# Chap 10-1 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-1 Chapter 10 Two-Sample Tests Basic Business Statistics 12 th Edition.

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Chap 10-1 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-1 Chapter 10 Two-Sample Tests Basic Business Statistics 12 th Edition

Chap 10-2 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-2 Learning Objectives In this chapter, you learn how to use hypothesis testing for comparing the difference between: The means of two independent populations The means of two related populations The proportions of two independent populations The variances of two independent populations by testing the ratio of the two variances

Chap 10-3 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-3 Two-Sample Tests Population Means, Independent Samples Population Means, Related Samples Population Variances Group 1 vs. Group 2 Same group before vs. after treatment Variance 1 vs. Variance 2 Examples: Population Proportions Proportion 1 vs. Proportion 2 DCOVA

Chap 10-4 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-4 Difference Between Two Means Population means, independent samples Goal: Test hypothesis or form a confidence interval for the difference between two population means, μ 1 – μ 2 The point estimate for the difference is X 1 – X 2 * σ 1 and σ 2 unknown, assumed equal σ 1 and σ 2 unknown, not assumed equal DCOVA

Chap 10-5 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-5 Difference Between Two Means: Independent Samples Population means, independent samples * Use S p to estimate unknown σ. Use a Pooled-Variance t test. σ 1 and σ 2 unknown, assumed equal σ 1 and σ 2 unknown, not assumed equal Use S 1 and S 2 to estimate unknown σ 1 and σ 2. Use a Separate-Variance t test. Different data sources Unrelated Independent Sample selected from one population has no effect on the sample selected from the other population DCOVA

Chap 10-6 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-6 Hypothesis Tests for Two Population Means Lower-tail test: H 0 : μ 1 μ 2 H 1 : μ 1 < μ 2 i.e., H 0 : μ 1 – μ 2 0 H 1 : μ 1 – μ 2 < 0 Upper-tail test: H 0 : μ 1 μ 2 H 1 : μ 1 > μ 2 i.e., H 0 : μ 1 – μ 2 0 H 1 : μ 1 – μ 2 > 0 Two-tail test: H 0 : μ 1 = μ 2 H 1 : μ 1 μ 2 i.e., H 0 : μ 1 – μ 2 = 0 H 1 : μ 1 – μ 2 0 Two Population Means, Independent Samples DCOVA

Chap 10-7 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-7 Two Population Means, Independent Samples Lower-tail test: H 0 : μ 1 – μ 2 0 H 1 : μ 1 – μ 2 < 0 Upper-tail test: H 0 : μ 1 – μ 2 0 H 1 : μ 1 – μ 2 > 0 Two-tail test: H 0 : μ 1 – μ 2 = 0 H 1 : μ 1 – μ 2 0 /2 -t -t /2 t t /2 Reject H 0 if t STAT < -t Reject H 0 if t STAT > t Reject H 0 if t STAT < -t /2 or t STAT > t /2 Hypothesis tests for μ 1 – μ 2 DCOVA

Chap 10-8 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-8 Population means, independent samples Hypothesis tests for µ 1 - µ 2 with σ 1 and σ 2 unknown and assumed equal Assumptions: Samples are randomly and independently drawn Populations are normally distributed or both sample sizes are at least 30 Population variances are unknown but assumed equal * σ 1 and σ 2 unknown, assumed equal σ 1 and σ 2 unknown, not assumed equal DCOVA

Chap 10-9 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-9 Population means, independent samples The pooled variance is: The test statistic is: Where t STAT has d.f. = (n 1 + n 2 – 2) (continued) * σ 1 and σ 2 unknown, assumed equal σ 1 and σ 2 unknown, not assumed equal Hypothesis tests for µ 1 - µ 2 with σ 1 and σ 2 unknown and assumed equal DCOVA

Chap 10-10 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-10 Population means, independent samples The confidence interval for μ 1 – μ 2 is: Where t α/2 has d.f. = n 1 + n 2 – 2 * Confidence interval for µ 1 - µ 2 with σ 1 and σ 2 unknown and assumed equal σ 1 and σ 2 unknown, assumed equal σ 1 and σ 2 unknown, not assumed equal DCOVA

Chap 10-11 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-11 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 Assuming both populations are approximately normal with equal variances, is there a difference in mean yield ( = 0.05)? DCOVA

