan.09539 0.16541 0.0080 0.85160 0.01153 0.55800 0.90461 0.13107 0.45621 0.29011 0.92297 0.35849 0.18209 0.1000 0.81791 0.14840 0.64151 0. with acceptance number c = 0.44200 0.86893 0.00000 Pr{reject} 0.0400 0. p 0.03925 0.12158 0.0200 0.0600 0.83459 0. 005.80 0. 14-12 .40 0.00 Probability of acceptance.00206)(5000 − 50) = 60 The c = 2 plan is preferred because the c = 0 plan will reject good lots 10% of the time. c = 0} = 0.005.1500 0.Chapter 14 Exercise Solutions 14-11 (d) continued OC Curve for n=20. Pa 0. p (e) Pr{reject | p = 0.09539)(5000 − 20) = 495 ATIc = 2 = 50 + (0.1000 0.09539 Pr{reject | p = 0. c = 2} = 0.2000 0.60 0.00 0.2500 Fraction defective.0500 0.20 0.00206 ATIc =0 = n + (1 − Pa )( N − n) = 20 + (0. c=0 1.0000 0. 9487 0.0001 0.5237 0.3922 0.0002 0.0692 0.0000 0.0330 0. c1=2.0138 0.100 0.0514 0.120 0.0004 0.6390 0.0001 0.0238 0.9414 0.0784 0.0166 0.0000 0.d2=0} 0.0000 0.120 0.0006 0.1605 0.0009 0.7740 0.0820 0.0001 0.0029 0.0000 0.0161 0.0001 0.0331 0.0000 0.0011 0.0000 0.0000 0.0818 0.9661 0.8296 0.060 0.9862 0.0791 0.0001 0.0000 0.0001 0.080 0.8928 0.0001 0.5405 0.0000 0.2770 0.0707 0.140 0. n1 = 50.Chapter 14 Exercise Solutions 14-12.040 0.070 0.9999 0.0001 0.0142 Primary and Supplementary OC Curves for n1=50.0004 0.0072 0.7342 0.0002 0.0120 0.6892 0.2548 0.0165 0.0238 0. n2 = 100.0028 0.0339 0.4595 0.9858 3 4 5 6 Pr{d1=3.0000 0.0000 0.5207 0.0035 0.0001 0.0708 0.0001 0.0000 PaII Pa 0.d3<=2} Pr{d1=5.000 0. c2=6 1.0763 0.0793 0.0098 0.0002 0.0002 0.00 0.9779 0.045 0.0000 0.9982 0.150 0.0002 0.005 0.0690 0.0000 0.0021 0.0567 0.1635 0.0521 0.20 0.9216 0.0000 0.7560 0.0019 0.0111 0.0011 0.9979 0.010 0.3233 0.035 0.0000 0.0020 0.0098 0.0051 0.0001 0.4162 0.065 0.090 0.0339 0.0000 0.0030 0.0019 0.080 0.0000 0.6767 0.0000 0.1892 0.0002 0.4763 0.0008 0. n2=100.0073 0.0221 0.0001 0.055 0. proportion defective Pr{accept on 1st sample} Pr{reject on 1st sample} Pr{accept lot} 14-13 .0513 0.0190 0.140 0.2658 0.130 0.0442 0.110 0. c2 = 6 P d1 = PaI PrI 0.8883 0.7452 0.2260 0.0029 0.6770 0.5974 0.0221 0.160 p.0844 0.8395 0.0000 0.075 0.3610 0.0444 0.0112 0.025 0.4494 0.0000 0.6078 0.0193 0.0000 0.d2<=1} Pr{d1=6.0522 0.1117 0.0025 0.020 0.60 0.0629 0.030 0.0000 0.060 0.0212 0.8108 0.3274 0.0000 0.0015 0.0019 0.0004 0.0011 0.040 0.0011 0.050 0.0767 0.0073 0.115 0.0568 0.100 0.8706 0.0000 0.0129 0.3848 0.0000 0.0627 0.5838 0.9373 0.9737 0.1128 0.0001 0.1294 0.0001 0.0209 0.0152 0.0000 0.0002 0.0024 0.d2<=3} Pr{d1=4.00 0.0142 0.0120 0.0000 0.0029 0.020 0.40 0. c1 = 2.80 Pr 0.9237 0.0000 0.2333 0.0004 0.0023 0.3108 0.0842 0.0013 0. 