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Chapter 8 Hypothesis Testing with Two Samples © 2012 Pearson Education, Inc. All rights reserved. 1 of 70 Chapter Outline • 8.1 Testing the Difference Between Means (Large Independent Samples) • 8.2 Testing the Difference Between Means (Small Independent Samples) • 8.3 Testing the Difference Between Means (Dependent Samples) • 8.4 Testing the Difference Between Proportions © 2012 Pearson Education, Inc. All rights reserved. 2 of 70 Section 8.1 Testing the Difference Between Means (Large Independent Samples) © 2012 Pearson Education, Inc. All rights reserved. 3 of 70 Section 8.1 Objectives • Determine whether two samples are independent or dependent • Perform a two-sample z-test for the difference between two means μ1 and μ2 using large independent samples © 2012 Pearson Education, Inc. All rights reserved. 4 of 70 Two Sample Hypothesis Test • • Compares two parameters from two populations. Sampling methods: Independent Samples • The sample selected from one population is not related to the sample selected from the second population. Dependent Samples (paired or matched samples) • Each member of one sample corresponds to a member of the other sample. © 2012 Pearson Education, Inc. All rights reserved. 5 of 70 Independent and Dependent Samples Independent Samples Dependent Samples Sample 1 Sample 1 Sample 2 © 2012 Pearson Education, Inc. All rights reserved. Sample 2 6 of 70 Example: Independent and Dependent Samples Classify the pair of samples as independent or dependent. • Sample 1: Weights of 65 college students before their freshman year begins. • Sample 2: Weights of the same 65 college students after their freshmen year. Solution: Dependent Samples (The samples can be paired with respect to each student) © 2012 Pearson Education, Inc. All rights reserved. 7 of 70 Example: Independent and Dependent Samples Classify the pair of samples as independent or dependent. • Sample 1: Scores for 38 adults males on a psychological screening test for attention-deficit hyperactivity disorder. • Sample 2: Scores for 50 adult females on a psychological screening test for attention-deficit hyperactivity disorder. Solution: Independent Samples (Not possible to form a pairing between the members of the samples; the sample sizes are different, and the data represent scores for different individuals.) © 2012 Pearson Education, Inc. All rights reserved. 8 of 70 Two Sample Hypothesis Test with Independent Samples 1. Null hypothesis H0 A statistical hypothesis that usually states there is no difference between the parameters of two populations. Always contains the symbol ≤, =, or ≥. 2. Alternative hypothesis Ha A statistical hypothesis that is true when H0 is false. Always contains the symbol >, ≠, or <. © 2012 Pearson Education, Inc. All rights reserved. 9 of 70 Two Sample Hypothesis Test with Independent Samples H0: μ1 = μ2 Ha: μ1 ≠ μ2 H0: μ1 ≤ μ2 Ha: μ1 > μ2 H0: μ1 ≥ μ2 Ha: μ1 < μ2 Regardless of which hypotheses you use, you always assume there is no difference between the population means, or μ1 = μ2. © 2012 Pearson Education, Inc. All rights reserved. 10 of 70 Two Sample z-Test for the Difference Between Means Three conditions are necessary to perform a z-test for the difference between two population means μ1 and μ2. 1. The samples must be randomly selected. 2. The samples must be independent. 3. Each sample size must be at least 30, or, if not, each population must have a normal distribution with a known standard deviation. © 2012 Pearson Education, Inc. All rights reserved. 11 of 70 Two Sample z-Test for the Difference Between Means If these requirements are met, the sampling distribution for x1 x2 (the difference of the sample means) is a normal distribution with Mean: x x x x 1 2 1 2 1 2 12 22 Standard error: x x x2 x2 n n 1 2 1 2 1 2 Sampling distribution for x1 x2 : © 2012 Pearson Education, Inc. All rights reserved. σ x x 1 2 1 2 σ x x 1 2 x1 x2 12 of 70 Two Sample z-Test for the Difference Between Means • Test statistic is x1 x2. • The standardized test statistic is x1 x2 1 2 12 22 z where x x . x x n1 n2 1 1 2 2 • When the samples are large, you can use s1 and s2 in place of σ1 and σ2. If the samples are not large, you can still use a two-sample z-test, provided the populations are normally distributed and the population standard deviations are known. © 2012 Pearson Education, Inc. All rights reserved. 