Chap 10-12 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-12 Pooled-Variance t Test Example: Calculating the Test Statistic The test statistic is: (continued) H0: μ 1 - μ 2 = 0 i.e. (μ 1 = μ 2 ) H1: μ 1 - μ 2 0 i.e. (μ 1 μ 2 ) DCOVA

Chap 10-13 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-13 Pooled-Variance t Test Example: Hypothesis Test Solution H 0 : μ 1 - μ 2 = 0 i.e. (μ 1 = μ 2 ) H 1 : μ 1 - μ 2 0 i.e. (μ 1 μ 2 ) = 0.05 df = 21 + 25 - 2 = 44 Critical Values: t = ± 2.0154 Test Statistic: Decision: Conclusion: Reject H 0 at = 0.05 There is evidence of a difference in means. t 0 2.0154-2.0154.025 Reject H 0.025 2.040 DCOVA

Chap 10-14 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Excel Pooled-Variance t test Comparing NYSE & NASDAQ Chap 10-14 Pooled-Variance t Test for the Difference Between Two Means (assumes equal population variances) Data Hypothesized Difference0 Level of Significance0.05 Population 1 Sample Sample Size21=COUNT(DATA!\$A:\$A) Sample Mean3.27=AVERAGE(DATA!\$A:\$A) Sample Standard Deviation1.3=STDEV(DATA!\$A:\$A) Population 2 Sample Sample Size25=COUNT(DATA!\$B:\$B) Sample Mean2.53=AVERAGE(DATA!\$B:\$B) Sample Standard Deviation1.16=STDEV(DATA!\$B:\$B) Intermediate Calculations Population 1 Sample Degrees of Freedom20=B7 - 1 Population 2 Sample Degrees of Freedom24=B11 - 1 Total Degrees of Freedom44=B16 + B17 Pooled Variance1.502=((B16 * B9^2) + (B17 * B13^2)) / B18 Standard Error0.363=SQRT(B19 * (1/B7 + 1/B11)) Difference in Sample Means0.74=B8 - B12 t Test Statistic2.040=(B21 - B4) / B20 Two-Tail Test Lower Critical Value-2.015=-TINV(B5, B18) Upper Critical Value2.015=TINV(B5, B18) p-value0.047=TDIST(ABS(B22),B18,2) Reject the null hypothesis=IF(B27 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/5/1597301/slides/slide_14.jpg", "name": "Chap 10-14 Copyright ©2012 Pearson Education, Inc.", "description": "publishing as Prentice Hall Excel Pooled-Variance t test Comparing NYSE & NASDAQ Chap 10-14 Pooled-Variance t Test for the Difference Between Two Means (assumes equal population variances) Data Hypothesized Difference0 Level of Significance0.05 Population 1 Sample Sample Size21=COUNT(DATA!\$A:\$A) Sample Mean3.27=AVERAGE(DATA!\$A:\$A) Sample Standard Deviation1.3=STDEV(DATA!\$A:\$A) Population 2 Sample Sample Size25=COUNT(DATA!\$B:\$B) Sample Mean2.53=AVERAGE(DATA!\$B:\$B) Sample Standard Deviation1.16=STDEV(DATA!\$B:\$B) Intermediate Calculations Population 1 Sample Degrees of Freedom20=B7 - 1 Population 2 Sample Degrees of Freedom24=B11 - 1 Total Degrees of Freedom44=B16 + B17 Pooled Variance1.502=((B16 * B9^2) + (B17 * B13^2)) / B18 Standard Error0.363=SQRT(B19 * (1/B7 + 1/B11)) Difference in Sample Means0.74=B8 - B12 t Test Statistic2.040=(B21 - B4) / B20 Two-Tail Test Lower Critical Value-2.015=-TINV(B5, B18) Upper Critical Value2.015=TINV(B5, B18) p-value0.047=TDIST(ABS(B22),B18,2) Reject the null hypothesis=IF(B27

Chap 10-15 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Minitab Pooled-Variance t test Comparing NYSE & NASDAQ Chap 10-15 Decision: Conclusion: Reject H 0 at = 0.05 There is evidence of a difference in means. Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 21 3.27 1.30 0.28 2 25 2.53 1.16 0.23 Difference = mu (1) - mu (2) Estimate for difference: 0.740 95% CI for difference: (0.009, 1.471) T-Test of difference = 0 (vs not =): T-Value = 2.04 P-Value = 0.047 DF = 44 Both use Pooled StDev = 1.2256 DCOVA