899 -0.01.049 XR 1. X R = 1.10 k = 1.113 3. h1 = 0.95 p = s = 0.0397.2054 = = 0.009 1.144 -0.01.015 0.364 1. Pa = 1 − α = 0.159 2.1 − α = 1 − 0.193 Acc n/a n/a n/a n/a n/a … n/a n/a n/a n/a 0 0 … 0 0 0 0 1 1 Rej 2 2 2 2 2 … 2 3 3 3 3 3 … 3 4 4 4 4 4 The sampling plan is n = 49.890 0. Pa = β = 0.9389 + 1.994 3.859 -0.969 1.104 -0.5621 h1 + h2 0.9389 + 0. s = 0.10.2054.2054 p2 = 0.120 2. Pa = h2 1.055 … 0.05 = 0.2054 + 0.153 3.10 14-14 .0397 X A = −0.000 2.820 -0.Chapter 14 Exercise Solutions 14-13.0414.780 -0. h2 = 1.040 2.740 … -0.080 2.025 0.245 1.95. Rej = 4.10.850 0.404 … 2.0397 n. p2 = 0. Acc = 1.199 … 2.325 1.034 3.929 0. β = 0.9389.285 1. (a) p1 = 0.074 3.0397n n 1 2 3 4 5 … 20 21 22 23 24 25 … 45 46 47 48 49 50 XA -0.064 -0. (b) Three points on the OC curve are: p1 = 0. 10 14-15 .604 1.375 4.189 2.0660.95.255 XR 1.0436.3399 p2 = 0.725 2.638 Acc n/a n/a n/a n/a n/a … n/a n/a n/a n/a 0 0 … 0 0 0 0 1 1 Rej 2 2 2 2 2 … 2 3 3 3 3 3 … 3 4 4 4 4 4 The sampling plan is n = 49.925 1. Pa = h2 1.5622 h1 + h2 1.791 2.714 … 0.857 2.538 1. s = 0.057 2.02.472 1.342 0.606 … 1.846 -0.9369.912 -0.10 k = 0. h1 = 1.309 4.0660 X A = −1.670 … 2.3399.0660n n 1 2 3 4 5 … 20 21 22 23 24 25 … 45 46 47 48 49 50 XA -0.3399 + 0. β = 0. Acc = 1 and Rej = 4.923 2.0436 + 0. (a) p1 = 0. X R = 1.572 4.540 0.05 = 0. (b) p1 = 0.276 0.0436 + 1.406 1.659 2.991 2.123 2. Pa = β = 0.02. Pa = α = 0.989 … 4.408 0.507 4. p2 = 0.15.474 0.Chapter 14 Exercise Solutions 14-14.978 -0.3399 = = 0.441 4.0660n.95 p = s = 0.15. h2 = 1.1 − α = 1 − 0.780 -0. AQL = 1% General level I Normal sampling plan: Sample size code letter = H.10%.000. AOQ = [ Pa × p × ( N − n)] [ N − Pa × (np) − (1 − Pa ) × ( Np) ] 14-16. Ac = 0. n = 50. Re = 4 14-17. Ac = 1. N = 3000. Ac = 1. n = 80. n = 125. Re = 1 14-16 . N = 10. Re = 2 14-18. N = 3000. Ac = 0. Re = 4 Tightened sampling plan: n = 125. n = 20. Re = 2 Reduced sampling plan: Sample size code letter = H. General inspection level II. Re = 1 Reduced: up to letter K. Re = 3 Reduced sampling plan: n = 50. Re = 1 Tightened: n = 200. Sample size code letter = L Normal: up to letter K. Ac = 0. AQL = 1% General level II Sample size code letter = K Normal sampling plan: n = 125. Ac = 1.Chapter 14 Exercise Solutions 14-15. Re = 2 Tightened sampling plan: Sample size code letter = J. Ac = 0. AQL = 0. Ac = 3. n = 50. Ac = 2. 0050 0.0800 0. AQL = 0.9756 0.9771 0.0090 0. Re = 4 (b) N = 5000 n= c= p 0. proportion defective Normal Tightened Reduced 14-17 .Chapter 14 Exercise Solutions 14-19.65%.0023 0.8377 0. Re = 3 Reduced sampling plan: n = 80. (a) N = 5000.0900 0.0022 OC Curves for N=2000.0001 0. Ac = 3.0001 0.6767 0.8800 0. Sample size code letter = L Normal sampling plan: n = 200.0000 0.0100 0.80 0.60 0.0101 0.8653 0. II.0200 0.9588 0.0400 0.0000 reduced 80 1 Pa=Pr{d<=1} 0.9529 0.0090 0.0000 tightened 