chap 8

Confidence Interval EstimationUSINGSTAilSTICS SaxonHome Improvement @ 8.1 CONFIDENCE INTERVAL ESTIMATION (o KNOWN) FORTHE MEAN 8.7 G)(CD-ROMTOPiC)ESTTMATTON AND SAMPLE SIZEDETERMINATION FORFINITEPOPULATIONS 8.2 CONFIDENCE INTERVAL ESTIMATION t t , tr I FORTHE MEAN (o UNKNOWN) Student's r Distribution Properties theI Distribution of TheConcept ofDegrees ofFreedom TheConfidence Interval Statement 8.3 CONFIDENCE INTERVAL ESTIMATION FORTHE PROPORTION 8.4 DETERMINING SAMPLE SIZE Sample Determination theMean Size for Sample Determination theProportion Size for 8.5 APPLICATIONS CONFIDENCE OF INTERVAL ESTIMATION AUDITING IN Estimating Population Amount the Total Difference Estimation One-Sided IntervalEstimation the Confidence of Rateof Noncompliance with InternalControls EXCEL COMPANION CHAPTER TO 8 E8.l Computing Confidence the Interval Estimate for the Mean (o Known) 88.2 Computing Confidence the IntervalEstimate for the Mean (o Unknown) E8.3 Computing Confidence the IntervalEstimate for the Proportion E8.4 Computing Sample the SizeNeeded for Estimating Mean the E8.5 Computing Sample the SizeNeeded for Estimating Proportion the E8.6 Computing Confidence the IntervalEstimate for the Population Total E8.7 Computing Confidence the IntervalEstimate for the Total Difference E8.8 Computing FinitePopulation Correction Factors 8.6 CONFIDENCE INTERVALESTIMATION AND ETHICALISSUES In this chapter, learn: you r To construct interpret and confidence intervalestimates the meanandtheproportion for I How to determine sample the sizenecessary develop confidence to a intervalfor the meanor proportion I How to useconfidence intervalestimates auditins in 284 EIGHT Confidence CHAPTER Interval Estimation Using Statistics Saxon Home Irprovement @ SaxonHome Improvement distributeshome improvement supplies in northeastern you UnitedStates. a company As accountant, are for the accuracy the integratedinventorymanagement sales of and in mation system. You could review the contents eachand everyrecord of checkthe accuracy this system, sucha detailedreviewwould of but time-consuming and costly.A better approach would be to usestatisti inference techniques draw conclusions to aboutthe population of recordsfrom a relativelvsmall samolecollected durins an audit. At end of eachmonth,you could selecta sampleof the salesinvoices determine followins: the r The meandollaramountlistedon the sales invoices themonth. for r The totaldollaramountlistedon the sales invoices themonth. for r Any differences between dollar amounts the salesinvoices the on the amounts entered into the sales information system. r The frequency occurrence errorsthatviolatethe internalcontrolpolicy of the wa of of whenthereis no authorized Sucherrorsincludemakinga shipment warehouse removal fai slip, number, shipping incorrect ureto includethe correctaccount and the homeimprovement item. How accurate the resultsfrom the samples are and how do you usethis information? Are the sample to you need? sizeslargeenough giveyou the information process is ofusing sample results drawconclusions to aboutthecharac. Q tatisticalinference the population. youto estimale population Inferential tJteristics of a statistics enables unknown charproportion. meanor a population acteristics suchasa population Twotypesof estimates used are population parameters: point estimates intervalestimates. point estimate the to estimate and A is valueof a singlesample A interval estimateis a rangeof numbers, statistic. confidence called an interval, the The constructed around point estimate. confidence intervalis constructed thatthe such probabilitythatthepopulation parameter located is somewhere within the intervalis known. you would like to estimatethe meanGPA of all the students your university. Suppose at The meanGPAfor all the students an unknownpopulation is mean,denoted p. Youselect by a sampleof students find that the samplemeanis 2.80.The samplemean,X = 2.80,is a and point estimate the population of mean,p. How accurate 2.80?To answer is you this question, mustconstruct confidence a intervalestimate. you In this chapter, will learnhow to construct interpret and confidence intervalestimates. Recallthat the sample mean, X, is a point estimate the population of mean,p. However, the meanvaries sample from sample sample to because depends theitemsselected thesamit on in ple.By takinginto account knownvariabilityfrom sample sample (seeSection onthe the to 7.4 sampling distribution the mean),you can developthe intervalestimate the population of for mean. The intervalconstructed shouldhavea specified confidence correctlyestimating the of parameter In otherwords,thereis a specified p. valueof the population confidence that p is somewhere the rangeof numbersdefinedby the interval. in you find that a 95ohconfidenceinterval for the that after studyingthis chapter, Suppose meanGPA at your universityis(2.75 < p < 2.85).Youcan interpret this intervalestimate by statingthat you are95o/o confidentthat the meanGPA at your universityis between 2.75 and 2.85.Thereis a 5o/o chance the meanGPAis below2.75or above that 2.85. After learning intervalfor themean, will learnhow to develop you abouttheconfidence an interval estimate the populationproportion.Then you will learn how large a sampleto for intervalsandhow to perform several selectwhenconstructing confidence importantestimation procedures accountants whenperforming use audits. 8.1: ConfidenceIntervalEstimationfor the Mean (o Known) 285 8.1 CONFIDENCEINTERVALESTIMATION FORTHE MEAN (o KNOWN) In Section 7 .4, you used the Central Limit Theorem and knowledge of the population distribution to determine the percentageof sample means that fall within certain distancesof the population mean. For instance,in the cereal-fill example used throughout Chapter 7 (seeExample 7 '6 on page 268), 95o/o all samplemeansare between362.12and 373.88grams.This of statement is basedon deductivereasoning.However, incluctivereasoning is what you need here. You need inductivereasoning because, statistical in inference, you use the resultsofa single sample to draw conclusionsabout the population, not vice versa.Supposethat in the cerealfill example, you wish to estimate the unknown population mean, using the information from o n l y a s a m p l e . h u s , r a t h e rt h a n t a k e p + ( 1 . 9 6 l ( o l 1 r n ) t o f i n d t h e u p p e ra n d l o w e r l i m i t s T f around p, as in Section 7 .4, you substitute the sample mean, X- for the unknown and , use I X t ( 1 . 9 6 ) ( o l ( . ' l n ) a s a n i n t e r v a l t o e s t i m a t et h e u n k n o w n p . A l t h o u g h i n p r a c t i c e y o u select a single sample of size n and compute the mean, X, in order to understandthe full meaning of the interval estimate,you need to examine a hypotheticalset of all possible samplesofr values. Supposethat a sample of n : 25 boxes has a mean of 362.3 grams. The interval developed t o e s t i m a t e i s 3 6 2 . 3I ( l . 9 6 X l 5 ) l ( ^ 1 2 5 ) , r 3 6 2 . 3+ 5 . g 9 .T h e e s t i r n a t e f p i s p o o 356.42(u(368.18 Because population the mean,p (equalto 368),is included within the interval, this sample results a correct in statement about (see p Figureg.l). FIGURE 8.1 Confidenceinterval estimatesfor five differentsamples of n = 25 taken from a p o p u l a t i o nw h e r e p=368ando:15 362.12 Xt = 362.s 356.42 362.3 JOJ.bZ 368 I 368.18 373.88 I =sos.s x. =:oo 360 Xa= 362.12 356.24 375.38 Xu= eZa.eA 362.12 398 To continuethis hypothetical example,suppose that for a differentsampleof n:25 the mean is 369.5.The interval developed from this samoleis boxes, + 36s.s (1.s6)(t bl2s) sy + or 369.5 5.88. Theestimate is 363.62SuS375.38 Because populationmean,p (equalto 368), is also includedwithin this interval.this statethe ment about u is correct. 286 Interval Estimation EIGHTConfidence CHAPTER aboutp are alwaysmadeby develNow, beforeyou begin to think that correctstatements oping a confidenceinterval estimate, suppose third hypothetical a sampleof n :25 boxes is selectedand the_ffrmple mean is equal to 360 grams.The interval developedhere is the 360 t (1.96)(15)l(a|25), 360+ 5.88.In thiscase, estimate p is or of 354.12(p(365.88 p, the mean, is not included theinterin This estimate not aconectstatement is because population (seeFigure8.1).Thus,for somesamples, intervalestimate p the val developed from this sample of the is correct,but for othersit is incorrect.In practice, only one sampleis selected, because and population you cannotdetermine is unknown, the intervalestimate correct. is mean whether havingan intervalthatprovides correctestimate a To resolve this dilemmaof sometimes and you needto determinethe prohaving an interval that providesan incorrect estimate, sometimes producing portionof samples mean, intervals resultin correctstatements that aboutthepopulation p. To do this, consider otherhypothetical two the samples: casein which X = 362.12 gramsrnd X X thecaseinwhich =373.88grams.If =362.12,theintervalis362.121(1.96X15)l(425), + Thisleads thefollowinsinterval: to or362.12 5.88. 356.24(uS368.00 is Because populationmeanof 368 is at the upperlimit of the interval,the statement a corthe rectone(seeFigure8.1). the t or When X = 373.88, intervalis 373.88 (1.96X15)l(425), 373.88+ 5.88.The meanis intervalfor the sample ap!319.76 368.00 meanof 368 is includedat the lowerlimit of the interval, In this case, because population the the statement correct. is In Figure 8.1, you seethat when the samplemeanfalls anywhere between362.12and grams, population somauhere within the interval. Example In 7.6 the meanis included 373.88 grams. page268,you foundthat 95o/o the sample means between fall of 362.12and373.88 on Therefore,95%o al| samples n : 25 boxeshavesamplemeansthat includethe population of of meanwithin the interval developed. Because, practice,you selectonly one sampleand p is unknown,you neverknow for in if surewhetheryour specificintervalincludesthe populationmean.However, you takeall pos95o/o the intervalswill includethe popof sible samples n and computetheir samplemeans, of that the ulation mean,and only 5o/o them will not. In otherwords,you have95% confidence of populationmeanis somewhere your interval. in in A of onceagain,the first sample discussed this section. sample n : 25 boxes Consider p hada sample meanof 362.3grams. The intervalconstructed estimate is: to 362.31.e6Xls) /( J25) t( + 362.35.88 356.42 u( 368.18 S interval. The intervalfrom356.42to 368.18 referred asa 95% confidence is to that "I am9ilo/oconfident the meanamountof cerealin the populationof boxesis somegrams." and wherebetween356.42 368.18 (o Interval E,stirnation theMean Known) 287 8.1: Confidencc for you might want a higher degreeof confidence(such as99{t/o) includIn somesituations, of ing the populationmean within the interval. In other cases, you rnight acceptlessconfidence (such as 90%) of correctlyestirnating populationmean. In general, the the level of confidence is symbolizedby ( I - o) x 100%o, where u is the proportion in the tails of the distributionthat is outsidethe confidenceinterval.The proportionin the uppertail of the distributionis ul2, and, the proportion in the lower tail of the distributionis ul2.You use Equation(8.1) to constructa ( I - ct) x 100% confidenceintervalestimateof the mean with o known. CONFIDENCE INTERVAL THEMEAN(o KNOWN) FOR o x t z --r= \l n F-zf .u <n+2ft (8.1) where Z: the value correspondingto a cumulative area of 1 - al2 from the standardized normal distribution (that is, an upper-tail probability of ul2). The value ofZ needed constructing confidenceinterval is calledthe critical value for for a the distribution.95o/o confidencecorresponds an o, value of 0.05.The critical Zvalue correto spondingto a cumulativeareaof 0.9750 is L96 because there is 0.025 in the uppertail of the distributionand the cumulativearealessthanZ: 1.96is 0.975. There is a differentcritical value for eachlevel ofconfidence. I - cr.A level ofconfidence of 95ohleadsto a Z value of | .96 (seeFigure 8.2). 99% confidencecorresponds an cr value to 2.58 because upper-tailareais 0.005 and the cumulaof 0.01.The Z value is approximately the t i v e a r e al e s st h a nZ : 2 . 5 8 i s 0 . 9 9 5( s e eF i c u r e8 . 3 ) . FIGURE 8.2 N o r m a c u r v ef o r l d e t e r m i n i n gh e t Z v a l u en e e d e d for 95% confidence 'p, -1.96 0 +1.96 X Z FIGURE 8.3 N o r m a c u r v ef o r l d e t e r m i n r ntg e h Zvalue eeded n for99% confidence x + 2 . 5 8Z 288 CHAPTER EIGHT Confidence lntervalEstimation Now that variouslevelsof confidence havebeenconsideredwhv not makethe confi level ascloseto I 00% aspossible? Beforedoing so,you needto realizethat any increase in level of confidenceis achievedonly by widening (and making lessprecise)the confi interval.There is no "free lunch" here.You would havemore confidencethat the meanis within a broaderrangeof values;however, this might make the interpretation of confidenceinterval lessuseful.The trade-offbetween width of the confidenceinterval the the level ofconfidence is discussed greaterdepthin the contextofdetermining the in sizein Section8.4. Example8.1 illustratesthe applicationof the confidenceinterval EXAMPLE 8.1 ESTIMATING MEANPAPER THE LENGTH WITH95% CONFIDENCE process operates A paper manufacturer a production has that continuously throughout an productionshift. The paper is expected havea meanlength of l1 inches,and the to deviationof the length is 0.02 inch. At periodic intervals,a sampleis selected to whetherthe meanpaperlengthis still equalto 1l inchesor whethersomething gone has in the productionprocess change lengthofthe paperproduced. to the You selecta random ple of 100 sheets, the meanpaperlengthis 10.998 and inches.Construct a95%o interval estimate the populationmeanpaperlength. for (8.1)on page287 with Z: |.96 for 95o/o SOLUTION UsingEquation confidence, , Xxz*=ro.essr(1.e6)g {100 = 10.998 0.00392 + 10.99408 ( 11.00192 Sp 4n and Thus, with 95% confidence,you concludethat the populationmeanis between10.99408 11.00192 inches.Because intervalincludes11, the value indicatingthat the production the process working properly,you haveno reason believethat anythingis wrong with the to is ductionprocess. To seethe effectof using a99o/o confidenceinterval,examineExample8.2. EXAMPLE 8.2 LENGTH WITH99% CONFIDENCE ESTIMATING MEAN PAPER THE paper mean length. interval estimate thepopulation for Construct a99%o confidence (8. Equation I ) onpage 287,withZ : 2.58for 99o/o confidence, SOLUTIONUsing Xtz*=lo.eest(2.58)g 4n {100 = 1 0 . 9 91 0 . 0 0 5 1 6 8 (p 10.99284 ( 11.00316 Once again,becauseI I is includedwithin this wider interval,you haveno reasonto beli that anythingis wrong with the productionprocess. 8.1: Confidence Interval Estimation for the Mean (o Known) 289 As discussed section in 7.4,the sampling distribution t is normallydistributed the of if populationofXis a normal distribution. And" if thepopulationofXis not a normaldistribution, the CentralLimit Theoremensuresthat X is normally distributedwhen r is large.However, when dealingwith a small samplesizeand a populationof X that is not a normal distribution, the samplingdistributionof X is not normally distributedand thereforethe confidenceinterin val discussed this sectionis inappropriate. practice,however, long asthe samplesizeis In as large enoughand the populationis not very skewed, you can use the confidenceinterval defined in Equation8.1 to estimatethe populationmeanwhen o is known. To assess the assumption normality,you can evaluate shapeof the sampledataby using a histogram, of the plot, or normal probabilityplot. stem-and-leaf display,box-and-whisker Learning Basics the 8.1 If X = 85, o : 8, andn:64, construct a 95% confidenceinterval estimateof the populap. tion mean, c. Must you assume that the populationamountof paint per can is normally distributedhere?Explain. d. Constructa 95ohconfidenceinterval estimate.How your answer (b)? doesthis change to o: and M ?:,r,x =t25, 24. n:ru,.:""::T.:,: ffi interval estimate the populaof ---- r-r"'-lAsslsTl99% confidence uon mean'p' 8.3 A marketresearcher statesthat shehas 95oh confidencethat the mean monthly sales of a product are between and $200,000. Explainthe meaningof $170,000 thisstatement. 