Research Hypothesis

ContentStatistics for Business and Economics 6. Large-Sample Test of Hypothesis about a Population Proportion 7. Test of Hypothesis about a Population Variance Chapter 7 Inferences Based on a Single Sample: Tests of Hypotheses © 2011 Pearson Education, Inc 8. Calculating Type II Error Probabilities: More about β* 1 Content 3 Learning Objectives 1. The Elements of a Test of Hypothesis 1. Test a specific value of a population parameter (mean or proportion), called a test of hypothesis 2. Formulating Hypotheses and Setting Up the Rejection Region 3. Observed Significance Levels: p-Values 2. Provide a measure of reliability for the hypothesis test, called the significance level of the test 4. Test of Hypothesis about a Population Mean: Normal (z) Statistic 5. Test of Hypothesis about a Population Mean: Student’s t-Statistic © 2011 Pearson Education, Inc © 2011 Pearson Education, Inc 2 © 2011 Pearson Education, Inc 4 Learning Objectives 3. 4. Hypothesis Testing Test a specific value of a population parameter (mean, proportion or variance) called a test of hypothesis Show how to estimate the reliability of a test Population J J J J J J I believe the population mean age is 50 (hypothesis). Reject hypothesis! Not close. J Random sample Mean JX = 20J © 2011 Pearson Education, Inc 5 © 2011 Pearson Education, Inc 7 What’s a Hypothesis? 7.1 A statistical hypothesis is a statement about the numerical value of a population parameter. The Elements of a Test of Hypothesis I believe the mean GPA of this class is 3.5! © 1984-1994 T/Maker Co. © 2011 Pearson Education, Inc 6 © 2011 Pearson Education, Inc 8 Null Hypothesis Alternative Hypothesis The null hypothesis, denoted H0, represents the hypothesis that will be accepted unless the data provide convincing evidence that it is false. This usually represents the “status quo” or some claim about the population parameter that the researcher wants to test. © 2011 Pearson Education, Inc 1. Opposite of null hypothesis 2. The hypothesis that will be accepted only if the data provide convincing evidence of its truth 3. Designated Ha 4. Stated in one of the following forms Ha: µ ≠ (some value) Ha: µ < (some value) Ha: µ > (some value) © 2011 Pearson Education, Inc 9 11 Identifying Hypotheses Alternative Hypothesis Example problem: Test that the population mean is not 3 Steps: The alternative (research) hypothesis, denoted Ha, represents the hypothesis that will be accepted only if the data provide convincing evidence of its truth. This usually represents the values of a population parameter for which the researcher wants to gather evidence to support. • State the question statistically (µ ≠ 3) • State the opposite statistically (µ = 3) — Must be mutually exclusive & exhaustive • Select the alternative hypothesis (µ ≠ 3) — Has the ≠, <, or > sign • State the null hypothesis (µ = 3) © 2011 Pearson Education, Inc 10 © 2011 Pearson Education, Inc 12 Inc 16 .What Are the Hypotheses? What Are the Hypotheses? Is the population average amount of TV viewing 12 hours? Is the average cost per hat less than or equal to $20? • State the question statistically: µ = 12 • State the question statistically: µ ≤ 20 • State the opposite statistically: µ ≠ 12 • State the opposite statistically: µ > 20 • Select the alternative hypothesis: Ha: µ ≠ 12 • Select the alternative hypothesis: Ha: µ > 20 • State the null hypothesis: H0: µ = 12 • State the null hypothesis: H0: µ = 20 © 2011 Pearson Education. Inc 15 14 © 2011 Pearson Education. Inc What Are the Hypotheses? Is the population average amount of TV viewing different from 12 hours? Is the average amount spent in the bookstore greater than $25? • State the question statistically: µ ≠ 12 • State the question statistically: µ > 25 • State the opposite statistically: µ = 12 • State the opposite statistically: µ ≤ 25 • Select the alternative hypothesis: Ha: µ ≠ 12 • Select the alternative hypothesis: Ha: µ > 25 • State the null hypothesis: H0: µ = 12 • State the null hypothesis: H0: µ ≤ 25 © 2011 Pearson Education. Inc 13 What Are the Hypotheses? © 2011 Pearson Education. Inc © 2011 Pearson Education.400. in fact.400 is only .645 standard deviations above 2. 