Chap 10-16 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-16 Pooled-Variance t Test Example: Confidence Interval for µ 1 - µ 2 Since we rejected H 0 can we be 95% confident that µ NYSE > µ NASDAQ ? 95% Confidence Interval for µ NYSE - µ NASDAQ Since 0 is less than the entire interval, we can be 95% confident that µ NYSE > µ NASDAQ DCOVA

Chap 10-17 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-17 Population means, independent samples Hypothesis tests for µ 1 - µ 2 with σ 1 and σ 2 unknown, not assumed equal Assumptions: Samples are randomly and independently drawn Populations are normally distributed or both sample sizes are at least 30 Population variances are unknown and cannot be assumed to be equal * σ 1 and σ 2 unknown, assumed equal σ 1 and σ 2 unknown, not assumed equal DCOVA

Chap 10-18 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-18 Population means, independent samples (continued) * σ 1 and σ 2 unknown, assumed equal σ 1 and σ 2 unknown, not assumed equal Hypothesis tests for µ 1 - µ 2 with σ 1 and σ 2 unknown and not assumed equal The test statistic is: t STAT has d.f. ν = DCOVA

Chap 10-19 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-19 Related Populations The Paired Difference Test Tests Means of 2 Related Populations Paired or matched samples Repeated measures (before/after) Use difference between paired values: Eliminates Variation Among Subjects Assumptions: Both Populations Are Normally Distributed Or, if not Normal, use large samples Related samples D i = X 1i - X 2i DCOVA

Chap 10-20 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-20 Related Populations The Paired Difference Test The i th paired difference is D i, where Related samples D i = X 1i - X 2i The point estimate for the paired difference population mean μ D is D : n is the number of pairs in the paired sample The sample standard deviation is S D (continued) DCOVA

Chap 10-21 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-21 The test statistic for μ D is: Paired samples Where t STAT has n - 1 d.f. The Paired Difference Test: Finding t STAT DCOVA

Chap 10-22 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-22 Lower-tail test: H 0 : μ D 0 H 1 : μ D < 0 Upper-tail test: H 0 : μ D 0 H 1 : μ D > 0 Two-tail test: H 0 : μ D = 0 H 1 : μ D 0 Paired Samples The Paired Difference Test: Possible Hypotheses /2 -t -t /2 t t /2 Reject H 0 if t STAT < -t Reject H 0 if t STAT > t Reject H 0 if t STAT < -t or t STAT > t Where t STAT has n - 1 d.f. DCOVA

Chap 10-23 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-23 The confidence interval for μ D is Paired samples where The Paired Difference Confidence Interval DCOVA

Chap 10-24 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-24 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: Paired Difference Test: Example Number of Complaints: (2) - (1) Salesperson Before (1) After (2) Difference, D i 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 D = DiDi n = -4.2 DCOVA

Chap 10-25 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-25 Has the training made a difference in the number of complaints (at the 0.01 level)? - 4.2D = H 0 : μ D = 0 H 1 : μ D 0 Test Statistic: t 0.005 = ± 4.604 d.f. = n - 1 = 4 Reject /2 - 4.604 4.604 Decision: Do not reject H 0 (t stat is not in the reject region) Conclusion: There is insufficient evidence there is significant change in the number of complaints. Paired Difference Test: Solution Reject /2 - 1.66 =.01 DCOVA

Chap 10-26 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Paired t Test In Excel Chap 10-26 Paired t Test Data Hypothesized Mean Diff.0 Level of Significance0.05 Intermediate Calculations Sample Size5=COUNT(I2:I6) Dbar-4.2=AVERAGE(I2:I6) Degrees of Freedom4=B8 - 1 SDSD 5.67=STDEV(I2:I6) Standard Error2.54=B11/SQRT(B8) t-Test Statistic-1.66=(B9 - B4)/B12 Two-Tail Test Lower Critical Value-2.776=-TINV(B5,B10) Upper Critical Value2.776=TINV(B5,B10) p-value0.173=TDIST(ABS(B13),B10,2) Do not reject the null Hypothesis =IF(B18 0.05 we do not reject the null hypothesis. Thus we conclude that there is Insufficient evidence to conclude there is a difference in the average number of complaints.