8.4 Why is it not possible Example8.1 on page288 to in have 100% confidence? Explain. 8.5 Fromthe results Example8.1 on page288 regardof ing paperproduction,is it true that 95%o the sample of means will fall between 10.99408 and 11.00192 inches? Explain. 8.6 Is it true in Example8.1 on page288 that you do not knowfor sure whetherthe populationmean is between 10.99408 11.00192 and inches? Explain. i;1"111.1H",'J.Til:11,[Tll"jJl of deviation is | z seul shipment lieht bulbs.The standard EU 100 hours.A-random sampleof 64 light bulbs indicated sample a meanlife of 350 hours. a. Constructa 95o/o confidenceinterval estimateof the populationmeanlife of light bulbs in this shipment. b. Do you think that the manufacturer the right to state has that the light bulbs last an averageof 400 hours? Explain. c. Must you assume the populationof light bulb life is that normally distributed? Explain. d. Suppose the standard that deviationchanges 80 hours. to What areyour answers (a) and (b)? in Applying the Concepts 8.7 The managerof a paint supply store wants to estimate actualamountof paint contained the in l-gallon cans purchasedfrom a nationally manufacturer. known The manufacturer's specifications state that the standarddeviationof the amountof paint is to 0.02 gallon. A random sample of 50 cans is meanamountof paint per l-gallon selected, the sample and is can 0.995sallon. a. Construct a99o/o confidence intervalestimate thepopof ulationmeanamountof paint includedin a l-gallon can. On the basisof theseresults,do you think the manager hasa right to complainto the manufacturer? Why? 8.9 The inspection division of the Lee County Weightsand Measures Department wantsto estimate the actual amount of soft drink in 2-Iiter bottles at the local bottling plant of a large nationally known soft-drink company. The bottling plant has informed the inspectiondivision that the populationstandard deviationfor 2-liter bottles is 0.05 liter. A random sample 1002-literbottlesat this bottlingplantindicates of a sample meanof 1.99liters. a. Constructa 95ohconfidenceinterval estimateof the populationmeanamountof soft drink in eachbottle. b. Must you assume the populationof soft-drinkfill is that normally distributed? Explain. c. Explainwhy a valueof 2.02liters for a singlebottleis not unusual,eventhough it is outsidethe confidence interval you calculated. d. Suppose that the samplemean is 1.97 liters. What is your answer (a)? to 290 CHAPTER EIGHT Confidencelnterval Estimarron 8.2 ESTIMATION INTERVAL CONFIDENCE FORTHE MEAN (o UNKNOWN) Just as the mean of the population, p, is usually unknown, you rarely know the actual standard deviation of the population,o. Therefore,you often need to constructa confidenceinterval estimateof p, using only the sample statistics X and S. Student'st Distribution At the beginning of the twentieth century, William S. Gosset, a statistician for Guinness Breweries in Ireland (see reference3) wanted to make inferencesabout the mean when o was unknown. BecauseGuinnessemployeeswere not permitted to publish researchwork under their own names,Gossetadoptedthe pseudonym"Student." The distribution that he developed is known as Student's r distribution and is commonly referred to as the t distribution. If the random variableX is normally distributed"then the following statistic has a t distribution with n - I degrees of freedom: Y -tt s ,FN This expressionhas the same form as the Z statistic in Equatron(7 .4) on page 266, except that is S is used to estimatethe unknown o. The concept of degreeso.f'.freedom discussedfurther on pages 291-292. Properties of the t Distribution the / distribution is very similar to the standardizednormal distribution. Both In appearance, distribrrtion.s are bell shapecl. Floweve-r, the t tlistribLrtiorr hits l77orcarea irt the tirils arrd less ln rfie ccnfcr lh:rn clocs thc sfandardized norntal distribufion (see Figtrre 8.4). Because Jis used to estirnate the unknown o. the values of 1 are more variable than those for Z. FIGURE 8.4 Standardized n o r m a ld i s t r i b u t i o n a n d t d i s t r i b u t i o nf o r 5 degrees of freedom S t a n d a r d i z en o r m a ld i s t r i b u t i o n d f distribution for 5 degrees of freedom The degreesof freedom, n - 1, aredirectly related to the sample size,n. As the samplesize and degreesof freedom increase,S becomesa better estimateof o, and the r distribution gradually approaches the standardized normal distribution, until the two are virtually identical. With a sample size of about 120 or more, S estimateso precisely enough that there is little differencebetweenthe t and Z distributions. As statedearlier, the t distribution assumesthat the random variableX is normally distributed. In practice, howeveq as long as the sample size is large enough and the population is not very skewed, you can use the I distribution to estimate the population mean when o is unknown. When dealing with a small sample size and a skewedpopulation distribution,the validity of the confidence interval is a concern.To assess assumptionof normality, you can the 8.2: Confidence Interval E,stimation theMean(o Unknown) 291 for evaluate the shape of the sample data by using a histogram, stem-and-leafdisplay, box-andwhisker plot, or normal probabilityplot. You find the critical values of r for the appropriatedegreesof freedom from the table of the r distribution (seeTable E.3). The columns of the table representthe area in the upper tail of the r distribution. The rows of the table representthe degreesof freedom.The cells of the table representthe particular / value for each specific degreeof freedom. For example, with 99 degrees of freedom, if you want 95% confidence, you find the appropriatevalue of l, as shown in Table 8. L The 95% confidencelevel meansthat2.5ohof the values(an areaof 0.025)are in eachtail of the distribution. Looking in the column for an upper-tail area of 0.025 and in the row corresponding to 99 degreesof freedom gives you a critical value for t of 1.9842.Becauser is a symmetrical distribution with a mean of 0, if the upper-tail value is +1.9842, the value for the lower-tail area (lower 0.025) is - l .9842.A I value of -1.9842 means that the probability that t is lessthan -1.9842 is 0.025,or2.5oh(seeFigure 8.5).Note that for a 95% confidenceinterval, you will always use an upper-tail area of 0.025. Similarly, for a99o/oconfidence interval, use 0 . 0 0 5 ,f o r 9 8 % o s e0 . 0 1 ,9 0 % u s e0 . 0 5 ,a n d 8 0 % u s e0 . 1 0 . u TABLE .1 8 of Determining Critical Degrees Freedom the Value from the t Table I foran Areaof 0.025 2 in Each with 99 Tail J Degrees Freedom of 4 5 ?{ Areas Upper-Tail .10 .05 .01 31.8207 6.9646 4.s407 3.1469 3.3.649 .005 63.6574 9.9248 5.8409 4.604r 4.0.322 1.0000 3.0777 6 . 3 1 3 8 1 2 . 7 0 . 8 1 6 5 1 . 8 8 s 6 2.9200 4 0.1649 1 . 6 3 1 7 2.3s34 3 . 824 8 2 . 164 0.7407 1.s332 2 . 1 l3 0.1267 1 . 4 1 s 9 2 . 0 1 s 0 2 706 96 97 98 100 Source; E.ytroctel ]inn 0.6771 0.6770 0.6770 0.6770 Tuhle E.3 1.2904 t.2903 t.2902 1.9840 2.3658 2.3654 2.36s0 2.3646 2.3642 2.6280 2.6275 2.6269 2.6264 2.6259 FIGURE 8.5 tdistribution ith 99 w degreesof freedom +1.9842 The Concept of Degrees of Freedom rt S In Chapter3 you learnedthat the numeratorof the samplevariance, [seeEquation(3.9) on 52 page 107],requiresthe computationof S -.) e n L(x,-xr 292 CHAPTER EIGHT ConfidenceIntervalEstimation In order to compute 52, you first need to know X . Therefore, only r - I of the samplevalues a are free to vary. This means that you have n - 1 degreesof freedom. For example, suppose sample of five values has a mean of 20. How many values do you need to know beforeyoucan determine the remainder of the values?The fact that n: 5 and X = 20 also tells you that il Tr .L"t =too because n \v i=l ; n Thus, when you know four of the values, the fifth one is not free to vary becausethe sum must add to 100. For example, if four of the values are 18, 24, 19, and 16, the fifth value must be 23 so that the sum equals100. The Confidence lnterval Statement Equation (8.2) defines the (l - a) x 100% confidence interval estimate for the mean with o unknown. INTERVAL THEMEAN(o UNKNOWN) FOR CONFIDENCE V +t.., s G (8.2) x-tn-r#.u <X+,,.# where /,_, is the critical value of the r distribution, with n - I degreesof freedom for an areaof al2 in the upper tail. To illustrate the application of the confidence interval estimate for the mean when the standarddeviation, o, is unknown, recall the Saxon Home Improvement Company Using Statisticsscenariopresentedon page 284.You wanted to estimatethe mean dollar amount listed on the salesinvoices for the month. You select a sample of 100 sales invoices from the population of salesinvoices during the month, and the sample mean of the 100 salesinvoicesrs $110.27,with a sample standarddeviationof $28.95.For95o/oconfidence,the critical value from the / distribution(as shown in Table 8.1) is L9842. Using Equation(8.2), it. ^ lln 'S I T 1n 2 = | t0.2+ (1.s842 8.95 7 ) {100 = 1 1 0 . 2+ 5 . 7 4 7 ( $ 1 0 4 . 5 3u < $ 1 1 6 . 0 1 A Microsoft Excel worksheet for these data is presentedin Figure 8.6. 8.2: Confidence Interval Estimation the Mean (o Unknown) for 293 FIGURE 8.6 Microsoft Excel worksheet compute to a confidence interval estimate the mean for i s a l e sn v o i c e m o u n t a for the Saxon Home lmprovement Company j 1 rEstimate the MeanSalesInvoiceI for f-A'*--q m1--_ j- Q Data ;4ffi 5ft 6 tSamnle ize S 7 ConfidenceLevel B 9 l n t e r m e d i a ta l c u l a t i o n s Ce 1 0 S t a n d a r dn o ro f t h eM e a n E s of Freedrlm 12 | Value SeeSectlon EB.2to create thrs. :84/SaRT{86} =85-1 =Trilv(l - 87,811) 57443 = 8 1 2" 8 1 0 Confidence Interval :85 - 813 :85 * 813 Thus, with 95% confidence,you concludethat the mean amount of all the salesinvoicesis between 104.53 and $ I 16.01 The 95% confidencelevel indicates . $ that if you selected possiall ble samplesof 100 (something that is never done in practice), 95o/o the intervals developed of would include the populationmean somewhere within the interval.The validity of this confidenceinterval estimatedepends the assumptionof normality for the distributionof the amount on of the salesinvoices.With a sampleof 100,the normality assumptionis not overly restrictive,and the useof the t distribution is likely appropriate. Example 8.3 further illustrateshow you construct the confidenceinterval for a mean when the population standarddeviation is unklown. E X A M P L E8 . 3 ESTIMATING MEANFORCE THE REOUIRED BREAK TO ELECTRIC INSULATORS A manufacturing companyproduceselectric insulators.If the insulatorsbreak when in use,a short circuit is likely. To test the strength ofthe insulators,you carry out destructivetesting to determine how rnuch,lbrzc is required to break the insulators.You measureforce by observing how many poundsare appliedto the insulatorbefore it breaks.Table 8.2 lists 30 valuesfrom this experiment, which are located in the file [![!fQ Construct a 95o/o confidence interval estimatefor the population mean force required to break the insulator. T A B L E8 . 2 F o r c e( i n P o u n d s ) R e q u i r e do B r e a k t theInsulator 1,870 1,128 1,656 I,610 1,634 t,784 1,522 1,696 1,592 t,662 1 , 8 6 6 1 , 7 6 4 1 , 7 3 4 1 . 6 6 2 1 , 7 3 4 1 , 7 7 4 1 , 5 5 0 | . t5 6 t . 7 2 r, 8 6 6 6 1 , 8 2 0 t , 7 4 4 1 , 7 8 8 1 , 6 8 8 1 , 8 1 0 1 , 7 5 2 1 , 6 8 0 I. 8 1 0 t . 6 5 2 t , 7 3 6 SOLUTION Figure8.7 shows thatthe sample meanis F = 1,723.4 pounds the sample and standard pounds. deviation S: 89.55 is (8.2)on page UsingEquation 292to construct conthe fidenceinterval, you needto determine criticalvaluefrom the / tablefor an area 0.025in the of FIGURE 8.7 Microsoft Excel confidence interval estimate the mean for amountof force required break to electric insulators q_ I i Intermediale Calculations =BIISORT{85) -86-l -Tlilv(l - 87, 811) -812 " 810 -85 - 813 -85 . 813 ___-_,,___ L llnlerval HalfWidlh SeeSection EB.2to create this. 294 EIGHT Confidence CHAPTER Intcrval Estin.ration each tail, with 29 degrees freedom.From Table 8.3, you seethat t),\:2.0452. Thus, usi of X = 1 , 1 2 3 . 4S : 8 9 . 5 5 n - 3 0 ,a n dt , o : 2 . 0 4 5 2 , , , x + t, t = | s G '60 89'55 -t)7 4 - 't- " r- "r' + ) o s) = 1.123.4 33.44 + 1,689.96{uS1.756.84 You concludewith95% confidencethat the meanbreakingforce requiredfor the population r insulators between1,689.96 is pounds. and 1,756.84 The validity of this confidence intervales mate dependson the assumption that the fbrce requiredis norrnally distributed.Remember, hou eveq that you can slightly relax this assumption large samplesizes. for Thus, with a sampleof 3( you can usethe r distributionevenif the amountof force requiredis only slightly left skewed. Fror the notmal probabilityplot displayedin Figure8.8 or the box-and-whisker plot displayedin Figur 8.9,the amountof force requiredappears slightly left skewed. Thus,the t distributionis appropria for thesedata. FIGURE 8.8 M i c r o s o f tE x c e ln o r m a l p r o b a b i l i t yp l o t f o r the amount of force ranr riraA tn hrorlz Force Requiredto Break ElectricalInsulators 2000 T 1800 1600 1400 1200 e l e c t r i ci n s u l a t o r s ! rooo 800 See Sectlon E6.2 to create this. 600 400 200 ---# -?.5 -1.5 .1 .0.5 0 Z Value 0.5 1 2 2.5 FIGURE 8.9 Microsoft Excelboxand-whisker for plot the amountof force required break to electric insulators Force Required to Break Electrical Insulators See Section E3.4 to create this. 1700 1750 1800 8.2: Confidence IntervalEstimation theMean(o Unknown) 295 for learningthe Basics 8.10 Determine criticalvalueof / in eachof the the following circumstances: o " I - 0 : 0 . 9 5 .n : l 0 b . I - o : 0 . 9 9n : l 0 . c.l-0:0.95,n:32 d . I - c r : 0 . 9 5n : 6 5 , e.l-0=0.90. n:16 8.11 If X = 75,S :24, and n:36. andassuming that the populationis normally distributed, constructa 95% confidenceinterval estimateof thepopulation mean,p. 8.',2 lf X = 50,,S:15,and n:16, andassuming that the populationis normally distributed, constructa 99Yoconfidenceinterval estimateof population mean,p. 8.13 Construct a95%o confidence intervalestimate the for iopulation mean,basedon eachof the followine setsof assuming the populationis normally distributed: that S e t1 S e t2 l, l, l, 1,8,8,8,8 1,2,3,4,5,6,7,8 aroundtime for stresstests.Turnaround time is defined as the time from when the test is orderedto when the radiologist signsoff on the test results.Initially, the meanturnaroundtime for a strbss was68 hours.After incorporattest ing changes into the stress-testprocess, the quality improvement team collecteda sampleof 50 turnaround times. In this sample,the meanturnaroundtime was 32 hours,with a standard deviationof 9 hours(Extracted from E. Godin, D. Raven,C. Sweetapple, F, R. Del Guidice, and "Faster Test Results," Quality Progress, January 2004, 37(l), pp. 33-39). a. Construct a95o/o confidence intervalfor the population meanturnaroundtime. b. Interpretthe intervalconstructed (a). in c. Do you think the quality improvement projectwasa success? Explain. 8.17 The U.S. Departmentof Transportation requirestire manufacturers providetire perforto manceinformation on the sidewallof the tire to betterinform prospective customers whenmakingpurchasing decisions.One very important measure tire perforof manceis the treadwear index, which indicates tire,s the resistance treadwearcompared to with a tire gradedwith a baseof 100.This meansthat a tire with a grade of 200 shouldlast twice as long, on average, a tire gradedwith as a baseof 100.A consumerorganization wantsto estimate the actualtreadwearindex of a brandnameof tires graded 200 that areproducedby a certainmanufacturer. random A sampleof n : l8 indicates samplemeantreadwearindex a of 195.3and a samplestandard deviationof 21.4. a. Assumingthat the populationof treadwear indicesis normallydistributed, construct a95% confidence interval estimateof the populationmeantreadwear index for tires produced by this manufacturerunder this brand name. b. Do you think that the consumerorganizationshould accusethe manufacturer producingtires that do not of meetthe performance informationprovidedon the sidewall of the tire? Explain. c. Explain why an observed treadwear index of 2 l0 for a particular tire is not unusual,eventhough it is outside the confidenceinterval developed (a). in 8.18 The following data (storedin the file !!