13 of 70 Using a Two-Sample z-Test for the Difference Between Means (Large Independent Samples) In Words 1. State the claim mathematically and verbally. Identify the null and alternative hypotheses. In Symbols State H0 and Ha. 2. Specify the level of significance. Identify α. 3. Determine the critical value(s). Use Table 4 in Appendix B. 4. Determine the rejection region(s). © 2012 Pearson Education, Inc. All rights reserved. 14 of 70 Using a Two-Sample z-Test for the Difference Between Means (Large Independent Samples) In Words 5. Find the standardized test statistic and sketch the sampling distribution. 6. Make a decision to reject or fail to reject the null hypothesis. In Symbols z x1 x2 1 2 x x 1 2 If z is in the rejection region, reject H0. Otherwise, fail to reject H0. 7. Interpret the decision in the context of the original claim. © 2012 Pearson Education, Inc. All rights reserved. 15 of 70 Example: Two-Sample z-Test for the Difference Between Means A credit card watchdog group claims that there is a difference in the mean credit card debts of households in New York and Texas. The results of a random survey of 250 households from each state are shown below. The two samples are independent. Do the results support the group’s claim? Use α = 0.05. (Adapted from PlasticRewards.com) New York Texas x1 $4446.25 x2 $4567.24 s1 = $1045.70 n1 = 250 s2 = $1361.95 n2 = 250 © 2012 Pearson Education, Inc. All rights reserved. 16 of 70 Solution: Two-Sample z-Test for the Difference Between Means • • • • • H0: μ 1 = μ 2 Ha: μ1 ≠ μ2 (claim) 0.05 n1= 250 , n2 = 250 Rejection Region: z = –1.11 © 2012 Pearson Education, Inc. All rights reserved. • Test Statistic: (4446.25 4567.24) 0 z 1.11 108.5983 • Decision: Fail to Reject H0. At the 5% level of significance, there is not enough evidence to support the group’s claim that there is a difference in the mean credit card debts of households in New York and Texas. 17 of 70 Example: Using Technology to Perform a Two-Sample z-Test A travel agency claims that the average daily cost of meals and lodging for vacationing in Texas is less than the same average costs for vacationing in Virginia. The table shows the results of a random survey of vacationers in each state. The two samples are independent. At α = 0.01, is there enough evidence to support the claim? (Adapted from American Automobile Association) Texas (1) Virginia (2) x1 $216 x2 $222 s2 = $24 s1 = $18 n1 = 50 © 2012 Pearson Education, Inc. All rights reserved. n2 = 35 18 of 70 Solution: Using Technology to Perform a Two-Sample z-Test TI-83/84 setup: • H0: μ 1 ≥ μ 2 • Ha : μ 1 < μ 2 (claim) Calculate: © 2012 Pearson Education, Inc. All rights reserved. Draw: 19 of 70 Solution: Using Technology to Perform a Two-Sample z-Test • Rejection Region: 0.01 –2.33 0 z • Decision: Fail to Reject H0 . At the 1% level of significance, there is not enough evidence to support the travel agency’s claim. –1.25 © 2012 Pearson Education, Inc. All rights reserved. 20 of 70 Section 8.1 Summary • Determined whether two samples are independent or dependent • Performed a two-sample z-test for the difference between two means μ1 and μ2 using large independent samples © 2012 Pearson Education, Inc. All rights reserved. 21 of 70 Section 8.2 Testing the Difference Between Means (Small Independent Samples) © 2012 Pearson Education, Inc. All rights reserved. 22 of 70 Section 8.2 Objectives • Perform a t-test for the difference between two population means μ1 and μ2 using small independent samples © 2012 Pearson Education, Inc. All rights reserved. 23 of 70 Two Sample t-Test for the Difference Between Means • If samples of size less than 30 are taken from normallydistributed populations, a t-test may be used to test the difference between the population means μ1 and μ2. • Three conditions are necessary to use a t-test for small independent samples. 1. The samples must be randomly selected. 