18 © 2011 Pearson Education.400. © 2011 Pearson Education. • The probability of committing a Type I error is denoted by α.Test Statistic Type I Error • A Type I error occurs if the researcher rejects the null hypothesis in favor of the alternative hypothesis when. The test statistic is a sample statistic. H0 is true.Example 19 Rejection Region The sampling distribution of assuming µ = 2. The chance of observing more than 1. Inc 17 Test Statistic . © 2011 Pearson Education.05 – if in fact the true mean µ is 2. that the researcher uses to decide between the null and alternative hypotheses. Inc The rejection region of a statistical test is the set of possible values of the test statistic for which the researcher will reject H0 in favor of Ha . Inc 20 . computed from information provided in the sample. in fact. – The theory generally represents the status quo. Inc 23 Elements of a Test of Hypothesis 3. Ha True Correct decision 22 © 2011 Pearson Education. . Rejection region: − The numerical values of the test statistic for which the null hypothesis will be rejected. • The probability of committing a Type II error is denoted by β.g. − The value of α is usually chosen to be small (e.. thereby leading to a Type I error.10) and is referred to as the level of significance of the test. Null hypothesis (H0): • A Type II error occurs if the researcher accepts the null hypothesis when. © 2011 Pearson Education. 2. 4. − The theory generally represents that which we will adopt only when sufficient evidence exists to establish its truth. . – A theory about the specific values of one or more population parameters. Inc © 2011 Pearson Education.Type II Error Elements of a Test of Hypothesis 1.01. Alternative (research) hypothesis (Ha): − A theory that contradicts the null hypothesis.05. or . H0 is false. − The rejection region is chosen so that the probability is α that it will contain the test statistic when the null hypothesis is true. which we adopt until it is proven false. Test statistic: A sample statistic used to decide whether to reject the null hypothesis. Inc 21 Conclusions and Consequences for a Test of Hypothesis True State of Nature Conclusion H0 True Accept H0 (Assume H0 True) Correct decision Type II error (probability β) Reject H0 (Assume Ha True) Type I error (probability α) © 2011 Pearson Education. Inc 24 . we do not reject H0. fraction. Thus. we reserve judgment about which hypothesis is true. spread Quantitative © 2011 Pearson Education. Inc 27 Determining the Target Parameter Elements of a Test of Hypothesis 7. b. Conclusion: a.Elements of a Test of Hypothesis Elements of a Test of Hypothesis 7. We do not conclude that the null hypothesis is true because we do not (in general) know the probability β that our test procedure will lead to an incorrect acceptance of H0 (Type II error). If the numerical value of the test statistic falls in the rejection region. Conclusion: 5. Inc 25 © 2011 Pearson Education. we reject the null hypothesis and conclude that the alternative hypothesis is true. We know that the hypothesis-testing process will lead to this conclusion incorrectly (Type I error) only 100α% of the time when H0 is true. variability. rate Qualitative σ2 Variance. Inc 26 Parameter Key Words or Phrases Type of Data µ Mean. Inc 28 . 