@@ represent bouncedcheck fee, in dollars, chargedby a the sampleof 23 banksfor direct-deposit customers who maintain a $100balance: 26 28 20 20 21 22 2s 2s 18 25 15 20 18 20 25 25 22 30 30 30 15 20 29 Source: Extractedfrom "The New Face of Banking," Consumer Reports,June 2000. lain why these data sets have different confidence eventhoughthey havethe samemeanandrange. 14 Construct 95o/o a confidence intervalfor the popumean, based the numbersl, 2, 3, 4, 5, 6, and20. on the number20 to 7 and recalculate confithe interval.Using theseresults,describe effectof the outlier(that is, an extremevalue) on the confidence the Concepts 15 A stationerystorewantsto estimatethe meanretail of greetingcardsthat it hasin its inventory. random A of 100 greetingcardsindicatesa meanvalue of 55anda standard deviation $0.44. of Assuming normal distribution,constructa95o/oconfia dence intervalestimate the meanvalueof all greeting of cards the store's in inventory. Suppose therewere2,500greetingcardsin the store's inventory. How arethe resultsin (a) usefulin assisting the store owner to estimate the total value of her inventory? 8.16 Southside Hospital in Bay Shore,New York,commonlyconducts stress teststo studythe heartmuscleafter a personhas a heart attack. of the diagnosticimaging department conducted lity improvementproject to try to reduce the turn- 296 EIGHTConfidence CHAPTER Interval Estimation pilation stagein which the policy pagesare generated and sentto thebank for delivery.The ability to deliverapproved policiesto customers a timely manneris criticalto the in profitabilityof this service the bank.During a period to of policieswas one month,a randomsample 27 approved of selected, the total processing and time, in days,wasas shownbelowandstored the file@, in t3 19 t6 64 28 28 3t 90 60 56 3l 45 48 t7 t7 lt s6 22 l8 91 92 63 50 sl 69 16 17 a. Constructa 95ohconfidenceinterval for the population meanbounced checkfee. b. Interpretthe intervalconstructed (a) in 8.19 The datain the file @ft[! represent total fat, the in gramsper serving,for a sampleof 20 chickensandwichesfrom fast-food chains. The dataareasfollows: 7 8 4 5 1620202419 30 23 30 25 t9 29 29 30 30 40 s6 "Fast Extractedfrom Food: Source: Adding Health theMenu," to Consumer Reports, September pp.28-31. 2004, a. Construct a95o/o confidence intervalfor the population meantotal fat, in gramsper serving. b. Interpret intervalconstructed (a). the in 8.20 Oneof the majormeasures the qualityof service of providedby any organization the speedwith which it is responds customer to A complaints. largefamily-held department storeselling furniture and flooring, including carpet, hadundergone majorexpansion thepastseveral in years. a In particular, flooring department the had expanded from 2 installationcrewsto an installationsupervisoq measurer, a Lastyear,therewere50 complaints and l5 installation crews. conceming The carpetinstallation. following data,in the file represent numberof daysbetween receipt the the @EE, of a complaint theresolution the complaint: and of 54 ll t2 5 35r37 3l 27 152 2 t23 81 74 21 114 13 t9 t26 ll0 110 2961359431265 4 16s 32 29 2 8 2 9 2 6 2 s 52 30 22 36 26 20 23 a. Constructa 95ohconfidenceinterval estimateof the meanprocessing time. b. What assumption must you make about the population distribution (a)? in c. Do you think that the assumption madein (b) is seriouslyviolated? Explain. 8.22 The data in the file [l$$@f!represent the battery life (in shots) three-pixel for digitalcameras: 300 180 85 170 380 460 260 35 380 t20 rr0 240 Source: Extractedfrom "Cameras: More Featuresin the Mix," ConsumerReports, July 2005,pp. 14 18. a. Constructa 95o/o confidenceinterval for the population meanbatterylife (in shots). b. What assumption you needto makeaboutthe popudo lationofinterestto construct intervalin (a)? the c. Giventhe datapresented, you think the assumption do needed (a) is valid?Explain. in 8.23 One operation a mill is to cut pieces steelinto of of partsthat are usedlater in the frame for front seats an in The automobile. steelis cut with a diamondsawandrequires the resultingpartsto be within +0.005inch of the length specifiedby the automobilecompany. The measurement reportedfrom a sampleof 100steelparts(andstoredin the file @ is the difference, inches,between actual in the lengthof the steelpart, asmeasured a lasermeasurement by device,andthe specifiedlengthofthe steelpart. For exam-0.002, represents steelpartthat ple, the first observation, a is 0.002inch shorter thanthe specified length. a. Construct 95% confidence intervalestimate the a of meandifference the between actuallengthof the steel part andthe specified lengthofthe steelpart. b. What assumption must you make aboutthe population distribution (a)? in c. Do you think that the assumption madein (b) is seriouslyviolated? Explain. d. Compare conclusions the reached (a) with thoseof in page53. Problem 2.23on 13105274 33 68 a. Constructa 95o/o confidenceinterval estimateof the meannumberof daysbetween receiptof a complaint the theresolution the complaint. and of mustyou makeaboutthe population b. What assumption in distribution (a)? c. Do you think that the assumption madein (b) is seriExplain. ouslyviolated? d. What effect might your conclusionin (c) haveon the validityof theresults (a)? in 8.21 In New York State,savings banksare permittedto sell a form of life insurance calledsavings bank life insurprocess ance(SBLI). The approval of consists underwriting, which includes reviewof the application, medical a a informationbureaucheck,possiblerequests additional for medical information medical and exams, a policy comand 8.3 INTERVAL ESTIMATION CONFIDENCE FORTHE PROPORTION This sectionextends concept the ofthe confidence intervalto categorical data. Hereyou are with estimating proportion itemsin a population the concerned of havinga certaincharacterisproportionis represented the Greeklettern. The The unknownpopulation tic of interest. by Interval Estirration thePronortion 297 fbr 8.3: Confidence point estirnate n is the sampleproportion,p - Xln, where ir is the sarrple size and X is the for (8.3)defines the nurnberof iter.t.rs the sarnplehaving the characteristic interest.E,quation in of confidenceinterval estir.nate the populationproportion. for coNFTDENCE TNTERVAL ESTTMATE TXSjROPORTTON FOX ,+ 7.lP(t- Pt IT or p-Z p(t - p) <n< p+z n X p(I - p) n (8.3) p: ; n : populationproportion Sampleproportion: = Number of items havins the characteristic Sr_pl. ,t^ Z : critical value from the standardizednorrnal distribution r = s a m p l es i z e assumingthat both X and n - X are greater than 5 You can usethe confidenceintervalestimate the proportiondefinedin Equation(8.3) to of estimatethe proportion of salesinvoicesthat contain errors (seethe Using Statistics scenario i l o n p a g e2 8 4 ) .S u p p o s eh a t i n a s a m p l e f 1 0 0s a l e s n v o i c e s . 0 c o n t a i ne r r o r s T h u s .f o r t h e s e t o . ( d a t a . p: X l n : 1 0 / 1 0 0 : 0 . 1 0 . U s i n gE q u a t i o n8 . 3 )a n d Z - 1 . 9 6f o r 9 5 ' %c o n f i d c n c e , rto an CS n*7 p(t- p) il ith )nt he ral 3nt mhat the ,eel =0.10t(1.96) ( 0r 0 x 0 . e 0 ) . 100 = 0.10 (1.96)(0.03) t : 0 .l 0 + 0 . 0 5 8 8 0.0412<n(0.1588 you Therefore, havc95% confidence betwecn that 4.12%and 15.88% all thesales of invoices Figure a Excelworksheet forthese data. contain errors. 8.10shows Microsoft FIGURE 8.10 Microsoft Excel worksheet construct to n ac o n f i d e n cie t e r v a l estimate the for proportion sales of invoices contain that errors 1 eri:of N , are srisThe 2 3 , 5 5 7 I I 10 11 S l a n d a r d n o r o f t h e E lntewalHalfWidth IJ =85/84 =NORMSTNV(I - 86)2) =sQRr(Be'(1- Bey84l =ABS(810'811) 1i 15 15 *89 - 812 =89'812 298 Estimation Interval EIGHT Confidence CHAPTER for intervalestimate the of anotherapplication a confidence Example8.4 illustrates proportion. E X A M P L E8 . 4 PRINTED OF THE PROPORTION NONCONFORMINGNEWSPAPERS ESTIMATING printedthathavea nonconof the wantsto estimate proportion newspapers A largenewspaper or missingpages, duplicate ruboff, improperpagesetup. suchas excessive forming attribute, printedduring from all the newspapers is of pages. randomsample 200 newspapers selected A and Construct of a singleday.For this sample 200,35 containsometype of nonconformance. printed during the day confidenceinterval for the proportion of newspapers interfret a 90o/o attribute. thathavea nonconforming (8.3), SOLUTION UsingEquation 8. VC: far (E of Jot a. 'D ?5 - ::- = 0.175. and with a90oh level of confidenceZ = 200 b. 8.2 p!z p(t - p) n -. ){ : = 0 . 1 7 t ( . ..6.4 s[ o-.;r 7 s ) ( 0 . 8 2 s ) 5 r = 0.175 (1.645X0.0269) r t75 !0.0442 3 0.1308 n30.2192 that between13.08%and2l.92ohof the newspapers with 90% confidence You conclude printedon that dayhavesometype of nonconformance. to you because canusethe normaldistribution approx' (8.3)contains Z statistic a Equation 8.4, sizeis sufficientlylarge.In Example the whenthe sample distribution imatethe binomial proportion for approximation the population provides excellent an intervalusingZ confidence than 5. However,if you do not havea sufficiently large both X andn - X are greater because (8.3)(see references ratherthanEquation use size,you should thebinomialdistribution sample proportions suc' of intervalsfor varioussamplesizesand 1.2, and6). The exactconfidence (reference 2). by havebeentabulated FisherandYates cesses sm the Ap use a.i I l I b.i I I I c.. 8.2 is: nol wi hei ph sai ag CI a. b. the Basics Learning I I Applying the Concepts wantsto estlmate company 8.24 lf n : 200 andX: 50, constrwt a 95o/o @E 8.26 The telephone @ STST that of the l A s s r sIT proportion households wouldpurchase of the population AS confidenceinterval estimate proportion. a 8.25 lf n = 400 andX : 25, construct 99oh of intervalestimate the population confidence proportion. 8. M ffi ffi availline telephone if it weremade an additional cost.A installation reduced ableat a substantially The is randomsampleof 500 households selected. results the would purchase the households indicatethat 135 of cost. installation line telephone at a reduced additional 8.4:Determining Sample Size 299 a. Constructa 99o/o confidenceinterval estimateof the proportionof households would purpopulation that line. the telephone chase additional proin of b. How would the manager charge promotional gramsconcerning residential use customers the results in (a)? 8.27 According the Centerfor Work-LifePolicy,a surto women who left careersfor vey of 500 highly educated family reasons found that 660/o wantedto return to work (Extracted from A. M. Chakerand H. Stout,'After Years Off, WomenStruggleto Revive Careers,"The Wall Street p. May 6, 2004, Al). Journal, intervalfor the population a. Construct a95ohconfidence proportionof highly educated womenwho left careers for family reasons who want to returnto work. the b. Interpret intervalin (a). 21o/o 8.28 In a surveyconducted AmericanExpress, for of indicated theynevercheckin with that smallbusiness owners ("Snapshots," theoffice when on vacation usatoday.com, April 18,2006.). The articledid not disclose sample the size used the study. in a. Suppose thesurvey that wasbased 500smallbusiness on intervalestimate Construct 95o/o a confidence owners. proportionof smallbusiness owners for the population who nevercheckin with the office whenon vacation. that on b. Suppose the surveywasbased 1,000smallbusiintervalestinessowners. Construct 95o/o a confidence proportion smallbusiness ownmatefor thepopulation of erswho nevercheckin with the office whenon vacation. c. Discuss the effect of samplesize on the confidence intervalestimate. in 8.29 The numberof olderconsumers the UnitedStates is growing, and they are becomingan even bigger ecowhen confronted nomic force. Many feel overwhelmed investments, bankingservices, with the task of selecting A health or careproviders, phoneserviceproviders. telephone found that 27oh 1,900older consumers surveyof theydidn't haveenough time to be goodmoneymansaid (Extracted from "SeniorsConfused Financial by agers msnbc.com, May 6, 2004). Choices-Study," intervalfor the population a. Construct a95o/o confidence proportion older consumers who don't think they of have enough time to be goodmoneymanagers. b. Interpret intervalin (a). the 8.30 A survey of 705 workers(USA TodaySnapshots, how muchthey used March 2006,p. lB) wereasked 21, JC theInternet work.423 saidtheyusedit within limits,and at 183saidthattheydid not usethe Internet work. at a. Construct a95ohconfidence intervalfor the proportion who usedthe Internet within limits. of all workers intervalfor the proportion b. Construct 95o/o a confidence of all workers who did not usethe Internet work. at 8.31 When do Americansdecidewhat to make for dinner?An online survey(N. Hellmich, 'AmericansGo for the Quick Fix for Dinner,"USAToday, February 2005, 14, p. 1B) indicated thatT4o/o Americans of decided eitherat the last minute or that day.Suppose that the surveywas based 500 respondents. on a. Construct 95o/o a confidence intervalfor the proportion of Americanswho decidedwhat to make for dinner eitherat the lastminuteor that day. intervalfor the proportion b. Construct 99o/o a confidence of Americanswho decidedwhat to make for dinner eitherat the last minuteor that day. c. Which intervalis wider?Explainwhy this is true. 8.32 In a surveyof 894 respondents with salaries below per that the primary reason $100,000 year,367 indicated job for staying theirjob was interesting responsibilities on ("What Is the PrimaryReason Stayingon Your Job?" for 5,2005,p. lB). USAToday Snapshots, October 95o/o confidence intervalfor the proportion a. Construct a primaryreason staying their for of all workers whose on job wasinteresting responsibilities. job the in b. Interpret intervalconstructed (a). 