2. The samples must be independent. 3. Each population must have a normal distribution. © 2012 Pearson Education, Inc. All rights reserved. 24 of 70 Two Sample t-Test for the Difference Between Means • The test statistic is x1 x2. • The standardized test statistic is x1 x2 1 2 t . sx x 1 2 • The standard error and the degrees of freedom of the sampling distribution depend on whether the population variances 12 and 22 are equal. © 2012 Pearson Education, Inc. All rights reserved. 25 of 70 Two Sample t-Test for the Difference Between Means • Variances are equal Information from the two samples is combined to calculate a pooled estimate of the standard deviation ˆ . n1 1 s12 n2 1 s22 ˆ n1 n2 2 The standard error for the sampling distribution of x1 x2 is sx x 1 2 1 1 ˆ n1 n2 d.f.= n1 + n2 – 2 © 2012 Pearson Education, Inc. All rights reserved. 26 of 70 Two Sample t-Test for the Difference Between Means • Variances are not equal If the population variances are not equal, then the standard error is sx x 2 s1 2 s2 . n1 n2 d.f = smaller of n1 – 1 and n2 – 1 1 2 © 2012 Pearson Education, Inc. All rights reserved. 27 of 70 Normal or t-Distribution? Are both sample sizes at least 30? Yes Use the z-test. No Are both populations normally distributed? No You cannot use the z-test or the t-test. Use the t-test with Yes Are both population standard deviations known? No Yes Use the z-test. Are the population variances equal? No sx x ˆ 1 2 1 1 n1 n2 d.f = n1 + n2 – 2. Use the t-test with sx x 1 © 2012 Pearson Education, Inc. All rights reserved. Yes 2 s12 s22 n1 n2 d.f = smaller of n1 – 1 or n2 – 1. 28 of 70 Two-Sample t-Test for the Difference Between Means (Small Independent Samples) In Words 1. State the claim mathematically and verbally. Identify the null and alternative hypotheses. 2. Specify the level of significance. 3. Determine the degrees of freedom. 4. Determine the critical value(s). © 2012 Pearson Education, Inc. All rights reserved. In Symbols State H0 and Ha. Identify α. d.f. = n1+ n2 – 2 or d.f. = smaller of n1 – 1 or n2 – 1. Use Table 5 in Appendix B. 29 of 70 Two-Sample t-Test for the Difference Between Means (Small Independent Samples) In Words In Symbols 5. Determine the rejection region(s). 6. Find the standardized test statistic and sketch the sampling distribution. x1 x2 1 2 t 7. Make a decision to reject or fail to reject the null hypothesis. If t is in the rejection region, reject H0. Otherwise, fail to reject H0. 8. Interpret the decision in the context of the original claim. © 2012 Pearson Education, Inc. All rights reserved. sx x 1 2 30 of 70 Example: Two-Sample t-Test for the Difference Between Means The results of a state mathematics test for random samples of students taught by two different teachers at the same school are shown below. Can you conclude that there is a difference in the mean mathematics test scores for the students of the two teachers? Use α = 0.01. Assume the populations are normally distributed and the population variances are not equal. Teacher 1 Teacher 2 x1 473 x2 459 s1 = 39.7 s2 = 24.5 n1 = 8 n2 = 18 © 2012 Pearson Education, Inc. All rights reserved. 31 of 70 Solution: Two-Sample t-Test for the Difference Between Means • • • • • H0: μ 1 = μ 2 Ha: μ1 ≠ μ2 (claim) α 0.10 d.f. = 8 – 1 = 7 Rejection Region: t ≈ 0.922 © 2012 Pearson Education, Inc. All rights reserved. • Test Statistic: t (473 459) 0 0.922 39.7 2 24.52 8 18 • Decision: Fail to Reject H0 . At the 10% level of significance, there is not enough evidence to support the claim that the mean mathematics test scores for the students of the two teachers are different. 32 of 70 Example: Two-Sample t-Test for the Difference Between Means A manufacturer claims that the mean calling range (in feet) of its 2.4-GHz cordless telephone is greater than that of its leading competitor. You perform a study using 14 randomly selected phones from the manufacturer and 16 randomly selected similar phones from its competitor. The results are shown below. At α = 0.05, can you support the manufacturer’s claim? Assume the populations are normally distributed and the population variances are equal. Manufacturer Competitor x1 1275ft x2 1250ft s1 = 45 ft s2 = 30 ft n1 = 14 n2 = 16 © 2012 Pearson Education, Inc. All rights reserved. 