6. © 2011 Pearson Education. percentage. average Quantitative p Proportion. Experiment and calculation of test statistic: Performance of the sampling experiment and determination of the numerical value of the test statistic. © 2011 Pearson Education. Assumptions: Clear statement (s) of any assumptions made about the population (s) being sampled. If the test statistic does not fall in the rejection region. that which will be presumed true unless the sampling experiment conclusively establishes the alternative hypothesis. lower-tailed (e. Ha: µ > 2. Two-tailed (e. Inc 31 One-Tailed Test A one-tailed test of hypothesis is one in which the alternative hypothesis is directional and includes the symbol “ < ” or “ >.g. Inc 2. One-tailed. H0: µ = 2.g.” 1..Steps for Selecting the Null and Alternative Hypotheses 7. Inc 32 . One-tailed.400) 30 © 2011 Pearson Education. upper-tailed (e. Select the alternative hypothesis as that which the sampling experiment is intended to establish.400) c. The null hypothesis will be specified as that parameter value closest to the alternative in one-tailed tests and as the complementary (or only unspecified) value in two-tailed tests.g. The alternative hypothesis will assume one of three forms: a..400) b.. Inc 29 Steps for Selecting the Null and Alternative Hypotheses © 2011 Pearson Education. Ha: µ ≠ 2. (e. Select the null hypothesis as the status quo. Ha: µ < 2.400) © 2011 Pearson Education.g..2 Formulating Hypotheses and Setting Up the Rejection Region © 2011 Pearson Education. .” Sampling Distribution Rejection Region H 0: µ = µ 0 1–α α Ho Value Ha: µ < 50 Sampling Distribution µ = 50 H0 © 2011 Pearson Education.. Inc 33 Basic Idea It is unlikely that we would get a sample mean of this value . Inc Sample Statistic 36 .Rejection Region (One-Tail Test) Two-Tailed Test A two-tailed test of hypothesis is one in which the alternative hypothesis does not specify departure from H0 in a particular direction and is written with the symbol “ ≠.... therefore. if in fact this were the population mean 20 35 Rejection Regions (Two-Tailed Test) Sampling Distribution H0: µ = 50 Sample Statistic © 2011 Pearson Education. Ha: µ < µ0 Fail to Reject Region Critical Value © 2011 Pearson Education. we reject the hypothesis that µ = 50. Rejection Region 1–α 1/2 α 1/2 α Fail to Reject Region .. Inc Level of Confidence Critical Value Sample Means 34 Ho Value Critical Value © 2011 Pearson Education. Inc H 0 : µ = µ0 Level of Confidence Ha: µ ≠ µ0 Rejection Region . 33 z > 2.28 Two-Tailed α = . for a specific statistical test is the probability (assuming H0 is true) of observing a value of the test statistic that is at least as contradictory to the null hypothesis.645 The observed significance level.10 z < –1.01 z < –2.28 UpperTailed z > 1.575 or z > 2.96 or z > 1. Inc 37 © 2011 Pearson Education.Rejection Regions p-Value Alternative Hypotheses LowerTailed α = . and supportive of the alternative hypothesis. or p-value. as the actual one computed from the sample data. reject H0 © 2011 Pearson Education. z < –2.96 α = .05 z < –1.645 or z > 1. Inc 39 If p-value ≥ α. Inc 40 . Inc p-Value 7. do not reject H0 If p-value < α.33 z < –1.645 z < –1.575 © 2011 Pearson Education.3 • • Observed Significance Levels: p-Values Probability of obtaining a test statistic more extreme (≤ or ≥) than actual sample value. given H0 is true Called observed level of significance • • 38 Smallest value of α for which H0 can be rejected Used to make rejection decision • • © 2011 Pearson Education.645 z > 1. 2. 1. Otherwise. if the alternative hypothesis is of the form > . Inc 43 Reporting Test Results as p-Values: How to Decide Whether to Reject H0 1. © 2011 Pearson Education. the p-value is equal to twice the tail area beyond the observed z-value in the direction of the sign of z – that is. If the observed significance level (p-value) of the test is less than the chosen value of α. do not reject the null hypothesis. the p-value is twice the area to the left of. If the test is two-tailed. if the alternative is of the form < . or above. the p-value is twice the area to the right of. Determine the value of the test statistic z corresponding to the result of the sampling experiment. Choose the maximum value of α that you are willing to tolerate. the p-value is the area to the left of. if z is negative. the observed z-value. the p-value is equal to the tail area beyond z in the same direction as the alternative hypothesis. the observed z-value. Conversely. reject the null hypothesis. Inc 41 Steps for Calculating the pValue for a Test of Hypothesis 2a.Steps for Calculating the pValue for a