Business and Economics 9th Edition

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Business and Economics 9th Edition Statistics for Business and Economics 9th Edition Chapter 10 Statistical Inferences About Means and Proportions with Two Populations

Two Sample Tests Two Sample Tests Population Means, Independent Samples Population Means, Matched Pairs 在第8章和第9章我们讲述了如何对一个总体均值及比例进行区间估计和假设检验。在本章中,我们将继续统计推断问题的讨论,讲述当两个总体均值或比例之差很重要的情况下,如何进行两个总体情形的区间估计和假设检验。 Population Proportions Interval Estimation Examples: Same group before vs. after treatment Group 1 vs. independent Group 2 Hypothesis Tests Proportion 1 vs. Proportion 2

Difference Between Two Means Population means, independent samples σx2 and σy2 known Test statistic is a z value σx2 and σy2 unknown 大样本 σx2 and σy2 assumed equal Test statistic is a a value from the Student’s t distribution 小样本 σx2 and σy2 assumed unequal

Difference Between Two Means Goal: Form a confidence interval for the difference between two population means, μx – μy Population means, independent samples 独立简单随机样本(independent simple random samples):总体1中的样本与总体2中的样本都是独立抽取的 Different data sources Unrelated Independent Sample selected from one population has no effect on the sample selected from the other population

Population means, independent samples σx2 and σy2 Known Population means, independent samples Assumptions: Samples are randomly and independently drawn both population distributions are normal Population variances are known * σx2 and σy2 known σx2 and σy2 unknown

Estimating the Difference Between Two Population Means Let 1 equal the mean of population 1 and 2 equal the mean of population 2. The difference between the two population means is 1 - 2. To estimate 1 - 2, we will select a simple random sample of size n1 from population 1 and a simple random sample of size n2 from population 2. Let equal the mean of sample 1 and equal the mean of sample 2. The point estimator of the difference between the means of the populations 1 and 2 is .

Sampling Distribution of Expected Value Standard Deviation (Standard Error) where: 1 = standard deviation of population 1 2 = standard deviation of population 2 n1 = sample size from population 1 n2 = sample size from population 2

Interval Estimation of 1 - 2: s 1 and s 2 Known Interval Estimate where: 1 -  is the confidence coefficient 在两个总体方差已知的情况下,大样本和小样本(假设两总体都是正态分布),都可以通过标准正态分布来求 样本区间。

Difference Between Two Means (continued) Population means, independent samples σx2 and σy2 known Test statistic is a z value σx2 and σy2 unknown 大样本 σx2 and σy2 assumed equal Test statistic is a a value from the Student’s t distribution 小样本 σx2 and σy2 assumed unequal

Population means, independent samples σx2 and σy2 Known Population means, independent samples Assumptions: Samples are randomly and independently drawn both population distributions are normal Population variances are known * σx2 and σy2 known σx2 and σy2 unknown

Population means, independent samples σx2 and σy2 Known (continued) When σx2 and σy2 are known and both populations are normal, the variance of X – Y is Population means, independent samples * σx2 and σy2 known …and the random variable has a standard normal distribution σx2 and σy2 unknown

Test Statistic, σx2 and σy2 Known Population means, independent samples The test statistic for μx – μy is: * σx2 and σy2 known σx2 and σy2 unknown

Interval Estimation of 1 - 2: s 1 and s 2 Known Interval Estimate where: 1 -  is the confidence coefficient 在两个总体方差已知的情况下,大样本和小样本(假设两总体都是正态分布),都可以通过标准正态分布来求 样本区间。

Hypothesis Tests for Two Population Means Two Population Means, Independent Samples Lower-tail test: H0: μx  μy H1: μx < μy i.e., H0: μx – μy  0 H1: μx – μy < 0 Upper-tail test: H0: μx ≤ μy H1: μx > μy i.e., H0: μx – μy ≤ 0 H1: μx – μy > 0 Two-tail test: H0: μx = μy H1: μx ≠ μy i.e., H0: μx – μy = 0 H1: μx – μy ≠ 0

Two Population Means, Independent Samples, Variances Known Decision Rules Two Population Means, Independent Samples, Variances Known Lower-tail test: H0: μ1 – μ2  0 H1: μ1 – μ2 < 0 Upper-tail test: H0: μ1 – μ2 ≤ 0 H1: μ1 – μ2 > 0 Two-tail test: H0: μ1 – μ2= 0 H1: μ1 – μy ≠ 0 a a a/2 a/2 -za za -za/2 za/2 Reject H0 if z < -za Reject H0 if z > za Reject H0 if z < -za/2 or z > za/2

Interval Estimation of 1 - 2: s 1 and s 2 Known Example: Par, Inc. Par, Inc. is a manufacturer of golf equipment and has developed a new golf ball that has been designed to provide “extra distance.” In a test of driving distance using a mechanical driving device, a sample of Par golf balls was compared with a sample of golf balls made by Rap, Ltd., a competitor. The sample statistics appear on the next slide.