200 2 Pa=Pr{d<=2} 0.0300 0.9886 0.9922 0.1000 normal 200 3 Pa=Pr{d<=3} 0.0040 0.0047 0.00 Pa.0200 0.0070 0.9667 0.0000 0.0080 0.9999 0.3038 0.0800 0.0000 0.0060 0.8092 0.0018 0.1000 p.0500 0.9163 0.9967 0.8340 0.7309 0.0000 0.0700 0.5230 0.8922 0.0395 0. Pr{acceptance} 0.2351 0.9389 0.0861 0.9813 0. General level II.20 0.0433 0.0004 0.0125 0.9911 0.9469 0.0003 0.9989 0.4315 0.7838 0.0600 0.9202 0.0010 0.1472 0.9220 0.0600 0.0030 0.0593 0.8580 0.0020 0.40 0. Ac = 2.0400 0. Re = 4 Tightened sampling plan: n = 200. AQL=0.65% 1.9970 0.1654 0.8916 0. Ac = 1.0211 0.9992 0.00 0. 002 0. 14-18 .00 2000 0.0102 0.020 p.002 0.0005 0.013 0.0002 0.018 0. without replacement.4361 0.0020 0.006 0. proportion defective 0.0010 0.5564 0.6875 0.0150 0.010 0.005 0.0021 0.c=2.004 0.016 0.25% n = 490.0001 0.0002 0.015 0. c = 2.020 D = N*p 2 4 6 8 10 12 14 15 16 18 20 22 24 26 28 30 32 34 36 38 40 Pa 0.n=490.1827 0.008 0.014 0.006 0.0007 0.009 0. AOQL = 0.002 0.0003 0.0031 ATI 511 605 767 962 1160 1341 1497 1564 1624 1724 1801 1858 1900 1930 1952 1967 1977 1985 1990 1993 1995 AOQ 0.00 0.008 0.0001 0.0942 0. A more precise solution would use the hypergeometric distribution to represent this sampling plan of n = 490 from N = 2000.0021 AOQL 0.0018 0.008 0.0014 0.007 0.008 0.0016 0.21%.004 0.006 0.2886 0.004 0.014 0.0012 0.018 0.0015 0.000 0.010 0.003 0.0464 0.000 0. N = 2000.016 0.8165 0. LTPD = 1%.2489 0.Chapter 14 Exercise Solutions 14-20.0321 0. p = 0.0018 0. n=490.017 0.0001 0.0006 0. c=2 1.0664 0.012 0.0008 0.20 500 0 0.1320 0.010 0.9864 0.0068 0.0220 0.012 0.018 0.21% ATI Curve for N=2000.014 0.020 0.3330 0.0000 OC Curve for N=2000.2% p 0. AOQL=0.011 0.019 0.80 Pa.001 0. Note that this solution uses the cumulative binomial distribution in a spreadsheet formulation.0046 0.012 p.9235 0. proportion defective The AOQL is 0.60 0.40 0.016 0. Pr{acceptance} 2500 ATI 1500 1000 0. 000 10. c = 3.040 0.40000 20.020 0.60000 0. c=3 1. 0.140 0. Dodge-Romig single sampling. c=3 60.000 0. the average inspection required will be 82 units.000 0.000 ATI Probability of acceptance. p 0.00000 0.3% 14-19 .180 0. average process fallout = p = 0.000 0.000 30.80000 40.150 0. n=65.000 1.001.000.160 Fraction defective.100 0. 001 Pa = Binom(3.001 ≤ N ≤ 100.200 0.080 0.Chapter 14 Exercise Solutions 14-21. n = 65.120 0.005) = 0. 001 − 65) = 82 On average.000 0.20000 0. Pa OC Curve for Dodge-Romig.200 0 0.99967)(50.100 0. (c) LTPD = 10.050 0.50% defective (a) Minimum sampling plan that meets the quality requirements is 50.00000 50. n=65. (b) ATI Curve for N=50. 65. if the vendor’s process operates close to process average. AOQL = 3%.99967 ATI = n + (1 − Pa )( N − n) = 65 + (1 − 0.060 0.250 p let N = 50.20000 0. 