Test of Hypothesis Steps for Calculating the pValue for a Test of Hypothesis 2b. If the test is one-tailed. © 2011 Pearson Education. or below. Inc 44 . or above. if z is positive. the p-value is the area to the right of. the observed z-value. © 2011 Pearson Education. Thus. Inc 42 © 2011 Pearson Education. Conversely. the observed z-value. or below. Find the p-Value.50 or z ≥ 1.44 • © 2011 Pearson Education.0749 0 1. Inc 45 Two-Tailed z Test p-Value Solution 0 1.44 • ƒ .5.0749 z z value of sample statistic (observed) © 2011 Pearson Education.44 368 gm.1498 1/2 p-Value . Inc 48 . Inc –1.44 z z value of sample statistic (observed) 46 © 2011 Pearson Education.44 z 1/2 p-Value .0749 0 1.50) = .Two-Tailed z Test p-Value Example Two-Tailed Z Test p-Value Solution Does an average box of cereal contain 368 grams of cereal? A random sample of 64 boxes showed x = 372.5000 – .44 From z table: lookup 1.44 or z ≥ 1.4251 .44) 1/2 p-Value 1/2 p-Value . ‚ © 2011 Pearson Education. Inc 47 Two-Tailed z Test p-Value Solution p-value is P(z ≤ –1. The company has specified σ to be 25 grams.4251 –1. p-value is P(z ≤ –1. How does it compare to α = .0749 1/2 p-Value = . 1/2 p-Value = .05? z 1.44) ‚ Use alternative hypothesis to find direction p-Value .44 0 1. Inc 49 © 2011 Pearson Education.5.1498 ≥ α = .Two-Tailed z Test p-Value Solution One-Tailed z Test p-Value Solution p-Value = .44 Test statistic is in ‘Do not reject’ region 0 z • © 2011 Pearson Education. Inc One-Tailed z Test p-Value Example Does an average box of cereal contain more than 368 grams of cereal? A random sample of 64 boxes showed x = 372.05 Do not reject H0. Inc z value of sample statistic © 2011 Pearson Education. Find the pValue.4251 .0749 z z value of sample statistic 52 .025 –1. ƒ 50 From z table: lookup 1. The company has specified σ to be 25 grams.44 1.4251 0 368 gm.5000 – .025 1/2 α = .0749 Reject H0 Reject H0 1/2 α = . Inc „ .44 • © 2011 Pearson Education.44 51 One-Tailed z Test p-Value Solution p-Value is P(z ≥1. 0749 Use alternative hypothesis to find direction Reject H0 α = .4251 .5000 – .65) = .44 „ . Do not reject H0. p-Value is P(z ≤ -2.0749) ≥ (α = .8 mpg.4251 ƒ 0 From z table: lookup 1. ‚ p-Value = .004 • 54 „ .44) = . Inc From z table: lookup 2. You want to find out if the average miles per gallon of Escorts is less than 32 mpg.One-Tailed z Test p-Value Solution p-Value Thinking Challenge p-Value is P(z ≥ 1.0749 Use alternative hypothesis to find direction . Inc 55 z ƒ z value of sample statistic © 2011 Pearson Education.5000 – .05 0 1.65 0 z Test statistic is in ‘Do not reject’ region © 2011 Pearson Education.44 • You’re an analyst for Ford.0749 z 1.4960 –2.004. Inc © 2011 Pearson Education. Similar models have a standard deviation of 3. Inc 53 One-Tailed z Test p-Value Solution p-Value Solution* (p-Value = .0749 ‚ p-Value .0040 . p-Value < (α = .65 56 .01).05).7 mpg. What is the value of the observed level of significance (p-Value)? z value of sample statistic © 2011 Pearson Education. You take a sample of 60 Escorts & compute a sample mean of 30.4960 .44 p-Value . Reject H0. Inc 57 Two-Tailed Test H 0 : µ = µ0 Ha : µ ≠ µ 0 Test Statistic: Test Statistic: © 2011 Pearson Education.Large-Sample Test of Hypothesis about µ Converting a Two-Tailed p-Value from a Printout to a OneTailed p-Value if Ha is of the form > and z is positive or Ha is of the form < and z is negative if Ha is of the form > and z is negative Ha is of the form < and z is positive © 2011 Pearson Education. Inc One-Tailed Test H0: µ = µ0 Ha: µ < µ0 (or Ha: µ > µ0) 58 © 2011 Pearson Education. Inc 60 .4 One-Tailed Test Rejection region: z < –zα (or z > zα when Ha: µ > µ0) where zα is chosen so that P(z > zα) = α Test of Hypotheses about a Population Mean: Normal (z) Statistic © 2011 Pearson Education. Inc 59 Large-Sample Test of Hypothesis about µ 7. A random sample is selected from the target population. Inc 62 64 . conclude that the sampling experiment does not provide sufficient evidence to reject H0 at the α level of significance..] © 2011 Pearson Education. not the particular result of a single test. Remember that the confidence is in the testing process. If the calculated test statistic falls in the rejection region.) 