33 of 70 Solution: Two-Sample t-Test for the Difference Between Means • • • • • H0: μ 1 ≤ μ 2 Ha: μ1 > μ2 (claim) α = 0.05 d.f. = 14 + 16 – 2 = 28 Rejection Region: 0.05 0 1.701 t © 2012 Pearson Education, Inc. All rights reserved. 34 of 70 Solution: Two-Sample t-Test for the Difference Between Means • Test Statistic: sx1 x2 n1 1 s12 n2 1 s2 2 n1 n2 2 14 1 45 16 1 30 2 1 1 n1 n2 14 16 2 2 1 1 13.8018 14 16 x1 x2 1 2 1275 1250 0 t 1.811 sx1 x2 © 2012 Pearson Education, Inc. All rights reserved. 13.8018 35 of 70 Solution: Two-Sample t-Test for the Difference Between Means • • • • • H0: μ 1 ≤ μ 2 Ha: μ1 > μ2 (claim) α = 0.05 d.f. = 14 + 16 – 2 = 28 Rejection Region: 0.05 0 1.701 t 1.811 © 2012 Pearson Education, Inc. All rights reserved. • Test Statistic: t 1.811 • Decision: Reject H0 . At the 5% level of significance, there is enough evidence to support the manufacturer’s claim that its phone has a greater calling range than its competitors. 36 of 70 Section 8.2 Summary • Performed a t-test for the difference between two means μ1 and μ2 using small independent samples © 2012 Pearson Education, Inc. All rights reserved. 37 of 70 Section 8.3 Testing the Difference Between Means (Dependent Samples) © 2012 Pearson Education, Inc. All rights reserved. 38 of 70 Section 8.3 Objectives • Perform a t-test to test the mean of the differences for a population of paired data © 2012 Pearson Education, Inc. All rights reserved. 39 of 70 t-Test for the Difference Between Means • To perform a two-sample hypothesis test with dependent samples, the difference between each data pair is first found: d = x1 – x2 Difference between entries for a data pair • The test statistic is the mean d of these differences. d d Mean of the differences between paired n data entries in the dependent samples © 2012 Pearson Education, Inc. All rights reserved. 40 of 70 t-Test for the Difference Between Means Three conditions are required to conduct the test. 1. The samples must be randomly selected. 2. The samples must be dependent (paired). 3. Both populations must be normally distributed. If these requirements are met, then the sampling distribution for d is approximated by a t-distribution with n – 1 degrees of freedom, where n is the number of data pairs. -t0 © 2012 Pearson Education, Inc. All rights reserved. μd t0 d 41 of 70 Symbols used for the t-Test for μd Symbol Description n The number of pairs of data d The difference between entries for a data pair, d = x1 – x2 d The hypothesized mean of the differences of paired data in the population © 2012 Pearson Education, Inc. All rights reserved. 42 of 70 Symbols used for the t-Test for μd Symbol d sd Description The mean of the differences between the paired data entries in the dependent samples d d n The standard deviation of the differences between the paired data entries in the dependent samples 2 ( d ) 2 2 d (d d ) n sd n 1 n 1 © 2012 Pearson Education, Inc. All rights reserved. 43 of 70 t-Test for the Difference Between Means • The test statistic is d d . n • The standardized test statistic is d d t . sd n • The degrees of freedom are d.f. = n – 1. © 2012 Pearson Education, Inc. All rights reserved. 44 of 70 t-Test for the Difference Between Means (Dependent Samples) In Words 1. State the claim mathematically and verbally. Identify the null and alternative hypotheses. In Symbols State H0 and Ha. 2. Specify the level of significance. Identify α. 3. Determine the degrees of freedom. d.f. = n – 1 4. Determine the critical value(s). © 2012 Pearson Education, Inc. All rights reserved. Use Table 5 in Appendix B if n > 29 use the last row (∞) . 45 of 70 t-Test for the Difference Between Means (Dependent Samples) In Words In Symbols 5. Determine the rejection region(s). 6. Calculate d and sd . d d n (d ) 2 d (d d ) n sd n 1 n 1 2 5. Find the standardized test statistic. © 2012 Pearson Education, Inc. All rights reserved. 2 d d t sd n 46 of 70 t-Test for the Difference Between Means (Dependent Samples) In Words 8. Make a decision to reject or fail to reject the null hypothesis. In Symbols If t is in the rejection region, reject H0. Otherwise, fail to reject H0. 