25% n = 46. AOQL = 3%. 0. (a) N = 8000. AOQL = 3%. LTPD = 11. p ≤ 0.01) = 0.Chapter 14 Exercise Solutions 14-22. p ≤ 1% n = 65. p ) = ∑ b(46. c = 2.6% c 2 d =0 d =0 Pa = ∑ binomial(n.9958 ATI = n + (1 − Pa )( N − n) = 65 + (1 − 0. 0.0025) = 0.9998 ATI = n + (1 − Pa )( N − n) = 46 + (1 − 0.9958)(8000 − 65) ≈ 98 (c) N = 8000. p) = ∑ b(65. c = 3.3% (b) c 3 d =0 d =0 Pa = ∑ binomial(n. LTPD = 10.9998)(8000 − 46) ≈ 48 14-20 . 150 0. β = 0.190 0.190 0.05 × 10−2 = 2. p2 = 0. k = 1.800 0.050 0.100 0.10.400 0.8571⎤ ≥ 1.160 0.020 0.05 (a) From the variables nomograph.05 × 10−2 ( ) ⎡ Z LSL = ( 0.7. LSL = 0.70 g/cm3. Calculate x and S.100 0.000 0.120 0.02. p1 = 0.010 0. (b) x = 0.730 0.988 0.060 0.001 0.025 0. the sampling plan is n = 35.900 1-alpha 0.73.600 Pr{accept} p1 p Pr{accept} 0. Accept the lot if ⎡⎣ Z LSL = ( x − LSL ) S ⎤⎦ ≥ 1.7 1.020 0.945 0. (c) Excel workbook Chap15.030 0.05} ≈ 0.7 .560 0.000 0.200 p Pa{p = 0. k=1.7 ⎣ ⎦ Accept the lot.38 (from nomograph) 15-1 .200 0.000 0.400 0.90. 1 – α = 1 – 0.70 ) 1.140 0.040 0.080 0.050 beta 0.xls : worksheet Ex15-1 From the variables nomograph at n = 35 and k = 1.7: p2 OC Curve for n=25.070 0.040 0.10 = 0.005 0.820 0. S = 1.73 − 0.Chapter 15 Exercise Solutions 15-1.016 0.180 0. Inspection level II.80. ktightened = 1. p2 = 0. The equations do not change: AOQ = Pa p (N – n) / N and ATI = n + (1 – Pa) (N – n). kreduced = 1. and generally involves some trial-and-error search. The design of a variables plan in rectifying inspection is somewhat different from the attribute plan design.3 15-3. Now. Then a k can be found to satisfy the AOQL equation. .02.005.Chapter 15 Exercise Solution 15-2. for a specified process average. σ = 5 p1 = 0. For example. LSL = 150. Sample size code letter = L: n Ac Re Normal 200 7 8 Tightened 200 5 6 Reduced 80 3 6 The MIL STD 414 sample sizes are considerably smaller than those for MIL STD 105E. N = 7000.5%. we know n and k are related. AQL = 1. knormal = 1. ATI is found from the above equation. for a given AOQL = Pa pm (N – n) / N (where pm is the value of p that maximizes AOQ). n and k will define Pa. Suppose n is arbitrarily specified.com/college/montgomery 15-5. Under MIL STD 105E.93 A reduced sampling (nreduced = 20. Inspection level IV From Table 15-1 (A-2): Sample size code letter = M From Table 15-2 (B-1): n = 50.Form 1.10 From variables nomograph. and special Romig tables are usually employed. because both Pa and pm are functions of n and k. Accept the lot if ⎡⎣ Z LSL = ( x − 150 ) S ⎤⎦ ≥ 2.wiley.3. Finally. No convenient mathematical method exists to do this. 15-4. standard deviation unknown Assume single specification limit . β = 0.1 − α = 1 − 0. The table required to do this