2. [Generally. © 2011 Pearson Education. this condition guarantees that the test statistic will be approximately normal regardless of the shape of the underlying probability distribution of the population. reject H0 and conclude that the alternative hypothesis Ha is true. If the test statistic does not fall in the rejection region. 2. Inc 61 Conditions Required for a Valid Large-Sample Hypothesis Test for µ 1. n ≥ 30). (Due to the Central Limit Theorem.e. The sample size n is large (i. Inc © 2011 Pearson Education. © 2011 Pearson Education. Inc 63 Possible Conclusions for a Test of Hypothesis 1.Large-Sample Test of Hypothesis about µ Possible Conclusions for a Test of Hypothesis Two-Tailed Test Rejection region: |z| > zα/2 where zα/2 is chosen so that P(|z| > zα/2) = α/2 Note: µ0 is the symbol for the numerical value assigned to µ under the null hypothesis. we will not “accept” the null hypothesis unless the probability β of a Type II error has been calculated. State that you are rejecting H0 at the α level of significance. 7 lb.05 level of significance.5 lb.51 <1.96 < z = -0. You’re a Q/C inspector.025 –1.Two-Tailed z Test Thinking Challenge Two-Tailed z Test Example Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes had x = 372.05 Conclusion: No evidence average is not 368 © 2011 Pearson Education.025 –1. Inc 68 .5. You take a sample of 36 cords & compute a sample mean of 69. © 2011 Pearson Education.05 n = 36 Critical Value(s): Decision: Reject H 0 Reject H 0 . Inc 67 H0: µ = 70 Ha: µ ≠ 70 α = . You want to find out if a new machine is making electrical cords to customer specification: average breaking strength of 70 lb.05 level of significance.96 Two-Tailed z Test Solution* • • • • • Test Statistic: Reject H 0 z Decision: Do not reject at α = .025 .96 Do not reject at α = . is there evidence that the machine is not meeting the average breaking strength? 368 gm.96 0 1. with σ = 3. Inc Two-Tailed z Test Solution H0: µ = 368 Ha: µ ≠ 368 α = .96 .05 n = 25 • Critical Value(s): • • • • Reject H 0 . The company has specified σ to be 25 grams. Test at the . At the .96 z 66 Test Statistic: Since -1.05 Conclusion: No evidence average is not 70 © 2011 Pearson Education.025 0 1. Inc 65 © 2011 Pearson Education. One-Tailed z Test Example One-Tailed z Test Thinking Challenge Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = 372. Inc 71 -2. Test at the . Inc 72 .05 H0: µ = 32 Ha: µ < 32 α = . Similar models have a standard deviation of 3. Inc 69 One-Tailed z Test Solution • • • • • H0: µ = 368 Ha: µ > 368 α = . © 2011 Pearson Education. is there evidence that the miles per gallon is less than 32? 368 gm.8 mpg.5.05 0 1.01 level of significance. At the . Inc © 2011 Pearson Education. You take a sample of 60 Escorts & compute a sample mean of 30. You’re an analyst for Ford.7 mpg. You want to find out if the average miles per gallon of Escorts is at least 32 mpg.01 n = 60 Critical Value(s): Reject .33 70 0 z Test Statistic: Decision: Reject at α = .05 n = 25 Critical Value(s): Reject .01 Conclusion: No evidence average is more than 368 © 2011 Pearson Education.05 level of significance.01 Conclusion: There is evidence average is less than 32 © 2011 Pearson Education.645 z One-Tailed z Test Solution* • • • • • Test Statistic: Decision: Do not reject at α = . The company has specified σ to be 25 grams. A random sample is selected from the target population. Inc 73 Small-Sample Test of Hypothesis about µ 75 Conditions Required for a Valid Small-Sample Hypothesis Test for µ One-Tailed Test H0: µ = µ0 Ha: µ < µ0 (or Ha: µ > µ0) 1.5 Two-Tailed Test H0: µ = µ0 H a: µ ≠ µ 0 Test of Hypothesis about a Population Mean: Student’s t-Statistic Test statistic: Rejection region: |t| > tα/2 © 2011 Pearson Education.Small-Sample Test of Hypothesis about µ 7. 