Interval Estimation of 1 - 2: s 1 and s 2 Known Example: Par, Inc. Sample #1 Par, Inc. Sample #2 Rap, Ltd. Sample Size 120 balls 80 balls Sample Mean 275 yards 258 yards Based on data from previous driving distance tests, the two population standard deviations are known with s 1 = 15 yards and s 2 = 20 yards.

Interval Estimation of 1 - 2: s 1 and s 2 Known Example: Par, Inc. Let us develop a 95% confidence interval estimate of the difference between the mean driving distances of the two brands of golf ball.

Estimating the Difference Between Two Population Means Par, Inc. Golf Balls m1 = mean driving distance of Par golf balls Population 2 Rap, Ltd. Golf Balls m2 = mean driving distance of Rap golf balls m1 – m2 = difference between the mean distances Simple random sample of n1 Par golf balls x1 = sample mean distance for the Par golf balls Simple random sample of n2 Rap golf balls x2 = sample mean distance for the Rap golf balls x1 - x2 = Point Estimate of m1 – m2

Point Estimate of 1 - 2 Point estimate of 1 - 2 = = 275 - 258 = 17 yards where: 1 = mean distance for the population of Par, Inc. golf balls 2 = mean distance for the population of Rap, Ltd. golf balls

Interval Estimation of 1 - 2:  1 and  2 Known 17 + 5.14 or 11.86 yards to 22.14 yards We are 95% confident that the difference between the mean driving distances of Par, Inc. balls and Rap, Ltd. balls is 11.86 to 22.14 yards.

Hypothesis Tests About m 1 - m 2: s 1 and s 2 Known Example: Par, Inc. Can we conclude, using a = .01, that the mean driving distance of Par, Inc. golf balls is greater than the mean driving distance of Rap, Ltd. golf balls?

Hypothesis Tests About m 1 - m 2: s 1 and s 2 Known p –Value and Critical Value Approaches 1. Develop the hypotheses. H0: 1 - 2 < 0  Ha: 1 - 2 > 0 where: 1 = mean distance for the population of Par, Inc. golf balls 2 = mean distance for the population of Rap, Ltd. golf balls 2. Specify the level of significance. a = .01

Hypothesis Tests About m 1 - m 2: s 1 and s 2 Known p –Value and Critical Value Approaches 3. Compute the value of the test statistic.

Hypothesis Tests About m 1 - m 2: s 1 and s 2 Known p –Value Approach 4. Compute the p–value. For z = 6.49, the p –value < .0001. 5. Determine whether to reject H0. Because p–value < a = .01, we reject H0. At the .01 level of significance, the sample evidence indicates the mean driving distance of Par, Inc. golf balls is greater than the mean driving distance of Rap, Ltd. golf balls.

Hypothesis Tests About m 1 - m 2: s 1 and s 2 Known Critical Value Approach 4. Determine the critical value and rejection rule. For a = .01, z.01 = 2.33 Reject H0 if z > 2.33 5. Determine whether to reject H0. Because z = 6.49 > 2.33, we reject H0. The sample evidence indicates the mean driving distance of Par, Inc. golf balls is greater than the mean driving distance of Rap, Ltd. golf balls.

Population means, independent samples σx2 and σy2 Unknown, Population means, independent samples σx2 and σy2 known σx2 and σy2 unknown * 大样本 小样本

Test Statistic, σx2 and σy2 Unknown, Equal 大样本 (nx>=30且ny>=30) * 合并方差估计量是两个样本方差的加权平均,其取值点介于s1和是之间 小样本(nx<30或ny<30) σx2 and σy2 assumed equal σx2 and σy2 assumed unequal

Test Statistic, σx2 and σy2 Unknown, Equal 且 小样本 σx2 and σy2 unknown Assumptions: Samples are randomly and independently drawn Populations are normally distributed Population variances are unknown but assumed equal * σx2 and σy2 assumed equal 合并方差估计量是两个样本方差的加权平均,其取值点介于s1和是之间 σx2 and σy2 assumed unequal

Test Statistic, σx2 and σy2 Unknown, Equal 且 小样本 σx2 and σy2 unknown Forming interval estimates: The population variances are assumed equal, so use the two sample standard deviations and pool them to estimate σ use a t value with (nx + ny – 2) degrees of freedom * σx2 and σy2 assumed equal 合并方差估计量是两个样本方差的加权平均,其取值点介于s1和是之间 σx2 and σy2 assumed unequal