8.33 A largenumberof companies trying to reduce are the cost of prescription drug benefitsby requiring employees purchase to drugsthrougha mandatory mailorder program.In a survey of 600 employers,126 indicated that they either had a mandatorymail-order program in place or were adoptingone by the end of 2004 (B. Martinez,"Forcing Employees Buy Drugs to via Mail," The Wall StreetJournal, February 18,2004, p. 1B). intervalfor the population a. Construct a95ohconfidence proportionof employers who hada mandatory mail-order programin placeor wereadopting by theendof 2004. one intervalfor the population b. Construct a99ohconfidence proportionof employers mail-order who hada mandatory program placeor wereadopting by theend of2004. in one the in c. Interpret intervals (a) and(b). the effect on the confidence intervalestimate d. Discuss whenyou change levelofconfidence. the n e 's I se ilA Its he 8.4 DETERMINING SAMPLESIZE In each example of confidence interval estimation so far in this chapter, the sample size was reported along with the results with little discussion with regard to the width of the resulting confidence interval. In the businessworld, sample sizes are determined prior to data 300 CHAPTER EIGHT ConfidenceIntervalEstimatron collection to ensure that the confidence interval is narrow enough to be useful in making decisions. Determining the proper sample size is a complicated procedure,subject to the constraints of budget, time, and the amount of acceptable sampling error. In the Saxon Home Improvement example, you want to estimate the mean dollar amount of the sales invoices,you must determine in advancehow large a sampling error to allow in estimating the population mean. You must also determine in advancethe level of confidence (that is, 90%,95%, or 99oh)to use in estimatingthe population pa'rameter. Sample Size Determination for the Mean To develop an equation for determining the appropriatesample size neededwhen constructing a confidenceintervalestimate the mean,recall Equation(8.1) on page287: of X o + z _T "\ln l ln this context, some refer to e as statisticians the "margin of error." The amount added to or subtractedfrom X is equal to half the width of the interval. This quantity represents amountof imprecisionin the estimatethat results from sampling error. the r.l T h e s a m p l i n ge r r o r . ' e . i s d e f i n e da s e=Z o _T 1n Solving for r gives the sample size needed to construct the appropriate confidence interval 'Appropriate" means that the resulting interval will have an acceptable estimatefor the mean. amount of sampling error. SAMPLESIZEDETERMINATION FOR THE MEAN The samplesize,n, is equalto the productof the Z valuesquared the variance, and dividedby the square the samplingerroq e, of o, squared, 22o2 n-""}:- e I (8.4) FIG 2YouuseZ insteadof t because,to determine the critical value of t, you need to know the samplesize,but you do not know it yet. For most studies, the sample size neededis large enough that the standardizednormal distribution a good is approximation of the t distribution. To determine the sample size, you must know three factors: 1. The desired confidence level, which determinesthe value of Z, the critical value from the s t a n d a r d i z en o r m a ld i s t r i b u t i o n 2 d 2. The acceptablesampling error, e deviation,o 3. The standard In some business-to-business relationships that require estimation of important parameters, legal contracts specify acceptablelevels of sampling error and the confidence level required. For companiesin the food or drug sectors,governmentregulationsoften specify sampling errors and confidence levels. In general, however, it is usually not easy to specify Micr won dete for e saleJ for t lmpr l= lG h See I this. 8 . 4 : D c t c r n r i n i Ss r r o lS i z c 3 0 1 na c the two factorsneededto determinethe samplesize.How can you determinethe level of corrfidence and sarnplingerror?Typically,thesequestionsare answered only by the subjectmatter expert(that is, the individual most farniliar with the variablesunder study).Although 95% is the most common confidencelevel used,if rnore confidenceis desired" then 99%,rnight be more appropriate;if less confidenceis deemedacceptable, then 90%,uright be used.For the samplingerror, you shouldthink not of how much samplingerror you would like to have(you really do not want any error) but of how much you'can toleratewhen drawing conclusions from the data. In additionto specifyingthe confidencelevel and the sarnpling error,you needan estimate of the standard deviation.Unfortunately, you rarely know the populationstandard deviation, o. In some instances, you can estimatethe standard deviationfrom past data.In other situations, you can make an educated guessby taking into accountthe rangeand distributionofthe variable. For example,if you assumea normal distribution,the rangeis approximately equalto 6o (that is, +3o aroundthe mean) so that you estimateo as the rangedivided by 6. If you cannot estintate in this way,you can conducta small-scale o study and estimate standard the deviatiorr f r o r nt h e r e s u l t i u g a l a . d To explore how to determinethe samplesize neededfor estirnating populationmean, the consideragainthe audit at SaxortHome lmprovement. Section8.2, yoLrselected sampleof In a 100 salesinvoicesand constructed 95o/o a confidenceintervalestimateof the populationrnearr salesinvoice amount.How was this samplesize detennined?Should you have selected difa ferentsamplesize'/ Supposethat, after consultationwith company officials, you determinethat a sarnpling e r r o r o f n o m o r e t h a n + $ 5 i s d e s i r e d ,a l o n g w i t h 9 5 o l c o n f i d e n c e .P a s td a t a i n d i c a t et h a t t h e s t a n d a r d e v i a t i o no f t h e s a l e sa r n o u n ti s a p p r o x i m a t e l y 2 5 . T h u s , e : $ 5 , o - $ 2 5 , a n d d $ Z : | . 9 6 ( f o r 9 5 o hc o n f i d e n c e ) . s i n g E q u a t i o n( 8 . 4 ) , U 22o2 e- ( I . 9 6) 2( 2 5 ) 2 (5)" = 96.04 Because generalrule is to slightly oversatisfy criteriaby roundingthe samplesize the the up to the next whole integer,you should selecta sampleof size 97. Thus, the sampleof 100 usedon page292 is closeto what is necessary satisfythe needsof the company, to basedon the estimated standard deviation,desiredconfidencelevel, and samplingerror. Because calcuthe lated samplestandarddeviationis slightly higher than expecte4 $28.95 comparedto $25.00, the corrfidence interval is slightly wider tharrdesired.Figure 8. I I illustrates Microsoft Excel a worksheet determinethe samnlesize. to FIGUR8.11 E Microsoft Excel worksheet for determining sample size torestimating mean the sales invoice amount forthe SaxonHome improvement Company Le- I 2 3 4 6 7 ct q t-r t--t I /el fv ul See SectionEB.4to create mls. 10 11 ;; IJ -NORTS[{V(l- 86}21 -((89'B4)/85r2 -ROUr{DUP(810, 0} tz ifv 302 CHAPTER EIGHT Confidence Interval Estimatron Example 8.5 illustratesanotherapplicationof determiningthe samole size neededto for develooa confidenceinterval estimate the mean. E X A M P L E8 . 5 THE SAMPLESIZEFOR THE MEAN DETERMINING mean want to estimate population the Returningto Example8.3 on page293, suppose.you force requiredto breakthe insulatorto within +25 poundswith 95% confidence.On thebasis Find deviationis 100pounds. ofa studytakenthe previousyear,you believethat the standard the sample needed. size SOLUTION Using Equation(8.4) on page300 and e : 25, o : 100, andZ : confidence, 'n- = - for 95% z2o2 e2 (1.e6)2(loo)2 Q'2 the Therefore,you shouldselecta samplesizeof 62 insulatorsbecause generalrule for determining samplesizeis to alwaysround up to the next integervalue in orderto slightly oversatisfy thecriteria desired. deviation calAn actualsamplingerror slightly largerthan25 will resultif the samplestandard if than 100andslightlysmaller the sample standard deviin of culated this sample 62 is greater ationis lessthan 100. Sample Size Determination for the Proportion for you havelearnedhow to determine samplesizeneeded est the So far in this section, for that you want to determinethe samplesize necessary the populationmean.Now suppose that containerrors. the estimating proportionof salesinvoicesat SaxonHome Improvement To determinethe samplesize neededto estimatea population proportion, fi, you use the methodsimilar to the methodfor a populationmean.Recallthat in developing sample for the mean,the samplingerror is definedby for a confidenceinterval e=Z o -T "ln you o a Whenestimating proportion, replace with 1G(1n(l- n) n " erroris ) Thus,the sampling to Solving for n, you havethe samplesizenecessary developa confidenceintervalestimate a proportion. FOR THE PROPORTION SAMPLESIZE DETERMINATION timesthe populationproportion,r, times The samplesizer is equalto theZvalne squared of minusthe populationproportion,n, dividedby the square the samplingerror,e, ,= Z2n(l - n) ", (8.s) 8.4:Determining Sample Size 303 To determine samplesize,you must know threefactors: the 1. The desiredconfidencelevel, which determines value of Z, the critical value from the the standardized normal distribution 2. The acceptable samplingerror,e 3. The populationproportion,n In practice,selecting thesequantities requiressomeplanning.Onceyou determine the you can find the appropriate valuefrom the standardized desiredlevel of confidence, Z normal distribution.The samplingerror,e, indicates amountof error that you arewilling to tolerate the in estimatingthe populationproportion.The third quantity,n, is actuallythe populationparumeterthat you want to estimate! Thus,how do you statea valuefor what you aretaking a sample in orderto determine? Hereyou havetwo alternatives. many situations, may havepastinformationor releIn you vant experience providean educated that estimate n. Or, if you do not havepastinformation of or relevantexperience, cantry to providea value for a that would neverunderestimate you the samplesize needed.Referring to Equation (8.5), you can seethat the quantity n(l - ru) appears the numerator. in Thus,you needto determinethe value of n that will makethe quantity n(l - n) as largeas possible. When n : 0.5, the productn(l - n) achieves maximum its result.To showthis, several valuesof n, alongwith the accompanying productsof a(l - n), are as follows Whenn : Whenn: Whenn : Whenn : 0.9,thenn(l 0.7,thenn(l 0.5,thenn(l 0.3,thenn(l W h e n : 0 . 1 , t h e n ( 1n n : n): (0.9X0.1) 0.09 :0.21 n) : (0.7X0.3) :0.25 n): (0.5X0.5) :0.21 (0.3X0.7) n): n): (0.1)(0.9):0.09 Therefore, whenyou haveno prior knowledgeor estimate the populationproportion,n, of you shoulduse7r: 0.5 for determining samplesize.This produces largest the possiblesamthe ple sizeandresultsin the highestpossiblecostof sampling. Using ru: 0.5 may overestimate the you samplesizeneeded because usethe actualsampleproportionin developing confidence the interval.You will get a confidenceintervalnarrowerthan originally intendedif the actualsample proportionis differentfrom 0.5. The increased precisioncomesat the cost of spending more time and moneyfor an increased samplesize. Returning to the SaxonHome ImprovementUsing Statisticsscenario,suppose that the auditingprocedures requireyou to have 95%o confidencein estimatingthe populationproportion of salesinvoices with errorsto within +0.07.The resultsfrom pastmonthsindicatethat proportion beenno morethan0.15.Thus,usingEquation the largest has (8.5)on page302 and e : 0 . 0 7 , n : 0 . 1 5 , a n d Z : 1 . 9 6 f o r 9 5 %c o n f i d e n c e . Z2n(l - n) ( 1.e6) 2( 0.1sxO.s5) (0.07)" = 99.96 Because generalrule is to round the samplesize up to the next whole integerto slightly the oversatisfii criteria,a samplesizeof 100is needed. the Thus,the samplesizeneeded satisf,i to the requirements the company, of proportion,desiredconfidencelevel, basedon the estimated and samplingerror,is equalto the samplesizetakenon page297.The actualconfidence interval is narrowerthan requiredbecause sampleproportionis 0.10,while 0.l5 wasusedfor n the in Equation(8.5).Figure8.12showsa Microsoft Excel worksheet determiningsamplesize. for "2 304 CHAPTEREIGHT Confidence IntervalEstimation A B or thc Proportlon of InInor Salcr Invrlecl Dd ofTru. ProDofdon FIGURE 8.12 MicrosoftExcel worksheetfor determining sample sizefor estimating the proportion sales of invoices with errors for the SaxonHome lmprovement Company 4 I|md. 0.rl 6 7 I I unlllno Enor funcr Lrvrl lnlermedieto Calculalions ! Value Celculated SamolaSize 1 t0 tl -lloRxsfirv(l - Bs]n] -(89^2. Bf. (1 - B4D/85^2 -ROUI{DUP(810,0) t-l I\N7T I-ll E t3 r Slr. See Section E8.5 to create this. Example8.6 provides second a application determining samplesize of the the populationproportion. EXAMPLE 8.6 DETERMINING SAMPLE THE SIZE FORTHEPOPULATION PROPORTION Youwantto have 90%confidence estimating proportion officeworkers of the of who to email within an hour to within +0.05. Because you havenot previouslyundertaken such study,thereis no informationavailable pastdata.Determinethe samplesizeneeded. from SOLUTION Because information available no is from pastdata,assume n: 0.50. that Usi (8.5)on page302 ande :0.05, n : 0.50,andZ : 1.645for 90% confidence, Equation (1.645)',(0.50X0.s0) (0.05)2 = 270.6 Therefore,you needa sampleof 27| office workersto estimatethe populationproportion within +0.05with 90% confidence. Learning the Basics 8.34 If you want to be 95%o confidentof esti@ meanto within a sampling lAsslsr I matingthepopulation error of +5 andthe standard deviationis assumed to be 15,whatsample sizeis required? 8.35 If you want to be 99o/o confidentof estimatingthe populationmeanto within a sampling error of +20 and the standard deviation is assumed be 100,what samplesizeis required? to that the populationproportionis approximately 0.40,what sample sizeis needed? Applying the Concepts 8.38 A surveyis plannedto determine mean the annualfamily medicalexpenses employees of of a large company.The management the comof pany wishesto be 95ohconfident that the samplemeanis correctto within +$50 of the populationmeanannualfamily medicalexpenses. previousstudy indicates A that the standard deviationis approximately $400. a. How largea samplesizeis necessary? b. If management wantsto be correctto within +$25,what sample sizeis necessary? 8.39 If the manager a paint supply storewantsto estiof matethe meanamountof paint in a l-gallon can to within ffi ffi 8.36 If you want to be 99Yo confidentof estimating the populationproportion to within an errorof t0.04,what sample sizeis needed? 8.37 If you want to be 95o/o confidentof estimating the populationproportion to within an error of +0.02 and thereis historicalevidence 8.4:Determining Sample Size 305 with95% confidenceand also assumes that is deviation 0.02 gallon.what samplesize is a quality control managerwants to estimatethe of light bulbsto within +20 hourswith 95% conalsoassumes thepopulationstandard that devi100 hours, what samplesizeis needed? If the inspectiondivision of a county weights and department wants to estimatethe mean amount fill soft-drink in 2-liter bottlesto within +0.01 liter with confidence also assumes the standard and that deviais 0.05liter,whatsample sizeis needed? group wantsto estimate 8.42 A consumer the meanelectric bill for the month of July for single-familyhomesin a largecity. Basedon studies in othercities,the standard deviationis assumed be $25.The group wants to estimatethe meanbill for to within+$5 with 99% confidence. Whatsample sizeis needed? lf 95% confidenceis desired,what samplesize is necessarv? An advertising agencythat serves major radio staa wants estimate meanamountof time that the stato the audience listeningto the radio daily.Frompast spends ies,the standard deviationis estimated 45 minutes. as Whatsamplesize is needed the agencywantsto be if 90%confidentof being correctto within +5 minutes? If 99% confidenceis desired,what samplesize is necessary? A growing niche in the restaurantbusinessis breakfast, lunch,andbrunch.Chainsin this includeLe Peep,Good Egg, Eggs& I, First Watch, EggsUp Grill. The meanper-person checkfor First is approximately and the meanper-person check $7, Eggs Up Grill is $6.50. (Extractedfrom J. Hayes, ition HeatsUp asBreakfast Eye Concepts Growth," pp. 8, 66.) b Restaurant News, April 24,2006, Assuming standard a deviationof $2.00,what sample sizeis needed estimate meanper-person to the checkfor Eggto within $0.25with 95% confidence? Good Assuming standard a deviationof $2.50,what sample sizeis needed estimate meanper-person to the checkfor GoodEggto within $0.25with 95% confidence? Assuming standard a deviationof $3.00,what sample sizeis needed estimate meanper-person to the checkfor GoodEggto within $0.25with 95% confidence? Discuss effect of variation on selectingthe sample the needed. size The U.S. Department Transportation of definesan ineflight asbeing"on time" if it landslessthan 15minafterthe scheduled time shownin the carrier'scomputeized reservationsystem.Cancelledand diverted flights are countedas late.A studyof the l0 largestU.S.domestic airlines found Southwest Airlines to havethe lowest proportion of late arrivals,at 0.1577 (Extractedfrom N. Tsikriktsisand J. Heineke,"The Impact of Process Variationon CustomerDissatisfaction: Evidencefrom the U S. Domestic Airline lndustry,"Decision Sciences, Winter you 2004,35(l),pp.129-142).Suppose wereasked perto form a follow-up study for Southwest Airlines in order to proportionof late arrivals.What samupdatethe estimated ple size would you use in order to estimate population the proportionto within an error of a. +0.06with95ohconfidence? b. +0.04 with95o/o confidence? c. +0.02 with95o/o confidence? 