2. The population from which the sample is selected has a distribution that is approximately normal. Test statistic: Rejection region: t < –tα (or t > tα when Ha: µ > µ0) where tα and tα/2 are based on (n – 1) degrees of freedom © 2011 Pearson Education. Inc 74 © 2011 Pearson Education. Inc © 2011 Pearson Education. Inc 76 . 77 © 2011 Pearson Education. is the manufacturer correct? 368 gm.25 α = .01 level of significance. with a standard deviation of .25 lb.656 0 2. You take a random sample of 64 containers.656 t Decision: Do not reject at α = .025 -2. Inc Two-Tailed t Test Solution* Two-Tailed t Test Solution • • • • • H0: µ = 368 Ha: µ ≠ 368 α = .005 .025 .25 © 2011 Pearson Education.Two-Tailed t Test Example Does an average box of cereal contain 368 grams of cereal? A random sample of 36 boxes had a mean of 372. Two-Tailed t Test Thinking Challenge You work for the FTC. Test at the .05 level of significance.25 lb.05 df = 36 – 1 = 35 Critical Value(s): Reject H0 .05 Conclusion: There is evidence population average is not 368 © 2011 Pearson Education.25 Ha: µ ≠ 3. © 2011 Pearson Education.01 df = 64 – 1 = 63 Critical Value(s): Test Statistic: Reject H 0 Reject H0 . Inc 79 78 H0: µ = 3. A manufacturer of detergent claims that the mean weight of detergent is 3. You calculate the sample average to be 3.5 and a standard deviation of 12 grams. Inc 3.238 lb.01 Conclusion: There is no evidence average is not 3.030 • • • • • Test Statistic: Reject H 0 t Decision: Reject at α = .030 0 2. At the .005 -2. Inc 80 .117 lb. 833 t Decision: Do not reject at α = .729.05 df = 20 – 1 = 19 Critical Value(s): . Inc 84 . Test at the . reject H0 at α = . Inc 83 82 0 1.47 and a standard deviation of 2. is there evidence that the average bear sales per store is more than 5 ($ 00)? 81 © 2011 Pearson Education. Wal-Mart had teddy bears on sale last week.05 df = 10 – 1 = 9 Critical Value(s): Test Statistic: Reject H0 . © 2011 Pearson Education. Inc One-Tailed t Test Solution* One-Tailed t Test Solution • • • • • H0: µ = 140 Ha: µ < 140 α = . Assume a normal distribution.05 Reject H0 H 0: µ = 5 H a: µ > 5 α = .05 level of significance.05 -1.729 • • • • • Test Statistic: Decision: Since t = -2.66. Inc You’re a marketing analyst for WalMart. The weekly sales ($ 00) of bears sold in 10 stores was: 8 11 0 4 7 8 10 5 8 3 At the .05 level of significance.05 Conclusion: 0 t There is evidence population average is less than 140 © 2011 Pearson Education.05 Conclusion: There is no evidence average is more than 5 © 2011 Pearson Education.One-Tailed t Test Example One-Tailed t Test Thinking Challenge Is the average capacity of batteries less than 140 ampere-hours? A random sample of 20 batteries had a mean of 138.57 < -1. Inc 88 . Inc 85 Large-Sample Test of Hypothesis about p 87 Conditions Required for a Valid Large-Sample Hypothesis Test for p One-Tailed Test H0: p = p0 Ha: p < p0 (or Ha: p > p0) 1. A random sample is selected from a binomial population. The sample size n is large.Large-Sample Test of Hypothesis about p 7.) Test statistic: Rejection region: z < –zα (or z > zα when Ha: p > p0) Note: p0 is the symbol for the numerical value of p assigned in the null hypothesis © 2011 Pearson Education. 2. Inc © 2011 Pearson Education.6 Two-Tailed Test H0: p = p0 H a: p ≠ p 0 Large-Sample Test of Hypothesis about a Population Proportion Test statistic: Rejection region: |z| > zα /2 Note: p0 is the symbol for the numerical value of p assigned in the null hypothesis © 2011 Pearson Education. Inc 86 © 2011 Pearson Education. (This condition will be satisfied if both np0 ≥ 15 and nq0 ≥ 15. 96 z Decision: Do not reject at α = . © 2011 Pearson Education.12 < -1.05 level of significance? 