is available on the Montgomery SQC website: www.51) can be obtained from the full set of tables in MIL-STD-414 using Table B-3. Calculate x and S. n = 120 and k = 2.05 = 0.95. Repeat until the n and k that minimize ATI are found. N = 500.005 g/cm3 x1 = 0. S = 10 ⎡⎣ Z LSL = ( x − LSL ) S = ( 255 − 225 ) 10 = 3.005 n x2 = 0.10 x A − x2 = Φ(β ) σ n x A − 0.33 15-7.000 Assume inspection level IV.95 = 0.282 0. N = 100.1 − α = 1 − 0.15 = +1. k = 1.15 Tightened sampling: n = 7. k = 2. LSL = 225psi.14 Assume normal sampling is in effect.145. β = 0.00 Tightened sampling: n = 100. σ = 0.1527 . inspection level II. k = 2. AQL = 1%.Chapter 15 Exercise Solution 15-6.005 n n ≈ 9 and the target x A = 0. AQL = 4% Sample size code letter = E Assume single specification limit Normal sampling: n = 7. x = 255. 15-8. so accept the lot.05 x A − x1 = Φ (1 − α ) σ n x A − 0. k = 1.645 0.145 = −1. sample size code letter = O Normal sampling: n = 100.00.000 ⎤⎦ > 2.15. 000 Assume inspection level II: sample size code letter = L Normal: n = 200.08) Reduced: n = 20. (c) 1 – α = 0.08 (a) 1 – α = 0. p1 = 0.95.95.8. p1 = 1% = 0. N = 5000.01. (d) AQL = 1%. Ac = 2.10ppm. p2 = 8% = 0. β = 1 – 0. M = 4.00 (k = 1. Re = 4 Reduced: n = 80.09 (k = 1. . c = 2 The sample size is slightly larger than required for the variables plan (a). Re = 5 The sample sizes required are much larger than for the other plans. the sampling plan is n = 30 and k = 1. Ac = 3. Variables sampling would be more efficient if σ were known.com/college/montgomery AQL = 1%. σ unknown Double specification limit. β = 0.93) Tightened: n = 50. M = 1.08 From nomograph (for attributes): n = 60. N = 5.71 (k = 2. M = 1.10.90 = 0. (b) Note: The tables from MIL-STD-414 required to complete this part of the exercise are available on the Montgomery SQC website: www.10 From the nomograph. assume inspection level IV From Table A-2: sample size code letter = M From Table A-3: Normal: n = 50. Re = 6 Tightened: n = 200. p2 = 0. σ = 0.69) σ known allows smaller sample sizes than σ unknown.wiley.01. target = 3ppm.Chapter 15 Exercise Solution 15-9. Ac = 5. 16 0.xls : worksheet Ex15-10 OC Curves for Various Plans with n=25. Excel workbook Chap15.00 0 0.20 0.80 0.08 0. c=0 1.06 0. chain sampling plans with c = 0 have slightly less steep OC curves.00 Pa 0.20 1.60 0.12 0.40 0.04 0. .1 0.02 0.2 p single I=1 I=2 I=5 I=7 Compared to single sampling with c = 0.14 0.Chapter 15 Exercise Solution 15-10.18 0. 