9. Interpret the decision in the context of the original claim. © 2012 Pearson Education, Inc. All rights reserved. 47 of 70 Example: t-Test for the Difference Between Means A shoe manufacturer claims that athletes can increase their vertical jump heights using the manufacturer’s new Strength Shoes®. The vertical jump heights of eight randomly selected athletes are measured. After the athletes have used the Strength Shoes® for 8 months, their vertical jump heights are measured again. The vertical jump heights (in inches) for each athlete are shown in the table. Assuming the vertical jump heights are normally distributed, is there enough evidence to support the manufacturer’s claim at α = 0.10? (Adapted from Coaches Sports Publishing) Athletes 1 2 3 4 5 6 7 8 Height (old) 24 22 25 28 35 32 30 27 Height (new) 26 25 25 29 33 34 35 30 © 2012 Pearson Education, Inc. All rights reserved. 48 of 70 Solution: Two-Sample t-Test for the Difference Between Means d = (jump height before shoes) – (jump height after shoes) • • • • • H0: μ d ≥ 0 Ha: μd < 0 (claim) α = 0.10 d.f. = 8 – 1 = 7 Rejection Region: © 2012 Pearson Education, Inc. All rights reserved. 49 of 70 Solution: Two-Sample t-Test for the Difference Between Means d = (jump height before shoes) – (jump height after shoes) Before 24 22 After 26 25 d –2 –3 d2 4 9 25 28 35 25 29 33 0 –1 2 0 1 4 32 30 34 35 27 30 –2 –5 4 25 –3 9 Σ = –14 Σ = 56 © 2012 Pearson Education, Inc. All rights reserved. d 14 d 1.75 n 8 2 ( d ) d 2 n sd n 1 (14) 56 8 8 1 2.1213 2 50 of 70 Solution: Two-Sample t-Test for the Difference Between Means d = (jump height before shoes) – (jump height after shoes) • Test Statistic: • H0: µd ≥ 0 • Ha: μd < 0 (claim) d d 1.75 0 t 2.333 • α = 0.10 sd n 2.1213 8 • d.f. = 8 – 1 = 7 • Decision: Reject H0. • Rejection Region: At the 10% level of significance, there is enough evidence to support the shoe manufacturer’s claim that athletes can increase their vertical jump heights using t ≈ –2.333 the new Strength Shoes®. © 2012 Pearson Education, Inc. All rights reserved. 51 of 70 Section 8.3 Summary • Performed a t-test to test the mean of the difference for a population of paired data © 2012 Pearson Education, Inc. All rights reserved. 52 of 70 Section 8.4 Testing the Difference Between Proportions © 2012 Pearson Education, Inc. All rights reserved. 53 of 70 Section 8.4 Objectives • Perform a z-test for the difference between two population proportions p1 and p2 © 2012 Pearson Education, Inc. All rights reserved. 54 of 70 Two-Sample z-Test for Proportions • Used to test the difference between two population proportions, p1 and p2. • Three conditions are required to conduct the test. 1. The samples must be randomly selected. 2. The samples must be independent. 3. The samples must be large enough to use a normal sampling distribution. That is, n1p1 ≥ 5, n1q1 ≥ 5, n2p2 ≥ 5, and n2q2 ≥ 5. © 2012 Pearson Education, Inc. All rights reserved. 55 of 70 Two-Sample z-Test for the Difference Between Proportions • If these conditions are met, then the sampling distribution for pˆ1 pˆ 2 is a normal distribution. • Mean: pˆ pˆ p1 p2 • A weighted estimate of p1 and p2 can be found by using x1 x2 p , where x1 n1 pˆ1 and x2 n2 pˆ 2 n1 n2 1 2 • Standard error: pˆ pˆ pq 1 1 , where q 1 p. n1 n2 1 2 © 2012 Pearson Education, Inc. All rights reserved. 56 of 70 Two-Sample z-Test for the Difference Between Proportions • The test statistic is pˆ1 pˆ 2 . • The standardized test statistic is z where p ( pˆ1 pˆ 2) ( p1 p2) 1 1 pq n1 n2 x1 x2 and q 1 p. n1 n2 Note: n1 p, n1q , n2 p, and n2q must be at least 5. © 2012 Pearson Education, Inc. All rights reserved. 57 of 70 Two-Sample z-Test for the Difference Between Proportions In Words In Symbols 1. State the claim. Identify the null and alternative hypotheses. State H0 and Ha. 2. Specify the level of significance. Identify α. 3. Determine the critical value(s). 4. Determine the rejection region(s). 