89 © 2011 Pearson Education. A random sample of 500 transactions had 25 errors.10 α = .10 Ha: p < .04 α = . Does the new system produce fewer defects? Test at the . Inc 92 . Inc 91 • • • • • H0: p = .05 n = 500 Critical Value(s): Test Statistic: Reject H 0 Reject H 0 .05 Conclusion: There is evidence new z system < 10% defective 90 © 2011 Pearson Education. A year-end audit showed 4% of transactions had errors.05 level of significance.05 -1.05 Conclusion: There is evidence proportion is not 4% © 2011 Pearson Education.96 0 1.04 Ha: p ≠ .025 -1. a random sample of 200 boxes had 11 defects.05 n = 200 Critical Value(s): Reject H0 . Using a new system.645 0 Test Statistic: Decision: Since z = -2. You implement new procedures.One-Proportion z Test Example One-Proportion z Test Thinking Challenge The present packaging system produces 10% defective cereal boxes. Inc You’re an accounting manager.025 .645 Reject at α = . Has the proportion of incorrect transactions changed at the . Inc One-Proportion z Test Solution* One-Proportion z Test Solution • • • • • H0: p = . 93 © 2011 Pearson Education.7 One-Tailed Test H0: σ 2 = σ02 Ha: σ 2 < σ02 (or Ha: σ 2 > σ02) Test of Hypothesis about a Population Variance © 2011 Pearson Education. σ2. Inc 95 Test of a Hypothesis about σ 2 Variance Two-Tailed Test H0: σ 2 = σ02 H a: σ 2 ≠ σ 0 2 Although many practical problems involve inferences about a population mean (or proportion).Test of a Hypothesis about σ 2 7. © 2011 Pearson Education. Test statistic: Rejection region: where σ02 is the hypothesized variance and the distribution of χ2 is a chi-square distribution with (n – 1) degrees of freedom. it is sometimes of interest to make an inference about a population variance. Inc Test statistic: Rejection region: (or χ 2 > χα2 when Ha: σ 2 > σ02 ) where σ02 is the hypothesized variance and the distribution of χ2 is a chi-square distribution with (n – 1) degrees of freedom. Inc 94 © 2011 Pearson Education. Inc 96 . 05? α = ..1 = 2 χ2 Table (Portion) © 2011 Pearson Education.05 3.103 … © 2011 Pearson Education. Inc 97 Several χ2 probability Distributions Finding Critical Value Example What is the critical χ2 value given: Ha: σ2 > 0. A random sample is selected from the target population.004 … 0.95 … … … 0.7 Reject n=3 α =.991 χ2 Upper Tail Area … . Inc 99 98 0 DF .991 100 . © 2011 Pearson Education. 2. Inc . The population from which the sample is selected has a distribution that is approximately normal.05 df = n .Conditions Required for a Valid Hypothesis Test for s 2 Part of Table VI: Critical Values of Chi Square 1.010 5.995 1 .. Inc © 2011 Pearson Education. 2 0.841 5. 95 … … 0. What is the critical χ2 value given: Ha: σ 2 < 0. measured by the variance.401<28. equal to 15 grams? A random sample of 25 boxes had a variance of 17. Inc © 2011 Pearson Education. 2 0. Inc 104 .32<39.05? if the rejection region is on the left? © 2011 Pearson Education.7 Reject H 0 n=3 α = .010 Upper Tail Area for Lower Critical Value = 1–.05 Conclusion: There is no evidence σ 2 is not 15 © 2011 Pearson Education.025 .995 1 .05 α =.7 n=3 What do you do α =.05 df = 25 – 1 = 24 Test Statistic: (n − 1) s 2 (25 − 1) 17.05 = .364 Do not reject at α = .401 102 39. Test at the .7 2 χ = = σ 02 152 = 28. Inc 101 Chi-Square (χ2) Test Solution Finding Critical Value Example What is the critical χ2 value given: Ha: σ 2 < 0.103 DF .991 0 12..05? df = n . Inc 103 • • • • • H0: σ 2 = 15 Ha: σ 2 ≠ 15 α = ..841 5.Chi-Square (χ2) Test Example Finding Critical Value Example Is the variation in boxes of cereal.004 … … 0.05 level of significance.364 χ2 Decision: Since 12.95 χ2 Upper Tail Area … .7 grams.32 2 Critical Value(s): α/2 = .05 3.103 … © 2011 Pearson Education.1 = 2 χ2 Table (Portion) 0 . rather than risking an error of unknown magnitude. Inc 105 107 Steps for Calculating β for a LargeSample Test about µ Type II Error • The Type II error probability β is calculated assuming that the null hypothesis is false because it is defined as the probability of accepting H0 when it is false. we adopted a policy of nonrejection of H0 when the test statistic does not fall in the rejection region. corresponding to a test with level of significance α: Calculating Type II Error Probabilities: More about β © 2011 Pearson Education. There will be one border value for a one-tailed test and two for a two-tailed test. Inc 108 . Calculate the value(s) of x corresponding to the border(s) of