8.46 In 2005, 34ohof workersreportedthat their jobs were more difficult, with more stress,and37%o reported that they worry aboutretiring comfortably. (Extracted from S. Armour, " Money WorriesHinder Job Performance," USAToday, October5,2005,p. Dl). Consider follow-up a studyto be conducted the nearfuture. in a. What samplesize is neededto estimatethe population proportionof workerswho reportedthat theirjobs were more difficult, with more stress, within +0.02 with to 95% confidence? b. What samplesize is neededto estimatethe population proportion of workerswho worried aboutretiring comfortably to within +0.02 with 95% confidence? c. Compare resultsof (a) and (b). Explain why these the resultsdiffer. d. If you wereto designthe follow-up study,would you use one sampleand ask the respondents both questions, or would you selecttwo separate samples? Explain the rationalebehindyour decision. 8.47 What proportion of people hit snagswith online transactions? According to a poll conductedby Harris Interactive,89% hit snagswith online transactions ("Top Online Transaction Trouble," USA TodaySnapshots, April 4,2006,p.lD). a. To conducta follow-up study that would provide 95% confidence that the point estimate correctto within is +0.04 of the populationproportion,how large a sample sizeis required? b. To conducta follow-up study that would provide 99% confidence that the point estimate correctto within is +0.04 of the populationproportion,how largea sample sizeis required? c. To conduct a follow-up study that would provide 95% confidence that the point estimate correctto within is +0.02 of the populationproportion,how large a sample sizeis required? d. To conduct a follow-up study that would provide 99oh confidencethat the point estimateis correct to within 306 CHAPTER EIGHT ConfidenceIntervalEstimatron +0.02of the population proportion, how largea sample sizeis required? the the confidence e. Discuss effectsof changing desired level and the acceptable sampling error on samplesize requirements. 8.48 A poll of 1,286youngadult cell phoneuserswas conducted March 2006.Thesecell phoneusers, in aged 78-29,wereactivelyengaged multipleusesof their cell in phones. The datasuggest that 707 took still pictureswith their phones, playedgames, 360 usedthe Internet 604 and (Extracted from "Poll: CellphonesAre Annoying but Invaluable," usatoday.com, April 3,2006). Construct a propor95%confidence intervalestimate thepopulation of tion of youngadultsthatusedtheir cell phoneto a. takestill pictures. b. play games. c. usethe Internet. d. You havebeenasked update results to the ofthis study, Determine samplesize necessary estimate the to the proportions (a)through(c) to within +0.02 population in with 95o/o confidence. 8.49 A studyof 658 CEOsconducted the Conference by greatBoard reportedthat 250 statedthat their company's est concern was sustainedand steadytop-line growth ("CEOs' Greatest Concerns," USA Today Snapshots, May 8 , 2 0 0 6p . 1 D ) . , a. Construct 95ohconfidence a intervalfor the proportion of CEOs whosegreatest concernwas sustained and steady top-linegrowth. b. Interpret intervalconstructed (a). the in c. To conduct follow-upstudyto estimate population a the proportionof CEOs whosegreatest concernwassustainedand steady top-linegrowthto within +0.01with 95% confidence, how manyCEOswouldyou survey? 8.5 APPLICATIONS CONFIDENCE INTERVAL OF ESTIMATION AUDITING IN proporThis chapter focused estimating has meanor the population on eitherthe population you have studiedapplication differentbusiness tion. In previouschapters, to scenarios. Auditing is one of the areasin business that makeswidespread of probabilitysampling use methods orderto construct in confidence intervalestimates. AUDITING Auditing is the collectionand evaluation evidence of aboutinformationrelatingto an proprietor,a partnership, corporation, a economicentity,suchas a solebusiness a or government agency, orderto determineand report on how well the information in iorrespondsto established criteria. t, population information. Auditorsrarelyexamine complete Instead, they rely on estia of you mationtechniques based theprobability sampling methods havestudied thistext. in The on followinglist contains sampling advantageous examining entire is to the someof the reasons population. . ' . is Sampling less timeconsuming. Sampling lesscostly. is Because samplesizeis the Samplingprovidesresultsthat are objectiveand defensible. principles, audit is defensible based demonstrable on statistical the beforeone'ssuperion and in a court of law. provides objective an way of estimating sample the sizein advance. Sampling provides estimate the sampling Sampling an of error. for Samplingis often more accurate drawingconclusions aboutlargepopulations than is othermethods. Examining largepopulations time-consuming therefore and oftensubject to morenonsampling errorthanstatistical sampling. Sampling allowsauditors combine, thenevaluate to and collectively, samples collected by differentindividuals. Sampling allowsauditors generalize to their findingsto the population with a knownsampling error. . ' ' . r 8.5: Applicationsof ConfidenceIntervalEstimationinAuditing 307 Estimatingthe PopulationTotalAmount In auditing applications,you are often more interestedin developing estimatesof the population total amount than the population mean. Equation (8.6) shows how to estimate a population total amount. ESTIMATING POPULATION THE TOTAL Thepointestimate thepopulation is equal thepopulation N, times for total to size, the sample mean. Total = NX (8.6) (8.7)defines confidence Equation the intervalestimate the population for total. CONFIDENCEINTERVALESTIMATE FOR THE TOTAL + "" ) rvx uu --, ': .8: ' ^ l nY N - l (8.7) To demonstrate applicationof the confidenceinterval estimate the populationtotal the for amount, returnto the Saxon HomeImprovement Using Statistics scenario page284.Oneof on the auditingtasksis to estimate total dollar amountof all salesinvoices the month.If the for there 5.000invoices thatmonthand x = $ | 10.27. are for thenusingEquation (8.6). = ryX = (5,000X$l 10.27) $551,350 If n : 100andS: $28.95, thenusingEquation (8.7)with tnr: 7.9842 95o/o for confidence, NX + N(tntl:"ln ,c N-n N-1 =55r,350 t (s,000x1.e 84D?J-/t'999 too ' ./tOO 5,000 I 1 = 551,350 28,721.295(0.99005) + = 551,350 28,436 + < total < $579,786 $522,914 Population Therefore, with95o/o you confidence, estimate thetotalamount sales that of invoices between is and$579,786. Figure8.13shows MicrosoftExcelworksheet these $522,914 a for data. 8.13 Excel for the interva I ofthetotal of all invoices TolalAmounl of All SalosInvoicet a 3 4 Data Populodon She amph teln StmDlG Standard Devla$on 5 fio^2l 6 Stmple Slze B fonffd**a ilx 2SJl 95t{ Home Saxon Company s Lsval 1B IntermediatsCalcularioffi t l rooulalionTotal 551358.tX*84 " 85 t2 'PCFactor 0.9s[ -sQRr(Fi-B6)r{81-l}t 1 3 itandard Enor of ths Toial E8.6to create t d eqrees Freedom of 1 5 Value l6 9g - 8 6 - t I 54: -tB{. 87" 812ySORTtB6l -illlvtt -m,814) r*srYalHafWidih m1x.7t -Bl5 * 813 17 18 Confidence lntcrval 1 S $tsffal Lowcr Llmlt 5,7,311Jf, -811- 816 nlsrvrl Uoosr Llmlt 5797&t.72 -Sl1 + 316 308 CHAPTER EIGHT ConfidenceIntervalEstimation Example furtherillustrates population the total. 8.7 E X A M P L E8 . 7 DEVELOPINGA CONFIDENCEINTERVALESTIMATE FOR THE POPULATIONTOTAL An auditoris facedwith a populationof 1,000vouchers wantsto estimate total value and the the population vouchers. sample 50 vouchers selected, the followingresults: A of is with of (X Meanvoucher amount ) = $1.076.39 (S) =$2'73.62 Standard deviation intervalestimate the total amountfor thepopulationof Construct a95o/o confidence of (8.6)on page307,thepoint estimate the population SOLUTION UsingEquation of totalis 7try = (1,000)(1,07 = $1,076,390 6.39) (8.7) on page307,a 95% confidence From Equation intervalestimate the population of amount is ( 1.000x 1,076.39) I ( 1.000x2.0 ogol2J3fr = 1,076,390 77,762.878 + (0.97517) = 1,076,390 75,832 + < total < $1,152,222 $1,000,558 Population you estimate Therefore,with95o/o confidence, that the total amountof the vouchers is and $ 1,000,558 $ l,l 52,222. DifferenceEstimation An auditoruses differenceestimationwhenhe or shebelieves errorsexistin a setof that and he or shewantsto estimate magnitudeof the errorsbasedonly on a sample. the The lowing stepsareusedin differenceestimation: the l . Determine sample sizerequired. , Calculate the differences between valuesreached the during the audit and the original uesrecorded. The differencein valuei, denoted is equalto 0 if the auditorfindsthat D,, positivevaluewhen the auditedvalueis largerthanthe original value is correct,is a nal value,and is negative whenthe auditedvalue is smallerthanthe original value. 3. Computethe meandifference the sample,D,by dividing the total differenceby in (8.8). sample size,as shownin Equation MEAN DIFFERENCE 2o, D-t=t whereD, : Audited value- Original value n (8.8) 8.5: Applicationsof ConfidenceInterval Estimationin Auditing 309 4. Computethe standard deviationof the differences, as shownin Equation(8.9). Sr, Remember any item that is not in error has a dilferencevalueof 0. that STANDARD DEVIATION THE DIFFERENCE OF Sr-^ :., L(ui-Dr SD= i=l n-l (8.e) (8.10)to construct confidence 5. Use Equation a intervalestimate of the total difference in thepopulation. INTERVAL STIMATE THETOTALDIFFERENCE CONFIDENCE E FOR N D + N ( t 'n ) l # " , l 4 n \ lN - l .s^ffi; (s.10) The auditingprocedures SaxonHomeImprovement for requirea 95o/o confidence interval estimate the difference of between actualdollar amounts the sales the on invoices and the amounts entered into the integrated inventory and sales informationsystem. Suppose in a that you have 12 invoicesin which the actualamounton the sales sampleof 100 salesinvoices, invoiceandthe amountentered into the integrated inventory management salesinformaand (stored the file[[@fift!) tion system different. is Thesel2 differences in are .32 $8.30 $s.21 $ r 0.80 $6.22 $5.63 $4.97 $7.43 52.99 54.63 $9.03 $1.41 $17 The other88 invoices not in error.Theirdifferences each0. Thus, are are D-t=t numerator, there differences. Each i,, -eo n ='" =0.90 100 and3 n last88 equal are 0.eF SD= \tn, ;-l -D)' ,-l (9.03 0.9)2+ (7.47- 0.9)2+ ... + (0 - 0.9)2 Sn = 2'752 Using Equation(8.10),constructthe95Voconfidenceinterval estimatefor the total difference in the population of 5,000 salesinvoices as follows: (s,000)(0.90) .s84D2+ r (s,000x1 = 4,500 2,702.91 + < 51,197.09 Total difference < $'7,202.91 3 10 CHAPTER EIGHT ConfidenceIntervalEstimation Thus,the auditorestimates with 95o/o that the total difference confidence between sales the invoices, determined as duringthe audit andthe amountoriginally entered into the accounting system between is Figure8.14shows Excelworksheet and$7,202.91. an for $1,797.09 these data. FIGURE 8.14 Microsoft Excel worksheet the tota for differencebetweenthe invoice found amounts duringauditand the amounts enteredinto the accounting system for the SaxonHome lmprovement Company otal Dlfforcnce Acfual lnd Enter.d In -SUl{olficrencod).t!!AAl -80/85 -Ba'B!0 -SARTGl6) -saRT{(Bl -85}/(8r - ll} -F4 - 812'8lrySQRT(Fq -B!t - | -Tll{\t(l - 86, 815} -815'Bll - 8 1 1. 8 1 7 -B1l + 817 See Section E8.7 to create this. In the previous example, 12 differences positivebecause actualamount the all are the on salesinvoiceis more than the amountentered into the accounting system. somecircumIn you stances, couldhavenegative errors.Example illustrates 8.8 sucha situation. E X A M P L E8 . 8 DIFFERENCE EST]MATION Returning Example on page308,suppose 14vouchers the sample 50 vouchers to 8.7 that in of containerrors.The valuesof the 14 errorsare listedbelow and storedin the file !@@. that two differences nesative: Observe are $75.41 $38.97 $108.54-$37.18 $62.7s $l18.32 -$88.84 $r27.74 $s5.42 $39.03 $29.4r $47.99 $28.73 $84.05 Constructa 950/o confidenceinterval estimate the total differencein the populationof 1,000 of vouchers. SOLUTION For these data, : ?i D-'=' tr, = n50 6e0.34 1 3 . 8 0 6 8 = SD= )= to, D)' j I n-l ( 7 5 . 4 r - 1 3 . 8 0 6 8 +2( 3 8 . 9 7 1 3 . 8 0 6 8 +2. ' + ( 0 - 1 3 . 8 0 6 8 ) 2 ) )' = 37.427 -] i I j 8.5: Applications Confidence Interval of EstinrationAuditing 3 I 1 in Using Equation(8. l0) on page309, constructthe confidenceintervalestirnate the total diffor ferencein the population as follows: I +( ( 1,000x l 3.8068) 1,000)(2.0096 ) lg = 13,806.8 10,372.4 + < $3,434.40 Totaldifference< 524,179.20 Therefore, with 95o/o confidence, you estimate that the vouchersis between 53.434.40 and524.179.20. differencen the population f i o One-SidedConfidence Interval Estimation of the Rate of Noncompliancewith Internal Controls Organizationsuse internal control mechanismsto ensurethat individuals act in accordance with company guidelines. For example, Saxon Home Improvement requiresthat an authorized warehouse-removal slip be completed before goods are removed from the warehouse.During the monthly audit of the company,the auditing team is chargedwith the task of estimating the proportion of times goods were removedwithout proper authorization.This is referred to as the rate of nonc'ompliancewith the internal c'ontrol.To estimate the rate of noncompliance,auditors take a random sample of salesinvoices and determine how often merchandisewas shipped without an authorized warehouse-removal slip. The auditors then compare their results with a previously established tolerable exception rate, which is the maximum allowable proportion of items in the populationnot in compliance.When estimatingthe rate of noncompliance, is it commonplace to use a one-sided confidence interval. That is, the auditors estimate an upper bound on the rate ofnoncompliance.Equation(8.1I ) definesa one-sided confidenceinterval for a proportion. ONE-SIDED CONFIDENCE INTERVAL A PROPORTION FOR Upperbound p+Z p(t - p) n (8.11) where Z : the value corresponding to a cumulative area of ( I - cr) from the standardized normal distribution (that is, a right-hand tailprobability of a). Ifthe tolerableexceptionrate is higher than the upperbound"the auditorconcludes that the company is in compliance with the internal control. If the upper bound is higher than the tolerable exception rate, the auditor concludesthat the control noncompliancerate is too high. The auditor may then requesta larger sample. Supposethat in the monthly audit, you select 400 sales invoices from a population of 10,000invoices.In the sampleof 400 salesinvoices,20are in violation of the internalcontrol. If the tolerableexceptionrate for this internalcontrol is 6%, what shouldyou conclude? Use a level of confidence. 95o/o The one-sidedconfidenceinterval is cornputedusingp :201400 - 0.05 and Z: 1.645. ( U s i n g E q u a t i o n 8 . 1I ) , = 0 . 0 5 1 . 6 4 5 ( 0 . 0 1 0 9 X 0=90 . ) 5 0 . 0 1 7 6 0 . 0 6 7 6 = + . 80 + 312 CHAPTEREIGHT Confidence IntervalEstimation Thus,you have95o/o that confidence the rateof noncompliance lessthan6.76%.Because is tolerableexceptionrateis 6Vo, rateof noncompliance the may be too high for this internal trol. In otherwords,it is possiblethat the noncompliance for the populationis higher rate the rate deemed you shouldrequestalarger sample. tolerable. Therefore, In manycases, auditoris ableto concludethat the rateof noncompliance the with the pany'sinternalcontrolsis acceptable. Example8.9 illustratessuchan occurrence. E X A M P L E8 . 9 ESTIMATING RATE NONCOMPLIANCE THE OF A large electronics writesI millionchecks firm policyfor the ayear. internal An control pany is that the authorization sign eachcheckis grantedonly after an invoicehasbeen to payablesupervisor. tialed by an accounts The company's tolerableexceptionrate for this trol is 4%. If control deviationsare found in 8 of the 400 invoicessampled, what should auditordo?To solvethis. usea95Yolevelofconfidence. SOLUTION Theauditorconstructs a95Yo one-sided confidence intervalfor the invoicesin noncompliance and compares this to the tolerableexceptionrate. Using (8.11) on page3 I l, p : 81 : 0.02,andZ : 1.645for 95olo 400 confidence, = 0.02 1.6450.02(l 0.02) + = 0.02 1.645(0.007X0.9998) + 0.01 = 0.0315 = 0.02 + l5 Theauditor concludes with95%confidence therateof noncomplianceless 3. that is than Because this is less than the tolerableexceptionrate, the auditor concludes that the i controlcompliance adequate. otherwords,the auditoris more thang5oh is In confident rateof noncompliance lessthan4%. is Learningthe Basics from a populationof 500 8.50 A sampleof 25 is selected items.The samplemean is 25.7, andthe samplestandard intervalestiis a99ohconfidence deviation 7.8. Construct mateof the populationtotal. 