Test Statistic, σx2 and σy2 Unknown, Equal 且 小样本 σx2 and σy2 unknown σ 2的合并估计量: * σx2 and σy2 assumed equal 合并方差估计量是两个样本方差的加权平均,其取值点介于s1和是之间 σx2 and σy2 assumed unequal

Test Statistic, σx2 and σy2 Unknown, Equal 且 小样本 σx2 and σy2 unknown The test statistic for μx – μy is: * σx2 and σy2 assumed equal T分布并不局限于小样本情形,只要两个分布都服从服从正态分布且总体方差相等,它都是可行的。只是在大样本情形下,我们不需要使用t分布,用标准正态分布就可以了。 σx2 and σy2 assumed unequal Where t has (n1 + n2 – 2) d.f., and

Interval Estimation, σx2 and σy2 Unknown, Equal 且 小样本 σx2 and σy2 unknown * σx2 and σy2 assumed equal T分布并不局限于小样本情形,只要两个分布都服从服从正态分布且总体方差相等,它都是可行的。只是在大样本情形下,我们不需要使用t分布,用标准正态分布就可以了。 σx2 and σy2 assumed unequal Where t has (n1 + n2 – 2) d.f., and

σx2 and σy2 Unknown, Assumed Unequal 且 小样本 σx2 and σy2 unknown Assumptions: Samples are randomly and independently drawn Populations are normally distributed Population variances are unknown and assumed unequal σx2 and σy2 assumed equal T分布并不局限于小样本情形,只要两个分布都服从服从正态分布且总体方差相等,它都是可行的。只是在大样本情形下,我们不需要使用t分布,用标准正态分布就可以了。 * σx2 and σy2 assumed unequal

σx2 and σy2 Unknown, Assumed Unequal 且 小样本 σx2 and σy2 unknown Forming interval estimates: The population variances are assumed unequal, so a pooled variance is not appropriate use a t value with  degrees of freedom, where σx2 and σy2 assumed equal T分布并不局限于小样本情形,只要两个分布都服从服从正态分布且总体方差相等,它都是可行的。只是在大样本情形下,我们不需要使用t分布,用标准正态分布就可以了。 * σx2 and σy2 assumed unequal

Test Statistic, σx2 and σy2 Unknown, Unequal 且 小样本 σx2 and σy2 unknown The test statistic for μx – μy is: σx2 and σy2 assumed equal * σx2 and σy2 assumed unequal Where t has  degrees of freedom:

Interval Estimation, σx2 and σy2 Unknown, Unequal 且 小样本 σx2 and σy2 unknown σx2 and σy2 assumed equal * σx2 and σy2 assumed unequal Where t has  degrees of freedom:

Two Population Means, Independent Samples, Variances Unknown Decision Rules Two Population Means, Independent Samples, Variances Unknown Lower-tail test: H0: μx – μy  0 H1: μx – μy < 0 Upper-tail test: H0: μx – μy ≤ 0 H1: μx – μy > 0 Two-tail test: H0: μx – μy = 0 H1: μx – μy ≠ 0 a a a/2 a/2 -ta ta -ta/2 ta/2 Reject H0 if t < -tn-1, a Reject H0 if t > tn-1, a Reject H0 if t < -tn-1 , a/2 or t > tn-1 , a/2 Where t has n - 1 d.f.

Pooled Variance t Test: Example You are a financial analyst for a brokerage firm. Is there a difference in dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data: NYSE NASDAQ Number 21 25 Sample mean 3.27 2.53 Sample std dev 1.30 1.16 纽约股票交易所(NYSE)纳斯达克(NASDAQ)是指美国"全国证券交易商协会自动报价系统", Assuming both populations are approximately normal with equal variances, is there a difference in average yield ( = 0.05)?

Calculating the Test Statistic The test statistic is:

Solution t Decision: Reject H0 at a = 0.05 Conclusion: H0: μ1 - μ2 = 0 i.e. (μ1 = μ2) H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)  = 0.05 df = 21 + 25 - 2 = 44 Critical Values: t = ± 2.0154 Test Statistic: .025 .025 -2.0154 2.0154 t 2.040 Decision: Conclusion: Reject H0 at a = 0.05 There is evidence of a difference in means.