0090 0.000. Pa 0.2512 0.6227 0. N = 30.9379 0.9657 0. average process fallout = 0.0040 0.20 .1482 0.04 0. c=0 1.7488 0.0060 0.8619 0.0030 0.3773 0.Chapter 15 Exercise Solution 15-11.0489 0.8063 0.0500 0.001.0400 0.0917 0.0343 0.5239 0.0981 0.14 0.0070 0.1937 0.7733 0.0 Probability of Acceptance. c = 0 Excel workbook Chap15.9019 0.9511 0.18 0.0700 0.10% = 0.9306 0.3000 Pa Pr{reject} 0. n = 32.4 0.16 0.2750 0.8518 0.8 0.1000 0.0315 0.0000 OC Chart for n=32.7250 0.0900 0.0200 0.0010 0.1381 0.0600 0.0300 0.10 0.7292 0.8796 0.0000 1.xls : worksheet Ex15-11 (a) p 0.8248 0.2 0.0621 0. p 0.2708 0.1752 0.00 0.08 0.02 0.12 Fraction Defective.06 0.2000 0.9083 0.0100 0.0800 0.2013 0.1204 0.0020 0.0050 0.0 0.6 0.0080 0.0008 0.0694 0.4761 0.7987 0.2267 0.9992 0.9685 0. 0310)(0.32) = 0.9967)(30000 − 32) = 131 Compared to conventional sampling.9958)(30000 − 32) = 158 . the Pa for chain sampling is slightly larger.001 Pa = P(0.9685 + (0.0310 Pa = 0. p = 0.9967 ATI = 32 + (1 − 0. c = 0. but the average number inspected is much smaller. there is little change in performance by increasing i.9685)3 = 0. n) = P (0. n) + P(1. ATI = 32 + (1 − 0. n)[ P(0. n) = P(1.9958.9685)(30000 − 32) = 976 (c) Chain-sampling: n = 32. n)]i P(0.Chapter 15 Exercise Solution 15-11 continued (b) ATI = n + (1 − Pa )( N − n) = 32 + (1 − 0.9685 P(1. (d) Pa = 0.32) = 0. i = 3. 1536 0.4000 0.0626 0.80 0.0900 0.7732 0. i = 3 Excel workbook Chap15.50 p 0.0000 OC Curve for ChSP-1 n=4.0081 0.0388 0.8145 0.9072 0.00 .2713 0.7807 0.0005 Pa 0.0016 0.2916 0.Chapter 15 Exercise Solution 15-12.0600 0.9999 0.0000 P(1.9500 P(0.00 0.10 0.4116 0.8080 0.2458 0.8000 0.8423 0.2000 0.4096 0.6000 0.30 0.00 0.9960 0.60 0.0500 0.4) 0.0081 0.9000 0.0001 0.4096 0.xls : worksheet Ex15-1 p 0.c=0 1.9950 0.1296 0.0100 0. c = 0.0200 0.7000 0.3000 0.0800 0.70 0.8756 0.0040 0.0625 0.0016 0.7164 0.2401 0.10 0.3456 0.4) 0.0256 0.2492 0.80 0.40 0.0256 0.60 0.00 0.1000 0.7385 0.20 0.20 0.9613 0.2252 0.0753 0.0001 0.40 0.1993 0.7481 0.6561 0.90 0.90 1.0756 0.9224 0.2500 0.9815 0.1304 0.0700 0.0010 0.1095 0.4377 0.0036 0.70 Pa 0.9606 0.0256 0.0300 0.8853 0.30 0.50 0. n = 4.5000 0.1715 0.6857 0. 0000 2.0150 0.8863 3.0000 3.3984E+04 44.7143 0.Chapter 15 Exercise Solution 15-13.6298 2.0000 0.9562E+02 444.0060 0.88584 + 0.0030 0.5523.0000 0.0000 8.0277 4.5444 2.8885 1.9590E+02 100.0024 3.0800 0.0001 4.5553E+02 6666.0852E+18 142.9429 6.0200 0.0025 0.2222 0.2749 6.0000 2.0000 3. 6)[ P(0.0070 0.0029 1.8284E+02 222.3333 0.0000 0.0015 9.0300 0.9985 1.0000 0.0008 4.8608 1.6331E+04 33333.000.2867E+08 333.0179E+30 2000.0000 0.0000 0. Pa = 0.7032 1. If c = 1.0251E+06 22.3395E+06 14285.6020 2.0000 0.0032 2. AFI = 0. Pa = 0.7157E+05 666.3333 0.0400 0.6667 0.0135 7.0000 1.6790E+03 100000.7266 1. accept if i = 4.9502E+03 2000.6355E+04 33. AFI = 0.3333 0.6195E+36 1666.0000 2.0000 0.0050 0.1000 f = 1/2 and i = 140 u v Pa 1.0000 0.6266E+04 40.8192E+07 10000.5592E+02 1333.3467E+06 500.1848E+03 4000.5536 