8.51 Suppose that a sample of 200 (see the file from a populationof 10,000items. @ft@Q is selected Of these,l0 items are found to haveerrors of the following amounts: 13.76 42.87 34.65 I 1.09 14.54 22.87 25.52 9.81 10.03 15.49 confidenceinterval estimateof the total Constructa 95o/o in the population. difference the 8.52 If p: 0.04,n : 300,and N: 5,000,calculate upper bound for a one-sidedconfidenceinterval estimate of the populationproportion,7t,using the following levels ofconfidence: a.90o/o b. 95% c. 99o/o Applying the Concepts 8.53 A stationery store wants to estimate the total value of the 1,000greetingcardsit has in its inventory. of confidenceinterval estimate the popuConstructa95%o greetingcardsthat are in inventory lation total value ofall if a randomsampleof 100 greetingcardsindicatesa mean deviationof $0.44. valueof $2.55and a standard of department a largecor8.54 The personnel poration employing3,000 workerswantsto estiof matethe family dentalexpenses its employees to determinethe feasibility of providing a dentalinsurance the reveals followplan.A randomsampleof 10 employees (in ing family dental expenses dollars) for the preceding year(seethe!!S@ file): ll0 362 246 85 510 208 173 425 316 r79 Constructa 90ohconfidenceinterval estimateof the total in for family dental expenses all employees the preceding yeat. 8.6: ConfidenceIntervalEstimationand Ethical Issues 3 l3 is A branch chainoflaree electronics ofa stores conin inventory of the merchandise ing an end-of-month Therewere 1,546items in inventoryat that time. A of 50 itemswas randomlvselectedand an audit with conducted, the following results: Value Merchandise of Sample Historical Audited Sample Historical Audited Number Cost ($) Value ($) Number Cost ($) Value ($) X = 5252.28 ,S: $93.67 intervalestimate the total a 95% confidence of in at ofthemerchandise inventory theendof themonth. in A customer the wholesalesarmenttrade is often to a discountfor a cashpaymentfor goods.The variesby vendor. sample 150items A of of discount at fromapopulation of4,000 invoices the endofa revealed in 13 that of time (seethe!@@file) thecustomer failed to take the discountto which he she entitled.The amounts(in dollars)of the 13 diswas thatwerenot takenwereasfollows: t5.32 97.36 230.63 r04.18 84.92 132.76 t2 26.55 129.43 88.32 47.81 89.01 interval estimate the popuof a 99ohconfidence not totalamountof discounts taken. Dresses a smallcompany is that manufactures Econe 's dresses saleto specialtystores. has 1,200 It for items,and the historicalcost is recordedon a in, first out (FIFO) basis.In the past,approximately of the inventory items were incorrectly priced. were usually not significant. ; any misstatements (seethe@$ file), of 120itemswas selected with the thehistorical cost of eachitem was compared The resultsindicatedthat l5 items differed value. ir historical costsand auditedvalues.Thesevalues asfollows: 5 987 17 t8 28 35 43 51 261 201 t2l 315 4tl 249 216 240 105 276 ll0 298 356 2tl 305 60 73 86 95 96 107 ll9 21 140 t29 340 341 135 228 2r0 r52 tlz 2t6 402 97 220 Constructa 95ohconfidenceinterval estimateof the total populationdifferencein the historical cost and audited value. 8.58 Tom and Brent'sAlpine Outfittersconductan annual An audit ofits financial records. internalcontrolpolicy for the companyis that a checkcan be issuedonly after the payable manager initials the invoice.The toleraaccounts rate for this internalcontrolis 0.04.During ble exception of is from a popan audit,a sample 300 invoices examined ulation of 10,000invoices,and 1l invoicesare found to violatethe internalcontrol. the confia. Calculate upperbound for a 95ohone-sided denceinterval estimate the rateof noncompliance. for b. Based (a),what shouldthe auditorconclude? on 8.59 An internal control policy for Rhonda'sOnline FashionAccessories requiresa quality assurance check before a shipmentis made.The tolerableexceptionrate for this internalcontrolis 0.05.During an audit,500 shipping recordswere sampled from a populationof 5,000 shippingrecords,and 12 were found that violatedthe internalcontrol. a. Calculate upperbound for a 95o/o the one-sided confiintervalestimate the rateof noncompliance. for dence the b. Based (a),what should auditorconclude? on 8.6 ESTIMATION AND ETHICAL ISSUES INTERVAL CONFIDENCE that themcan relatingto the selection samples the inferences accompany of and Ethicalissues occur in several ways.The major ethical issuerelatesto whetherconfidenceinterval estimates To areprovidedalongwith the samplestatistics. providea samplestatisticwithout alsoincludsamplesizeused,and an interpreing the confidenceinterval limits (typically set at 95o/o),the raises can tation of the meaningof the confidenceintervalin termsthat a layperson understand intervalestimate might misleadthe userof the Failureto includea confidence ethicalissues. is to resultsinto thinking that the point estimate all that is needed predictthe populationcharin Thus,it is importantthat you indicatethe intervalestimate a promiacteristicwith certainty. of along with a simple explanation the meaningof nent place in any written communication, you shouldhighlightthe sample size. interval.In addition, the confidence estimation occurs in the is the mostcommonareas whereethicalissues concerning Oneof politicalpolls.Often,theresults the polls arehighlighted the publication the results on of of of 314 Interval Estimatron CHAPTER EIGHT Confidence front page of the newspapeq and the sampling error involved along with the methodology used is printed on the page where the article is typically continued,often in the middle of the newspaper.To ensurean ethical presentationofstatistical results,the confidence levels,samplesize, and confidence limits should be made availablefor all surveysand other statisticalstudies. 8.7 TopiclESTIMATION AND SAMPLE SlzE o (CD-ROM DETERMINATION FOR FINITEPOPULATIONS In this section,confidence intervals are developedand the sample size is determinedfor situations in which sampling is done without replacement from a finite population. For further discussion,seeE@EEEEUon the StudentCD-ROM that accompanies this book. This chapter discussesconfidence intervals for estimating the characteristicsof a population, along with how you can determine the necessarysample size. You learned how an accountantat Saxon Home Improvement can use the sample paramedatafrom an auditto estimate importantpopulation ters suchas the total dollar amounton invoicesandtheproportionof shipments madewithout theproperauthorization. Table8.3provides list oftopicscovered thischapter. a in Type ofData TABLE 8.3 Summary Topics of in Chapter B Tlpe ofAnalysis Confidenceinterval for a populationparameter Numerical Categorical Confidenceinterval estimate (Section for theproportion 8.3). One-sided confidenceinterval for estimate theproportion (Section 8.5) Confidenceinterval estimate for the mean(Sections 8.1 and8.2) Confidenceinterval estimate the total and for the meandifference (Section 8.5) To determine what equation usefor a particularsituto questions: ation,you needto askseveral . Are you developinga confidenceinterval or are you determining sample size? . Do you havea numericalvariableor do you havea categorical variable? . Ifyou havea numericalvariable,do you know thepopulationstandard deviation? you do, usethe normaldisIf tribution.If you do not,usethe t distribution. The next four chapters developa hypothesis-testing parameten. approach makingdecisions to aboutpopulation Confidence Interval for the Mean (o Known) Confidence Interval for the Mean (o Unknown) o X + Z -T= \ln T, (8.1) X-tn-t--r<tt<X+tn-'# tln ,s X -'* \ln , <X+z* s.s .ln Chapter Review Problems 315 Confidence Interval Estimatefor the Proportion L. Sample SizeDetermination the Mean for ll -----=- P = zl!!!:-!) \n p(r- p) < 1 r < p+z n 22o2 e' (8.4) Sample SizeDetermination for the Proportion p-Z p(r- p) n (8.3) Z2n(l - n) e- (8.s) auditing 306 intervalestimate 284 confidence criticalvalue 287 degrees offreedom 290 differenceestimation308 levelofconfidence 287 one-sided confidence interval 3l I point estimate 284 sampling error 300 Student's distribution 290 I total amount 307 Your Understanding Checking 8.60 Why can you neverreally have 100%confidenceof the estimating population characteristic interest? of conectly 1 When do you use the / distributionto developthe intervalestimate the mean? for Why is it true that for a given samplesize,n. an in confidence achieved widening(andmakis by precise) confidence less the interval? Underwhat circumstances you use a one-sided do idenceinterval insteadof a two-sidedconfidence When would you want to estimatethe population instead thepopulation mean? of How doesdifferenceestimationdiffer from estimaof themean? the nationalaverage 900. After four weeks,the sample of stores stabilizeat a meancustomer countof 974 and a standard deviationof 96. This increase like a substantial seems amountto you, but it also seems like a pretty small sample. Is theresomeway to get a feel for what the meanper-store countin all the stores will be if you cut coffeepricesnationyou think reducingcoffeepricesis a good stratwide? Do egy for increasing meancustomer the count? 8.67 Companies spending are moretime screening applicantsthan in the past.A study of 102 recruitersconducttld by ExecuNetfound that 77 did Internetresearch candion dates.(Extractedfrom P.Kitchen, "Don't Let Any 'Digital Dirt'BuryYour JobProspects," Nousday, August21,2005, p. A5e). a. Constructa 95Voconfidenceinterval estimateof the population proportion of recruiterswho do Internet research candidates. on b. Basedon (a), is it correctto concludethat more than 70o/o ofrecruitersdo Internetresearch candidates? on c. Supposethat the study uses a sample size of 400 recruiters and 302 did Internet research candidates. on Constructa 95o/o confidenceinterval estimateof the population proportion of recruiterswho do Internet research candidates. on d. Basedon (c), is it correctto concludethat more than 700/o recrtiters do Internetresearch candidates? of on e. Discussthe effect of samplesize on your answers to (a) through(d). 8.68 High-fructose corn syrup(HFCS)wascreated the in 1970sand is usedtodav in a wide varietvof foodsand the Concepts Youwork in the corporateoffice for a nationwide nearly 10,000 store franchisethat operates The per-store daily customer count has beensteady 900for sometime (thatis, the meannumberof customers astore onedayis 900).To increase customer in the count, franchise considering is cuttingcoffeepricesby approxhalf.The l2-ouncesizewill now be $.59instead of size of , andthe 16-ounce will be $.69instead $1.19. with this reductionin price, the franchise will havea grossmargin on coffee.To test the new initiative,the isehasreduced coffeepricesin a sample 34 stores, of customer countshavebeenrunningalmostexactlyat 316 CHAPTER EIGHT ConfidenceIntervalEstimation HFCS is cheaper than sugarand is about75% beverages. thansucrose. think that HFCSis sweeter Someresearchers linkedto the growingobesityproblemin the UnitedStates (Extracted from P. Lempert, "War of the Sugars," Progressive Grocer,April 15, 2006,p.20). The following views are from a nationwide survevof 1.1l4 consumer responses: Viewson HFCS Yes No Are you concerned HFCS? 80o/o20o/o aboutconsuming Do you think HFCSshouldbe banned in foodsoldto schools? 88% 12% Do you think HFCSshould banned be inallfoods? 56% 44% intervalestimate the popuConstruct a95o/o confidence of who lationproportion people of HFCS. a. areconcerned aboutconsuming in b. think HFCSshould banned food soldto schools. be in c. think HFCSshould banned all foods. be d. You are in chargeof a follow-up survey.Determinethe the in sample sizenecessary estimate proportions (a) to (c) to within +0.02with 95olo through confidence. Hotels conducted surveyof 401 top 8.69 Starwood a from D. Jones, "Many executives play golf (Extracted who CEOs Bend the Rules (of Golf)," USA Today,June26, 2002).Among the resultswerethe following: . 329cheatat golf. . 329 hateotherswho cheatat golf. . 289believe parallel. business golfbehavior and . 80 wouldlet a clientwin to getbusiness. . 40 would call in sick to play golf. intervalestimate eachof for Construct 95o/o a confidence thesequestions. Basedon theseresults,what conclusions canyou reachaboutCEOs'attitudes towardgolfl 8.70 A marketresearcher a consumer for electronics companywantsto studythe televisionviewing habitsof residentsof a particular area.A random sampleof 40 respondents selected, eachrespondent instructed is and is recordof all television viewingin a parto keepa detailed ticularweek.The results asfollows: are . Viewingtime per week: X = 15.3hours,.t: 3.8 hours. . 27 respondents newson at least3 watchthe evening weeknights. a. Constructa 95o/o confidenceinterval estimatefor the meanamountof television watchedper week in this city. b. Constructa 95ohconfidenceinterval estimatefor the population proportionwho watchthe eveningnewson per at least3 weeknights week. Supposethat the market researcherwants to take questions: another surveyin a differentcity.Answerthese c. What samplesize is requiredto be 95o/o confidentof estimating population the meanto within *2 hoursand assumes the population that standard deviation equal is to 5 hours? d. What samplesize is needed be 95o/o to confident of proportion being within +0.035of the population who watchthe eveningnewson at least3 weeknights no if previous estimate available? is e. Basedon (c) and (d), what sample sizeshouldthemarket researcher selectif a singlesurveyis beingconducted? for 8.71 The real estateassessor a county government wants to study various characteristics single-family of housesin the county. A random sample of 70 houses reveals following the . Heated (in areaof the houses square feet): X = 1,759, ^s:380. . 42 houses havecentral air-conditioning. a. Constructa 99o/o confidenceinterval estimate the of population meanheated areaof thehouses. b. Constructa 95o/o confidenceinterval estimate the of populationproportionof houses that havecentral airconditioning. 8.72 Thepersonnel directorof a largecorporation wishes to study absenteeism amongclerical workersat the corpo ration'scentraloffice during the year.A randomsample of 25 clericalworkers reveals following: the . Absenteeism: = 9.7 days, 4.0days. X S: . l2 clericalworkers wereabsent morethan l0 days. a. Constructa 95o/o confidenceinterval estimate t of meannumberof absences clerical workersduri for the year. b. Construct a 95o/oconfidence interval estimate of the populationproportionof clericalworkersabsent than 10 daysduring the year. that Suppose thepersonnel directoralsowishes take to questions: in a branch Answerthese survey office. c. What sample size is neededto have 95% confidence i estimating population the meanto within +1.5days the population standard deviation 4.5 days? is d. What sample sizeis needed have90% confidence to proportion within +0.075 estimating population the to no previous estimate available? is if e. Based (c) and(d), whatsample on sizeis needed a gle surveyis beingconducted? 