5. Find the weighted estimate of p1 and p2. Verify that n1 p, n1q, n2 p, and n2q are at least 5. © 2012 Pearson Education, Inc. All rights reserved. Use Table 4 in Appendix B. x1 x2 p n1 n2 58 of 70 Two-Sample z-Test for the Difference Between Proportions In Words 6. Find the standardized test statistic and sketch the sampling distribution. 7. Make a decision to reject or fail to reject the null hypothesis. In Symbols z ( pˆ1 pˆ 2) ( p1 p2) 1 1 pq n1 n2 If z is in the rejection region, reject H0. Otherwise, fail to reject H0. 8. Interpret the decision in the context of the original claim. © 2012 Pearson Education, Inc. All rights reserved. 59 of 70 Example: Two-Sample z-Test for the Difference Between Proportions In a study of 150 randomly selected occupants in passenger cars and 200 randomly selected occupants in pickup trucks, 86% of occupants in passenger cars and 74% of occupants in pickup trucks wear seat belts. At α = 0.10 can you reject the claim that the proportion of occupants who wear seat belts is the same for passenger cars and pickup trucks? (Adapted from National Highway Traffic Safety Administration) Solution: 1 = Passenger cars © 2012 Pearson Education, Inc. All rights reserved. 2 = Pickup trucks 60 of 70 Solution: Two-Sample z-Test for the Difference Between Means • • • • • H0: p1 = p2 (claim) Ha : p 1 ≠ p 2 α = 0.10 n1= 150 , n2 = 200 Rejection Region: © 2012 Pearson Education, Inc. All rights reserved. 61 of 70 Solution: Two-Sample z-Test for the Difference Between Means x1 n1 pˆ1 129 x2 n2 pˆ 2 148 x1 x2 129 148 p 0.7914 n1 n2 150 200 q 1 p 1 0.7914 0.2086 Note: n1 p 200(0.2086) 5 n1q 200(0.6556) 5 n2 p 200(0.2086) 5 n2q 200(0.6556) 5 © 2012 Pearson Education, Inc. All rights reserved. 62 of 70 Solution: Two-Sample z-Test for the Difference Between Means pˆ1 pˆ 2 p1 p2 z 1 1 pq n1 n2 0.86 0.74 0 1 1 0.7914 0.2086 150 200 2.73 © 2012 Pearson Education, Inc. All rights reserved. 63 of 70 Solution: Two-Sample z-Test for the Difference Between Means • • • • • H0: p 1 = p 2 Ha : p 1 ≠ p 2 α 0.10 n1= 150 , n2 = 200 Rejection Region: © 2012 Pearson Education, Inc. All rights reserved. • Test Statistic: z 2.73 • Decision: Reject H0 At the 10% level of significance, there is enough evidence to reject the claim that the proportion of occupants who wear seat belts is the same for passenger cars and pickup trucks. 64 of 70 Example: Two-Sample z-Test for the Difference Between Proportions A medical research team conducted a study to test the effect of a cholesterol-reducing medication. At the end of the study, the researchers found that of the 4700 randomly selected subjects who took the medication, 301 died of heart disease. Of the 4300 randomly selected subjects who took a placebo, 357 died of heart disease. At α = 0.01 can you conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo? (Adapted from New England Journal of Medicine) Solution: 1 = Medication 2 = Placebo © 2012 Pearson Education, Inc. All rights reserved. 65 of 70 Solution: Two-Sample z-Test for the Difference Between Means • • • • • H0: p 1 ≥ p 2 Ha: p1 < p2 (claim) α 0.01 n1= 4700, n2 = 4300 Rejection Region: 0.01 –2.33 0 z © 2012 Pearson Education, Inc. All rights reserved. 66 of 70 Solution: Two-Sample z-Test for the Difference Between Means pˆ1 x1 301 0.064 n1 4700 pˆ 2 x2 357 0.083 n2 4300 x1 x2 301 357 p 0.0731 n1 n2 4700 4300 q 1 p 1 0.0731 0.9269 Note: n1 p 4700(0.0731) 5 n1q 4700(0.9269) 5 n2 p 4300(0.0731) 5 n2q 4300(0.9269) 5 © 2012 Pearson Education, Inc. All rights reserved. 67 of 70 Solution: Two-Sample z-Test for the Difference Between Means z pˆ1 pˆ 2 p1 p2 1 1 pq n1 n2 3.46 © 2012 Pearson Education, Inc. All rights reserved. 0.064 0.083 0 1 1 0.0731 0.9269 4700 4300 68 of 70 Solution: Two-Sample z-Test for the Difference Between Means • • • • • H0: p 1 ≥ p 2 Ha: p1 < p2 (claim) α 0.01 n1= 4700 , n2 = 4300 Rejection Region: 0.01 –2.33 0 z –3.46 © 2012 Pearson Education, Inc. All rights reserved. • Test Statistic: z 3.46 • Decision: Reject H0 At the 1% level of significance, there is enough evidence to conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo. 69 of 70 Section 8.4 Summary • Performed a z-test for the difference between two population proportions p1 and p2 © 2012 Pearson Education, Inc. All rights reserved. 70 of 70