2.8953 1.198% would be: 1.0080 0.4774 2.2451E+09 285.5731E+13 200.2222 0.8106E+02 571. Average process fallout.0000 0.4066E+03 3333.2202 1.0000 2.0000 1.7586E+48 1250.1088 1.915.3934 4.3449E+03 80.0000 0.0035 0.7927 1.0000 0.7431E+02 666.1116 1. f = 1/100 and i = 1302: u = 4040.5984E+02 250.2024.95264 15-14. v = 66.0000 0.4286 0.0010E+04 40000.0000 6.6619E+03 1428.0250 0.3371E+03 66.6921E+05 28.0000 0. accept.9316 8.530.8648E+02 133.0000 1.3333 0.0000 0.0090 0. 6) + P(1.0100 0.3909E+02 285.0066 1.0318E+21 125.0666.0000 0.4619E+07 400.0000 0.0900 0.1766 1. AFI = 0.0000 1.5714 0.6944 2.0000 .1111 0.0000 0.2037E+02 333.0350 0.0000 0.7.6175E+02 1000.4040 4.3373E+02 10000.7410E+23 111.8816E+02 500.0600 0.0045 0.3521E+06 12500.6628 2.4121E+27 2222.0000 3.7715 1. f = ½ and i = 140 2.6667 0.2222 0.0000 5.0000 0.0000 3.1111 0.4444 0.0000 0.0003 3.1279 4.0450 0.3972E+03 1666. N = 500.0000 0. p = 0.7143 0.0000 9.8675E+04 22222.2222 0.2131E+05 16666.2151 7. v = 1333.9301 1. 6)]4 = 0.8266 1.88584) 4 = 0.0020 0.5056E+04 1000.0000 0.2765E+03 50000.5035E+02 2000.0000 0.0000 2.8571 0.0652 2.0238E+04 1250.0000 1.0015 0.0106 4.0000 0.0000 0.02} Pa = P(0.5714 0.0002 2.1429 0.1111 0.0000 0.8953 2.4904 2.0255E+24 2500.9429 p 0.4444 0. n = 6 If c = 0.6712 2. f = 1/10 and i = 550: u = 855.0000 3.0015 and q = 1 – p = 0.3333 0.2804E+15 4000.0000 f = 1/100 and i = 1302 u v Pa 2.6667 0.6788E+02 800.7161E+04 28571.3333 0.0839E+02 200.9936E+21 2857.1111 0.1604E+03 57.7594 1.0000 1.0000 f = 1/10 and i = 550 u v Pa 7.7.2128E+12 222.0010 0. v = 6666.4085E+11 250.1429 0.8863 1.7143 0.4383E+07 11111.0500 0.4676E+06 25.0000 0.15% = 0.6667 0.0000 0.4286 0.6667 0.2222 0.7692E+60 1000.5912E+04 25000. f = 1/100 and i = 1302 15-15.0162E+03 2500. f = 1/10 and i = 550 3.0700 0.0000 0.0401E+03 66666.0000 0.7998 1.6667 0.3439E+10 6666.3262E+13 5000.0000 6.6667 0. Need to find Pa{p = 0. f = ½ and i = 140: u = 155.6667 0.5471E+07 20.3.5930E+04 1111.0000 0.5492E+42 1428.3333 0.0000 0.0000 3.5602E+03 50.0000 0.0000 0.0000 0.0561 2.8328 1.5127 4.6795E+03 2857.5729E+18 3333.0346E+02 400.5714 0.0384 2.0381 1.0000 1. Three different CSP-1 plans with AOQL = 0.2222 0.3526 7.0000 0.0037E+03 5000.9739 4.3333 0.6667 0.10847(0.0001 1. Pa = 0.0040 0.1429 0.4400 3.0000 0.3652E+54 1111.0035E+16 166.6667 0.4329E+03 2222.0000 0.3638E+05 20000.0000 1.666.4676E+26 100. 5272 and Pa{p = 0.Chapter 15 Exercise Solution 15-16.0375} = 0.6043 Plan B: AFI = 0.90% Plan A: f = 1/5 and i = 38 Plan B: f = 1/25 and i = 86 15-17.5165 and Pa{p = 0. Plan A: AFI = 0. CSP-1 with AOQL = 1. .0375} = 0.4925 Prefer Plan B over Plan A since it has a lower Pa at the unacceptable level of p.
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