the rejection region.Steps for Calculating β for a LargeSample Test about µ 7. © 2011 Pearson Education. Inc © 2011 Pearson Education. • The situation corresponding to accepting the null hypothesis. is not generally as controllable. Inc 106 Upper-tailed test: Lower-tailed test: Two-tailed test: © 2011 Pearson Education. • For that reason. The formula is one of the following.8 1. and thereby risking a Type II error. The company has specified σ to be 15 grams. Inc Power of Test © 2011 Pearson Education.Steps for Calculating β for a LargeSample Test about µ • 2. Then convert the border value(s) of to z-value(s) using the alternative distribution with mean µa. 112 . Inc 368 gm. Inc Probability of rejecting false H0 • Correct decision • Equal to 1 – β • Used in determining test adequacy • Affected by • • • 109 Steps for Calculating β for a LargeSample Test about µ True value of population parameter Significance level α Standard deviation & sample size n © 2011 Pearson Education. Sketch the alternative distribution (centered at µa) and shade the area in the acceptance (nonrejection) region. Test at the . Inc 110 111 Two-Tailed z Test Example Does an average box of cereal contain less than 368 grams of cereal? A random sample of 25 boxes had x = 372. Use the z-statistic(s) and Table IV in Appendix B to find the shaded area. © 2011 Pearson Education. Specify the value of µa in the alternative hypothesis for which the value of β is to be calculated. which is β.05 level of significance. The general formula for the z-value is © 2011 Pearson Education.5. 05 µ0 = 368 α = .065 µ© a2011=Pearson Inc Finding Power Steps 2 & 3 • Reject H0 Do Not Draw Reject H0 ‚ Specify ƒ Draw 1–β 360 µ© a2011= Pearson Education.846 β = . Inc • Reject H0 Hypothesis: H0: µ ≥ 368 Ha: µ < 368 114 ƒ x Draw … z Table 1–β =.05 µ0 = 368 x ‘True’ Situation: µa = 360 (Ha) β ‚ Specify x x Finding Power Step 5 µ0 = 368 ‘True’ Situation: µa = 360 (Ha) „ 1–β 113 α = .05 ƒ ‘True’ Situation: µa = 360 (Ha) ‚ 360Education.05 x Draw Specify Hypothesis: H0: µ ≥ 368 Ha: µ < 368 Do Not Draw Reject H0 µ0 = 368 x © 2011 Pearson Education.065 360Education. 363.154 363. Inc 115 • Reject H0 Hypothesis: H0: µ ≥ 368 Ha: µ < 368 Do Not Draw Reject H0 α = . µ© a2011=Pearson Inc x „ 116 .Finding Power Step 1 Finding Power Step 4 • Reject H0 Hypothesis: H0: µ ≥ 368 Ha: µ < 368 Do Not Draw Reject H0 α = . Inc © 2011 Pearson Education. For fixed n and α. 1. the value of β increases (the power decreases) as the value of α is decreased. Rate. the value of β decreases (the power increases) as the sample size n is increased. For fixed α and values of µ0 and µa. Probability σ 2 – Variance. Average 2. Variability. Spread 118 © 2011 Pearson Education. Inc 117 Properties of β and Power 119 Key Ideas Key Words for Identifying the Target Parameter µ – Mean. Inc p – Proportion. Fraction. For fixed n and values of µ0 and µa. © 2011 Pearson Education. the value of β decreases (the power increases) as the distance between the specified null value µ0 and the specified alternative value µa increases. Inc 120 . Percentage. © 2011 Pearson Education.Properties of β and Power Properties of β and Power 3. Test statistic (z. Significance level (α) Two-tailed : Ha : µ ≠ 50 5. Alternative hypothesis (Ha) 3. Inc 123 Key Ideas Errors in Hypothesis Testing Using p-values to Decide Type I Error = Reject H0 when H0 is true (occurs with probability α) 1. Conclusion © 2011 Pearson Education. If α > p-value. or χ2) Upper-tailed : Ha : µ > 50 4. Inc 124 . Null hypothesis (H0) Lower-tailed : Ha : µ < 50 2. Inc 122 © 2011 Pearson Education. Inc 121 Key Ideas © 2011 Pearson Education. Obtain p-value of the test Power of Test = P(Reject H0 when H0 is false) =1–β © 2011 Pearson Education. Choose significance level (α ) Type II Error = Accept H0 when H0 is false (occurs with probability β ) 3. t. reject H0 2.Key Ideas Key Ideas Elements of a Hypothesis Test Forms of Alternative Hypothesis 1. p-value 6. #133.Assignments • #97.#121.#130. #142 © 2011 Pearson Education. #132. Inc 125 .