8.73 The marketresearch directorfor Dotfy's Store wants to study women's spendingon A surveyof thestore's is in creditcardholders designed to estimate proportionof womenwho purchase the Problems 317 Chapter Review primarily from Dotty's DepartmentStore and the yearly A amount thatwomenspend cosmetics. preon foundthat the standard deviationof the amount survev in on spend cosmetics a yearis approximately 18. $ sample sizeis needed have99o/o to confidence of imating populationmeanto within +$5? the samplesize is neededto have 90% confidence ,-What estimatingthe population proportion to within a. Constructa 95ohconfidenceinterval estimateof the populationmeanamountspentin the pet supplystore. b. Constructa 90ohconfidenceinterval estimateof the population proportion of customers who own only a cat. The branchmanagerof anotheroutlet (Store2) wishes to conducta similar surveyin his store.The manager does not have access the information generated the manto by agerof Storel. Answerthe followingquestions: c. What sample sizeis needed have95% confidence to of estimatingthe populationmean amount spentin his is storeto within +$1.50if the standard deviation $10? d. What samplesize is needed have90% confidenceof to estimatingthe populationproportion of customers who own only a cat to within +0.045? to e. Basedon your answers (c) and (d), how large a sample shouldthe manager take? that servescontinental 8.76 The owner of a restaurant food wantsto study characteristics his customers. He of the decidesto focus on two variables: amountof money spentby customers whethercustomers and orderdessert. The results from a sample of 60 customersare as follows: . Amountspent:X = $38.54,S= $7.26. . l8 customers purchased dessert. a. Constructa 95o/o confidenceinterval estimateof the population mean amount spent per customerin the restaurant. b. Constructa 90ohconfidenceinterval estimateof the population proportion of customerswho purchase dessert. The ownerof a competingrestaurant wantsto conducta This owner doesnot have similar surveyin her restaurant. to access the information of the owner of the first restaurant.Answerthe following questions: c. What samplesizeis needed have95% confidence to of estimatingthe populationmean amount spentin her restaurant within +$1.50,assuming to that the standard is deviation $8? to d. What samplesize is needed have90% confidenceof estimatingthe populationproportion of customers who purchase dessert within +0.04? to to e. Basedon your answers (c) and (d), how large a sample shouldthe ownertake? of 8.77 The manufacturer "Ice Melt" claimsits product as will melt snow and ice at temperatures low as 0o A for Fahrenheit. representative a large chain of hardware in storesis interested testingthis claim. The chain purchases largeshipment 5-poundbagsfor distribution. a of The representative wants to know with 95% confidence, within +0.05,what proportionof bagsof Ice Melt perform thejob as claimedby the manufacturer. .045? on the resultsin (a) and (b), how many of the 's credit card holders should be sampled? The branch managerof a nationwide bookstore of wantsto study characteristics her store'scusto Shedecides focus on two variables:the amount spentby customersand whetherthe customers purchasingeducationalDVDs relating to consider preparation exams,such as the GMAT, GRE, or . Theresultsfrom a sampleof 70 customersare as spent: = $28.52,,S: X Amount $11.39. 28 customers statedthat they would considerpurthe DVDs. chasins educational a 95%o confidenceinterval estimateof the meanamountspentin the bookstore. a 90% confidenceinterval estimateof the proportionof customers who would consider Lon ine educational DVDs. that the branchmanaserof anotherstorein the wantsto conduct a similar survey in his store. thefollowing questions: size is needed have95% confidenceof to sample ing the populationmean amount spent in his deviationis assumed to within +$2 if the standard $10? to of sample sizeis needed have90% confidence populationproportionwho would coning the purchasingthe educational DVDs to within ,| on your answers (c) and (d), how large a samto the take? should manager The branchmanagerof an outlet (Store l) of a chainof pet supply storeswantsto study charto In icsof her customers. particular,shedecides the twovariables: amountof moneyspentby cusown only one dog, only andwhetherthe customers ormorethanonedog and/orcat.The resultsfrom a are of 70 customers as follows: t of moneyspent:X = $21.34, S:59.22. own customers only a dog. own only a cat. customers mersown more than one dos and/or cat. 318 EIGHTConfidence CHAPTER Interval Estimation amountthat the auditor determined shouldhavebeen reimbursed. 8.80 A home furnishingsstore that sells bedroomfurniture is conductingan end-of-monthinventory of the beds (mattress, bed spring, and frame) in stock.An auditorfor the store wants to estimatethe meanvalue of the bedsin stock at that time. Shewantsto have99% confidence that her estimate the meanvalueis correctto within +$100. of On the basisof pastexperience, estimates the stanshe that darddeviation ofthe valueofa bed is $200. a. What sample sizeshouldsheselect? b. Using the samplesize selected (a), an audit wasconin ducte4 with the following results: a. How many bagsdoesthe representative needto test? What assumption shouldbe madeconcerningthe population proportion?(This is called destructivetesting; that is, the productbeing testedis destroyed the test by andis thenunavailable be sold.) to b. The representative 50 bags,and42 ofthem do the tests job as claimed.Constructa 95o/o confidenceinterval estimatefor the populationproportion that will do the job asclaimed. c. How can the representative the results of (b) to use determine whetherto sell the Ice Melt product? 8.78 An auditorneeds estimate percentage timesa to the of fails to follow an internalcontrolprocedure. samcompany A ple of 50 from a population 1,000itemsis selecte4 in of and 7 instances, internalcontrolprocedure not followed. the was a. Construct a90% one-sided confidence intervalestimate proportion itemsin which the interof thepopulation of nal controlprocedure wasnot followed. b. If the tolerable rate is 0.15, what shouldthe exception auditorconclude? 8.79 An auditorfor a government agencyneedsto evalupayments doctors'officevisitspaid by Medicare ate for in a particularzip code during the month of June.A total of 25,056 visitsoccurred duringJunein this area. The auditor wantsto estimatethe total amountpaid by Medicareto within +$5 with 95% confidence. the basisof past On experience, she believesthat the standarddeviation is approximately $30. a. What samplesizeshouldsheselect? in Using the samplesizeselected (a), an audit is conducted with the following results. Amountof Reimbursement X = $1,654.27 S = $184.62 Constructa 99o/o confidenceinterval estimateof the total value of the bedsin stock at the end of the month if there were258 bedsin stock. 8.81 A quality characteristic interestfor a tea-bagof filling processis the weight of the tea in the individual bags.In this example, labelweight on the package the indithat the meanamountis 5.5 gramsof tea in a bag.If cates the bags are underfille4 two problemsarise.First, customersmay not be able to brew the tea to be as strongas they wish. Second, companymay be in violation of the the truth-in-labelinglaws. On the other hand, if the mean amountof tea in a bag exceeds label weight, the comthe pany is giving awayproduct. Getting an exact amountof of tea in a bag is problematicbecause variationin the temin peratureand humidity insidethe factory,differences the densityof the tea, and the extremelyfast filling operation of the machine(approximately170bagsper minute).The of following dataarethe weights,in grams,of a sample 50 tea bagsproducedin one hour by a singlemachine(the dataarestoredin the file S$!!@: Weight of TeaBags, in Grams X = $93.70 ,l = $34.55 In l2 of the office visits, an incorrectamountof reimbursement was provided.For the 12 office visits in which there the was an incorrect reimbursement, differencesbetween and the amountthat the auditor the amountreimbursed determinedshouldhavebeen reimbursedwere as follows (andare storedin the file@!@ $17 $2s $14 -$10 $20 $40 $3s $30 $28 $22 $15 $5 of intervalestimate the popb. Construct 90% confidence a that ulationproportionof reimbursements containerrors. confidenceinterval estimateof the c. Constructa 95o/o per populationmeanreimbursement office visit. interval estimateof the confidence d. Constructa 95%o for populationtotal amountof reimbursements this geographicareain June. confidenceinterval estimateof the e. Constructa 95%o and the the total differencebetween amountreimbursed 5.65 s.44 5.42 5.40 s.53 5.34 5.54 s.4s 5.52 5.41 5.57 5.40 5.53 5.54 5.s5 5.62 5.56 5.46 5.44 5.5Ii 5.47 5.40 s.47 5.61 5.53 5.32 5.67 5.29 5.49 5.55 5.77 5.57 5.42 5.58 5.58 5.50 5.32 5.50 5.53 5.61 5.45 5.44 5.25 5.56 s.63 5.50 5.57 5.67 of confidenceinterval estimate a. Constructa 99o/o populationmeanweight of the teabags. set b. Is the companymeetingthe requirement forth on labelthatthe meanamountof teain a bag is 5.5 8.82 A manufacturingcompanyproducessteel part of The main component for electricalequipment. housingis a steeltroughthat is madeout of a l4-gauge Chapter Review Problems 3I9 punchpress It is produced usinga 250-tonprogressive that puts two 90-degree forms a wipe-down operation flat steelto makethe troush. The distancefrom one of theform to the otheris critical because weatherof in outdoorapplications. The data(storedin the from a sampleof 49 troughsfollows: Width of Trough, in Inches 8.3438.3178.383 8.348 8.410 8.351 8.313 8.481 8.422 8.3828.4848.403 8.414 8.419 8.385 8.465 8.498 8.447 8.4138.489 8.414 8.481 8.4t5 8.479 8.429 8.458 8.462 8.M 8.4298.460 8.412 8.420 8.410 8.40s 8.323 8.420 8.4478.4058.439 8.411 8.427 8.420 8.498 8.409 warranty period. The data file !@ containsa sample madeon the company's of 170 measurements Bostonshinglesand 140measurements madeon Vermontshingles. a. For the Boston shingles,constructa 95ohconfidence interval estimate the meangranuleloss. of b. For the Vermont shingles,constructa 95o/o confidence interval estimate the meangranuleloss. of c. Evaluatewhetherthe assumption needed (a) and (b) for hasbeenseriouslyviolated. d. Based theresults (a) and(b), whatconclusions on of can you reach concerningthe mean granule loss of the BostonandVermontshingles? Report Writing Exercises to in 8.85 Referring the results Problem 8.82on page318 concerningthe width of a steeltrough, write a report that your conclusions. summarizes a confidenceinterval estimateof the Construct 95o/o mean width of the troughs. Interpret intervaldeveloped (a). the in The manufacturer Boston and Vermontasphalt of know that productweight is a major factor in the 's perceptionof quality. The last stageof the line packages shinglesbeforethey areplaced the pallets.Once a pallet is full (a pallet for most wooden it holdsl6 squares shingles), is weighed, the of and t is recorded.The data file [@ contains weight(in pounds)from a sampleof 368 pallets of and shingles 330palletsof Vermontshingles. Forthe Boston shingles,constructa 95ohconfidence intervalestimate the meanweight. of Forthe Vermont shingles,constructa 95Voconfidence intervalestimate the meanweight. of Evaluate whetherthe assumption neededfor (a) and (b) has beenseriously violated. can Based theresults on of(a) and(b), whatconclusions you reachconcerningthe meanweight of the Boston andVermontshingles? I ll Team Proiect 8.86 Refer to the team project on page 73 (see the file). Constructall appropriate confidence Eil@@@ interval estimates the populationcharacteristics lowof of risk, average-risk, high-risk mutual funds. Include and theseestimates a reportto the vice president research in for at the financial investment service. Student Survey Database 8.87 Problem1.27on page 15 describes surveyof 50 a (seethe file GEEEEffiE!E$. undergraduate students a. For thesedata,for eachvariable,constructa95%o confidenceinterval estimate the populationcharacteristic. of your conclusions. b. Write a reportthat summarizes 8.88 Problem1.27on page15 describes surveyof 50 a (seethe fileEEE@). undergraduate students students your at a. Selecta sampleof 50 undergraduate schooland conducta similar surveyfor thosestudents. b. For the data collected in (a), repeat (a) and (b) of Problem 8.87. the of 8.87. c. Compare results (b) to thoseof Problem a 8.89 Problem1.28on page 15 describes surveyof 50 (seethe fileEEElffiElldH). MBA students a. For thesedata,for eachvariable,constructa95o/o confiinterval estimate the populationcharacteristic. dence of your conclusions. b. Write a reportthat summarizes 8.90 Problem1.28on page 15 describes surveyof 50 a (seethe file!@@l!@). MBA students a. Selecta sampleof 50 graduatestudentsin your MBA programandconducta similar surveyfor thosestudents. b. For the data collected in (a), repeat (a) and (b) of Problem 8.89. c. Compare resultsof (b) to thoseof Problem8.89. the 5 fs r' )6 h" I hr gs he rcl br The manufacturerof Boston and Vermont asphalt providesits customerswith a 20-yearwarranty on whether shinglewill last a To of its products. determine testingis conlongasthewarrantyperiod" acceleratedlife plant. Accelerated-life testing at the manufacturing to it the shingleto the stresses would be subject in a of normal usevia a laboratoryexperimentthat takes a few minutesto conduct.In this test, a shingleis scraped with a brush for a short period oftime, granules removed by the brushing are andthe shingle (in that experience amounts low weighed grams).Shingles lossareexpected lastlongerin normalusethan to ofgranule that high of shingles experience amounts granuleloss.In this no a situation, shingleshouldexperience more than 0.8 grams granulelossif it is expected lastthe lengh of the to of 320 CHAPTER EIGHT ConfidenceIntervalEstimatron Managingthe SpringvilleHerald The marketing departmenthas been consideringways to increasethe number of new subscriptionsand increasethe rate of retention among customerswho agreed to a trial subscription. Following the suggestion of Assistant Manager Lauren Alfonso, the department staff designed a survey to help determine various characteristicsof readers of the newspaperwho were not home-delivery subscribers. The surveyconsists the following 10 questions: of 1. Do you or a memberofyour household everpurchase the Springville Hera ld? (l)Yes (2)No [ f t h e r e s p o n d e n ta n s w e r s n o , t h e i n t e r v i e w i s terminated.] 2. Do you receive the Springville Herald via home delivery? (l)Yes (2)No The group agreed to use a random-digit dialing method to poll 500 local households telephone. by Using this approach,the last four digits of a telephonenumberare randomly selected to go with an area code and exchange (the first 6 digits of a l0-digit telephonenumber). Only those pairs of area codes and exchangesthat were for the Springville city areawere used for this survey. Of the 500 households selected"94 householdseither refusedto participate,could not be contactedafter repeated attempts, or representedtelephone numbers that were not in service.The summary results are as follows: Households That Purchase Springville the Herald Frequency Yes No Householdswith Home Delivery 352 JA Frequency 4.] [f no,skipto question 3. Do you receivetheSpringvilleHerald: ( I ) Monday-Saturday(2) Sunday only (3) Everyday 9.] [f everyday,skip to question 4. How often during the Monday-Saturday period do you purchase SpringvilleHerald? the (l) Everyday (2) Mostdays (3) Occasionally never or 5 . How often do you purchase SpringvilleHerald on the Sundays? ( l) EverySunday (2) 2-3 Sundays month per (3) No morethanoncea month 6 . Whereareyou mostlikely to purchase Springville the Herald? (l) Convenience store (2) Newsstand/candy store (3) Vending machine (4) Supermarket (5) Other 7 . Woufdyou consider subscribing the Springville to Heraldfor a trial periodif a discount wereoffered? (l) Yes (2)No 9.] [f no, skipto question 8 . The Springville Herald currently costs $0.50 Monday-Saturday $1.50on Sunday, a totalof and for per Howmuchwouldyoubewillingto pay $4.50 week. perweekto gethomedeliveryfor a 90-day tnal period? 