Two Sample Tests Two Sample Tests Population Means, Independent Samples Population Means, Matched Pairs 在第8章和第9章我们讲述了如何对一个总体均值及比例进行区间估计和假设检验。在本章中,我们将继续统计推断问题的讨论,讲述当两个总体均值或比例之差很重要的情况下,如何进行两个总体情形的区间估计和假设检验。 Population Proportions Interval Estimation Examples: Same group before vs. after treatment Group 1 vs. independent Group 2 Hypothesis Tests Proportion 1 vs. Proportion 2

Matched Pairs Tests Means of 2 Related Populations di = xi - yi Paired or matched samples Repeated measures (before/after) Use difference between paired values: Assumptions: Both Populations Are Normally Distributed Matched Pairs 匹配样本方案:在匹配样本方案中,两种生产方法是在相同条件下接受检验的,的以该方案会比独立样本方案产生更小的误差。 di = xi - yi

Test Statistic: Matched Pairs The test statistic for the mean difference is a t value, with n – 1 degrees of freedom: Matched Pairs Where D0 = hypothesized mean difference sd = sample standard dev. of differences n = the sample size (number of pairs)

Decision Rules: Matched Pairs Paired Samples Lower-tail test: H0: μx – μy  0 H1: μx – μy < 0 Upper-tail test: H0: μx – μy ≤ 0 H1: μx – μy > 0 Two-tail test: H0: μx – μy = 0 H1: μx – μy ≠ 0 a a a/2 a/2 -ta ta -ta/2 ta/2 Reject H0 if t < -tn-1, a Reject H0 if t > tn-1, a Reject H0 if t < -tn-1 , a/2 or t > tn-1 , a/2 has n - 1 d.f. Where

Matched Pairs Example  d = = - 4.2 Assume you send your salespeople to a “customer service” training workshop. Has the training made a difference in the number of complaints? You collect the following data:  di d = Number of Complaints: (2) - (1) Salesperson Before (1) After (2) Difference, di C.B. 6 4 - 2 T.F. 20 6 -14 M.H. 3 2 - 1 R.K. 0 0 0 M.O. 4 0 - 4 -21 n = - 4.2

Critical Value = ± 4.604 d.f. = n - 1 = 4 Matched Pairs: Solution Has the training made a difference in the number of complaints (at the  = 0.01 level)? Reject Reject H0: μx – μy = 0 H1: μx – μy  0 /2 /2  = .01 d = - 4.2 - 4.604 4.604 - 1.66 Critical Value = ± 4.604 d.f. = n - 1 = 4 Decision: Do not reject H0 (t stat is not in the reject region) Test Statistic: Conclusion: There is not a significant change in the number of complaints.

Two Sample Tests Two Sample Tests Population Means, Independent Samples Population Means, Matched Pairs 在第8章和第9章我们讲述了如何对一个总体均值及比例进行区间估计和假设检验。在本章中,我们将继续统计推断问题的讨论,讲述当两个总体均值或比例之差很重要的情况下,如何进行两个总体情形的区间估计和假设检验。 Population Proportions Interval Estimation Examples: Same group before vs. after treatment Group 1 vs. independent Group 2 Hypothesis Tests Proportion 1 vs. Proportion 2

Two Population Proportions Goal: Test hypotheses for the difference between two population proportions, P1 – P2 Population proportions Assumptions: Both sample sizes are large,

Two Population Proportions

Interval Estimation for Two Population Proportions

Test Statistic for Two Population Proportions The test statistic for H0: P1 – P2 = 0 is a z value: Population proportions Where

Decision Rules: Proportions Population proportions Lower-tail test: H0: p1 – p2  0 H1: p1 – p2 < 0 Upper-tail test: H0: p1 – p2 ≤ 0 H1: p1 – p2 > 0 Two-tail test: H0: p1– p2 = 0 H1: p1 – p2 ≠ 0 a a a/2 a/2 -za za -za/2 za/2 Reject H0 if z < -za Reject H0 if z > za Reject H0 if z < -za/2 or z > za/2

Example: Two Population Proportions Is there a significant difference between the proportion of men and the proportion of women who will vote Yes on Proposition A? In a random sample, 36 of 72 men and 31 of 50 women indicated they would vote Yes Test at the .05 level of significance

Example: Two Population Proportions (continued) The hypothesis test is: H0: PM – PW = 0 (the two proportions are equal) H1: PM – PW ≠ 0 (there is a significant difference between proportions) The sample proportions are: Men: = 36/72 = .50 Women: = 31/50 = .62 The estimate for the common overall proportion is:

Example: Two Population Proportions (continued) Reject H0 Reject H0 The test statistic for PM – PW = 0 is: .025 .025 -1.96 1.96 -1.31 Decision: Do not reject H0 Conclusion: There is not significant evidence of a difference between men and women in proportions who will vote yes. Critical Values = ±1.96 For  = .05

Chapter Summary Compared two dependent samples (paired samples) Performed paired sample t test for the mean difference Compared two independent samples Performed z test for the differences in two means Performed pooled variance t test for the differences in two means Compared two population proportions Performed z-test for two population proportions