9 . Do you read a daily newspaper other than the SpringvilleHerald? (l)Yes (2)No t 0 .As an incentive long-termsubscribers, newsfor the paperis considering possibility the ofoflering a card thatwouldprovide discounts certain at restaurants in the Springvilleareato all subscribers who pay in advance six monthsof homedelivery. for Wouldyou wantto get sucha cardunderthe termsof this offer? ( l ) Y e s ( 2 )N o Yes No Type of Home Delivery Subscription 136 216 Frequency Monday-Saturday Sunday only 7 daysa week PurchaseBehavior of Nonsubscribers for Monday-Saturday Editions l8 25 93 Frequency Every day Most days Occasionally never or PurchaseBehavior of Nonsubscribers for Sunday Editions 78 95 43 Frequency Every Sunday 2-3 Sundays month a No more than oncea month Nonsubscribers'PurchaseLocation 138 54 )4 Frequency Convenience store Newsstand/candy store Vending machine Supermarket Otherlocations Would Consider Trial Subscription if Offered a Discount 74 95 21 13 IJ Frequency +0 Yes No 170 Rcf-crenccs 321 RateWilling to Payper Week(in Dollars) Data file EIjEEE Trial Subscription Home-Delivery for a 90-Day EXERCISES SH8.1 Some members of the marketing departrrentare concernedabout the random-digitdialing method Prepare memoa usedto collect surveyresponses. the randumthat examines following issues: . The advantages of and disadvantages using the git dialing rnethod. random-di ' P o s s i b l ea l t e r n a t i v ea p p r o a c h e s o r c o n d u c t f ing the survey and their advantagesand disadvantages, S H 8 . 2 A n a l y z e t h e r e s u l t so f t h e s u r v e y o f S p r i n g v i l l e Write a report that discusses urarthe households. k e t i n g i m p l i c a t i o n so f t h e survey results for the Springville Heruld. 4.00 4.403.ts 4.003.75 4.103.60 3.60 3.60 4.15 3.60 4.00 4.r0 3.90 3.753.00 3.65 3.50 3.25 3.75 3.30 3.75 4.304.20 4.253.50 4.003.80 4.10 3.903.95 3.40 3.50 2.853.75 3.20 4.403.80 3.403.50 3.50 3.75 3.30 3.85 3.80 3.90 Reada Daily Newspaper Other Than the Springville Herald Frequency Yes No Would Prepal' Six Months to Receive a RestaurantDiscountCard r38 214 Frequency Yes No 66 286 Web Case Applt' your knov'ledge about confidenc'eintervctl estimation in this Weh Case. whic'h extends the OurCamptrs! Web Case.from Chapter 6. Among its other features,the OurCampus! Web site allowscustomers purchase to OurCampus!LifeStylesmerpaymentprocessing, chandise online. To handle the management OurCampus!has contracted of with the following firrns: . P a y A F r i e n d( P A F ) : a n o n l i n e p a y m e n t s y s t e m w i t h w h i c h c u s t o m e r s n d b u s i n e s s es u c h a s O u r C a m p u s ! a registerin order to exchangepaymentsin a secureand convenientlranner without the need for a credit card. . ContinentalBanking Company (Conbanco):a processi n g s e r v i c e sp r o v i d e r t h a t a l l o w s O u r C a m p u s ! c u s tomers to pay for merchandise using nationally recognizedcredit cardsissuedby a financial institution. To reducecosts,the management considering is eliminating one of these two payment systems. However, Virginia Duffy of the salesdepartmentsuspects that customersuse the two forms of payment in unequal numbers a n d t h a t c u s t o m e r sd i s p l a y d i f f e r e n t b u y i n g b e h a v i o r s when using the two forrns of payment. Therefore, she would like to first determrne a. the proportion of customersusing PAF and the proportion of customers using a credit card to pay for their purchases. b. the mean purchaseamount when using PAF and the meanpurchase amountwhen using a credit card. A s s i s t M s . D u f f y b y p r e p a r i r r ga n a p p r o p r i a t ea n a l y s i s basedon a random sampleof 50 transactions that she has preparedand placedin an internalfile on the OurCampus! Web site, www.prenhall.com/Springville/OurCampus_ P y m t S a m p l e . h t m . S u m m a r i z ey o u r f i n d i n g s a n d d e t e r mine whether Ms. Duffy's conjectures about OurCampus! customerpurchasing behaviors correct.If you want the are samplingerror to be no more than $3 when estimatingthe mean purchase amount, is Ms. Duffy's sanrple large enoughto perform a valid analysis? 1. Cochran,W. G., Sumplingkc'hnique.r, ed. (NewYork: 3rd Wiley, 1977). 2. Fisher, R. A., and F. Yates, Sraristical Tables fiir Biological, Agricultural and Medic'al Researc'h,5thed. (Edinburgh:Oliver & Boyd, 1957). 3. Kirk, R. E., ed., Statistical Issues: Reader /br the A (Beln-ront, BehavinraI Sc'ienc'es CA: Wadsworth, 1972). 4. Larsen, L., and M. L. Marx, An Introduction MqthR. to ematical Statistic's and lts Applic'ations,4th ed. (Upper SaddleRiver,NJ: PrenticeHall, 2006). 5. Microsoft Excel 2007 (Redmond,WA: Microsoft Corp., 2007). 6. Snedecor, W., and W G. Cochran" G. StatisticulMethods, IA: 7th ed. (Arrres, Iowa StateUniversityPress, 1980). 322 EXCEL coMpANIoN to chaDter 8 E8.1 COMPUTING THE CONFIDENCE INTERVAL ESTIMATE FOR THE MEAN (o KNOWN) You compute the confidence interval estimatefor the mean (o known) either by using the PHStat2 Estimate for the Mean, sigma known procedureor by making entries in the If you know the sarnplesize and sample mean of your sample, click Sample Statistics Known and enter those values.Otherwise,click Sample Statistics Unknown and enter the cell range of your sample as the Sample Cell Range. g@@EEworkbook. Usingthe CIE_SK Worksheet Open to the CIE_SK worksheet of the workbook. This worksheet uses the NORMSINV(P<E and CONFIDENCE(l-conJidence level,population standard deviution, sample siee) functions to compute the Z value and interval half-width for the Example 8.1 mean paper length problem on page 288. To adapt this worksheet to other problems, enter the appropriatepopulation standard deviation, sample mean, sample size, and confidence level values in the tinted cells 84 throush 87 and entera n e w t i t l e i n c e l lA l . Usinq PHStat2 Estimate for tFe Mean, Sigma Known SelectPHStat ) Confidence Intervals ) Estimate for the Mean, sigma known. ln the Estimate for the Mean, sigma known dialog box (shown below), enter values for the Population Standard Deviation and the Confidence Level. Click one of the input optionsand make the required entries. Enter a title as the Title and click OK. D*A Psplatim *ardsd Daviatbnl Cmfdencs Lwd: Inpt O$ions (? funpbstdisticJ Kno$"n 5ar* Siae: ffi*r E8.2 COMPUTING THE CONFIDENCE INTERVAL ESTIMATE FORTHE MEAN (o UNKNOWN) You compute the confidence interval estimatefor the mean (o unknown) either by using the PHStat2 Estimate for the Mean, sigma unknown procedure or by making entriesin the rkbook. EIEEEEEEEEIEwo Usinq PHStat2 Estimate for tFe Mean, Sigma Unknown SelectPHStat ) Confidence Intervals ) Estimate for the Mean, sigma unknown. In the Estimate for the Mean, sigma unknown dialog box (shown on page 323), entera Confidence Level value, click one of the input options, and make the required entries.Enter a title as the Title and click OK. If you know the sample size, sample mean, and sample standarddeviation of your sample,click Sample Statistics Known and enter those values.Otherwise,click Sample Statistics Unknown and enter the cell range of your sample as the Sample Cell Range. Sarndefthanl (^ San& Sf*isticE Ur*nwun t':t I I Ttle: I l* fir*e PafxJdbn Correction : '. r-bb I { ti - | r.'*.,-..."---.a| oK tl Csrcd I I E8.4: Computing Sanrple the SizeNeeded Estimating Mean for the 323 Data CmfklencsLevdl Inrut O$bns {3 5end6 Stati*ics Known 5ampbSiea: SanphMeanr Sarnph Std. Devi*ion: f Sstple StatistksUr{qrown box (shownbelow),entervaluesfor the SampleSize,the Number of Successes, the Confidence and Level.Entera title astheTitle andclick OK. lG-v" D€tE Sarpla Siza: Nunberaf $iccmcesr CmfidenceLevd: ls-"a J l. **pt O$ions er$rt OSions Ttle: I f finite PoprJation Correctbn I | ---....-.,,....,-*.,*,.1 [* fir*e Popr.dation Csrrection Heb I li € ol< il I Using the CIE_Pworksheet Heb II I oK li t-:::! il the Worksheet Using CIE_SU Open the clE_su worksheetof th.EIEEEE@IE to workbook. The worksheet(seeFigure 8.6 on page 293) uses theTINV( 1-conJid en ce I eveI, d egrees of fre ed om) function to determine the critical value from the r distribution and the compute interval half-width for the Section 8.2 Saxon ImprovementCompany example.To adaptthis workHome sheet other problems,changethe sample statisticsand to level values in the tinted cells 84 throush 87 confidence and entera new title in cell Al. Open to the CIE_P worksheet of the Q@EEE workbook. The worksheet(see Figure 8. I 0 on page 297) uses the NORMSINV(P<,\') function to determine the Z value and uses the square root function to compute the standarderror ofthe proportion for the Section8.3 Saxon H o m e I m p r o v e m e n t C o m p a n y e x a r n p l e .T o a d a p t t h i s worksheetto other problerns.enterthe appropriate sample s i z e , n u m b e r o f s u c c e s s e sa n d c o n f i d e n c e l e v e l v a l u e s , i n t h e t i n t e d c e l l s 8 4 , 8 5 , a n d 8 6 a n d e n t e ra n e w t i t l e i n cellAl. E8.4 COMPUTING THE SAMPLES/,ZE NEEDEDFOR ESTIMATING THE MEAN You computethe samplesize neededfor estimatingthe mean either by using the PHStat2 Determination for the Mean THE CONFIDENCE E8.3 COMPUTING INTERVAL ESTIMATE FORTHE PROPORTION You compute the confidence interval estimate for the proportion either by using the PHStat2 Estimate for the Proportion procedure or by making entries in the procedure by makingentries theEEEEEEEIEEEEE in or workbook. Using PHStat2 Determination for the Mean ) Select PHStat Sample ) Determination the Size for M e a n . I n t h e S a m p l e S i z e D e t e r m i n a t i o nf o r t h e M e a n dialog box (shown on page 324), enter values for the Population Standard Deviation, the Sampling Error, and the Confidence Level. Enter a title as the Title and click OK. q[f[[fis[[!workbook. UsinqPHStat2 Estimate for tf,e Proportion Select PHStat ) Confidence Intervals ) Estimate for theProportion. In the Estimatefor the Proportiondialog 324 EXCELcoMPANIoNro chanter 8 Dd6 Poprlatim $ardard DcviaHon: Sanphg Errsn CmfirJence Lenrd: G*pnt Ofrims t-- Data Estimatc Trrr ProporHml of 5anphrg Error: Cmf*Jence Lwd: O-tp.t O$iors r- lgs*qe Using the SampleSize*MWorksheet Open to the SampleSize_M worksheet of the Worksheet Using the SampleSize_P SampleSize_P worksheet of the The worksheet(see Figure @file. 8.12 on page304) usesthe NORMSINV and ROUNDUP functions,discussed SectionE8.3, for the SaxonHome in ImprovementCompany example in Section 8.3. To adapt this worksheetto other problems,enter the appropriate estimate of true proportion, sampling error, and confidence level valuesin the tinted cells 84 throush 86 and enter a new title in cell A l. Open to the (see workbook. worksheet Figure The f,!ffi!f@fl!![[! I 8.1 onpage 301) uses NORMSINV@<$ the function to compute the Z value and usesthe ROUNDUP(value) function to round up the sample size neededto the next higher integer for the Section 8.4 Saxon Home Improvement Company example.To adapt this worksheet to other problems, enter the appropriate population standarddeviation, sampling error, and confidence level values in the tinted cells 84" 85. and 86 and entera new title in cell A l . I E8.5 COMPUTING THE SAMPLESIZE NEEDEDFOR ESTIMATING THE PROPORTION You compute the sample size needed for estimating the proportion either by using the PHStat2 Determinationfor the Proportion procedure or by making entries in the @workbook. E8.6 COMPUTING THE CONFIDENCE INTERVAL ESTIMATE FORTHE POPULATION TOTAL Y o u c o r n p u t et h e c o n f i d e n c e i n t e r v a l e s t i m a t ef o r t h e population total either by using the PHStat2 Estimate for the Population Total procedure or by making entries in the ( *orkbook. EIEEIE[IE Using PHStat2 Determination for the Proportion SelectPHStat ) $n6p1e Size ) Determination for the P r o p o r t i o n . I n t h e S a m p l e S i z e D e t e r m i n a t i o nf o r t h e Proportion dialog box (shown at right), enter values for the Estimate of True Proportion, the Sampling Error, and the Confidence Level. Enter a title as the Title and click OK. Using PHStat2 Estimate for the Population Total Select PHStat ) Confidence Intervals) Estimate for the Population Total. In the Estimate the Population for Totaldialogbox (shown page325),entervalues on forthe PopulationSizeandthe ConfidenceLevel. Click one of the inputoptions andmakethe required entries. t I ! E8.7: Computingthe ConfidenceIntervalEstimatefor theTotal Difference 325 Usinq PHStat2 Estimate for tEe Total Difference Foptd*lmSa: Cmffdcncc Lcnd: - brtrlt O$iont {i Sarpte5t6tirtks Knonrn Sarpla Size: Sarplafvban: sefiph sttrtdtd Dwi*hnr Intervals) Estimatefor PHStat ) Confidence Select Difference. In the Estimatefor the Total the Total Difference dialogbox (shownbelow),entervaluesfor the SampleSize,the PopulationSize,and the Confidence Level. Enter the cell range of the differences the as Differences Cell Range.If the first cell in the columnof label, click First cell contains differences contains label, a enter title astheTitle. andclick OK. a l* ab (^ Sarpte $atlsticsurfoowr J " Dda ri , Sarn$ Sal PotrJationSza: Corfiderre Levd: lE-s. , Of,fsroncesCdlRanga: I fiZ rtstcdcor*ahsl# O"tput@tims J li or il cercd I TllEr I lf.*.-."'el mean, andsample If youknowthesample size,sample of click SampleStatistics deviation your sample, Otherwise, click Sample andenterthosevalues. Unknown and enterthe cell rangeof your samasthe Sample Cell Range. Enter a title as the Title and li OK ll Cancd I I oK. Worksheet Usingthe CIE_TD to Open gtr@file. the CIE_TD worksheet of This worksheet(seeFigure 8.14 on page 310) uses the TINV(I-conJidence level, degrees of freedom) function to determine the critical value from the I distribution and the interval half-width for the Section 8.5 Saxon Home Improvement Company total difference example.The worksheet also contains a calculation area in c e l l r a n g eD 9 : E 1 6 ,a s s h o w n i n F i g u r e E 8 . 1 , t h a t c o u n t s and sums the differences listed on a DifferencesData worksheet.Figure EB.2 illustratesthe first 6 of the 13 rows in the Difference Data worksheet). the Worksheet the CIE_T workbook. totheCIE-T worksheet theEIEIEEIEIE of (see the 307)uses worksheet Figure 8.13on page l-conjidence level, degrees offreedom) function to and the inethe criticalvaluefrom the I distribution half-width for the Section 8.5 Saxon Home Company population total example.To adapt to enterthe appropriate worksheet otherproblems, ationsize, sample mean, sample size, sample standeviation.and confidence level values in the tinted 84 through 88 and enter a new title in cell A I . THE 7 COMPUTING CONFIDENCE FOR ESTIMATE THE INTERVAL TOTAL DIFFERENCE the comoute confidence interval estimate for the total of or standarddeviation difierences Nol of 0i6erences = 0 ofDifersnces=B for DiferencesNol = 0 678 12 -COUtlT(DlFerencoellalalAAl 88 -85 - El1 -SUt(0lfieronc.doata!B:B) forDifisrences=O ofSouares of Differences 71.n - E r 2 ' ( 4 1 0 1 2 719,6664E l 3 + E l f 7. -El5/815 for nceeitherby usingthe PHStat2Estimate the Difference or orocedure bv makins entriesin the workbook. E8.1 FIGURE Calculationsarea in the Difference Data worksheet 326 ExcEL coMPANIoNto chapter 8 A Dlficrcncer B 2 3 4 5 5 -{A2 - gE_TDltB$10}^2 7.11 €.14rc-(/[3 . gE_riDltBtl0fz 17.fr mt.6161-0[4 - CIE_Tl]lf B$10]^ Z 8. 51.78-!45 - qE_T,-DltBtl0l^2 -(A5 - CtE-T:DttB$10r 5"21 18.5751 2 9.tr -IlBail^i 66.C formula in cell Bl3 to all rows with differencedata.if havemore than 12 differences; by deleting or columnB formulas,if you havefewerthan 12 di FIGURE E8.2 (partia DifferencesData worksheet l) E8.8 COMPUTING FINITE POPULATION CORRECTION FACTORS The workbooksfor confidenceinterval estimations of mean and proportion and for computing the sample needed estimatingthe meanor proportion includea for worksheetthat calculates confidenceinterval estimate the samplesize, using a finite population conection factor Section8.7 on the Student CD-ROM).(Opento those sheetsfor further information.) If you use PHStat2,you includethesecomputations clicking the Finite by Correction output option and enteringthe Population beforeclicking OK in the appropriate dialogboxes. To adaptthis worksheet otherproblems, you needto to changeboth the CIE_TD and DifferencesData worksheet. In the CIE_TD worksheet,enterthe appropriatepopulation size,samplesize,and confidencelevel valuesin the tinted cells 84 through86 and entera new title in cell Al. In the DiffererencesData worksheet, enterthe differencesin column A. Then adjustcolumn B by either copying down the