Chap 10-27 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Paired t Test In Minitab Yields The Same Conclusions Chap 10-27 DCOVA Paired T-Test and CI: After, Before Paired T for After - Before N Mean StDev SE Mean After 5 2.40 2.61 1.17 Before 5 6.60 7.80 3.49 Difference 5 -4.20 5.67 2.54 95% CI for mean difference: (-11.25, 2.85) T-Test of mean difference = 0 (vs not = 0): T-Value = -1.66 P-Value = 0.173

Chap 10-28 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-28 Two Population Proportions Goal: test a hypothesis or form a confidence interval for the difference between two population proportions, π 1 – π 2 The point estimate for the difference is Population proportions DCOVA

Chap 10-29 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-29 Two Population Proportions Population proportions The pooled estimate for the overall proportion is: where X 1 and X 2 are the number of items of interest in samples 1 and 2 In the null hypothesis we assume the null hypothesis is true, so we assume π 1 = π 2 and pool the two sample estimates DCOVA

Chap 10-30 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-30 Two Population Proportions Population proportions The test statistic for π 1 – π 2 is a Z statistic: (continued) where DCOVA

Chap 10-31 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-31 Hypothesis Tests for Two Population Proportions Population proportions Lower-tail test: H 0 : π 1 π 2 H 1 : π 1 < π 2 i.e., H 0 : π 1 – π 2 0 H 1 : π 1 – π 2 < 0 Upper-tail test: H 0 : π 1 π 2 H 1 : π 1 > π 2 i.e., H 0 : π 1 – π 2 0 H 1 : π 1 – π 2 > 0 Two-tail test: H 0 : π 1 = π 2 H 1 : π 1 π 2 i.e., H 0 : π 1 – π 2 = 0 H 1 : π 1 – π 2 0 DCOVA

Chap 10-32 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-32 Hypothesis Tests for Two Population Proportions Population proportions Lower-tail test: H 0 : π 1 – π 2 0 H 1 : π 1 – π 2 < 0 Upper-tail test: H 0 : π 1 – π 2 0 H 1 : π 1 – π 2 > 0 Two-tail test: H 0 : π 1 – π 2 = 0 H 1 : π 1 – π 2 0 /2 -z -z /2 z z /2 Reject H 0 if Z STAT < -Z Reject H 0 if Z STAT > Z Reject H 0 if Z STAT < -Z or Z STAT > Z (continued) DCOVA

Chap 10-33 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-33 Hypothesis Test 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 35 of 50 women indicated they would vote Yes Test at the.05 level of significance DCOVA

Chap 10-34 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-34 The hypothesis test is: H 0 : π 1 – π 2 = 0 (the two proportions are equal) H 1 : π 1 – π 2 0 (there is a significant difference between proportions) The sample proportions are: Men: p 1 = 36/72 = 0.50 Women: p 2 = 35/50 = 0.70 The pooled estimate for the overall proportion is: Hypothesis Test Example: Two population Proportions (continued) DCOVA

Chap 10-35 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-35 The test statistic for π 1 – π 2 is: Hypothesis Test Example: Two population Proportions (continued).025 -1.961.96.025 -2.20 Decision: Reject H 0 Conclusion: There is evidence of a difference in proportions who will vote yes between men and women. Reject H 0 Critical Values = ±1.96 For =.05 DCOVA

Chap 10-36 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Two Proportion Test In Excel Chap 10-36 DCOVA Z Test for Differences in Two Proportions Data Hypothesized Difference0 Level of Significance0.05 Group 1 Number of items of interest36 Sample Size72 Group 2 Number of items of interest35 Sample Size50 Intermediate Calculations Group 1 Proportion0.5=B7/B8 Group 2 Proportion0.7=B10/B11 Difference in Two Proportions-0.2=B14 - B15 Average Proportion0.582=(B7 + B10)/(B8 + B11) Z Test Statistic-2.20=(B16-B4)/SQRT((B17*(1-B17))*(1/B8+1/B11)) Two-Tail Test Lower Critical Value-1.96=NORMSINV(B5/2) Upper Critical Value1.96=NORMSINV(1 - B5/2) p-value0.028=2*(1 - NORMSDIST(ABS(B18))) Reject the null hypothesis=IF(B23 < B5,"Reject the null hypothesis", "Do not reject the null hypothesis") Decision: Reject H 0 Conclusion: There is evidence of a difference in proportions who will vote yes between men and women. Since -2.20 < -1.96 Or Since p-value = 0.028 < 0.05 We reject the null hypothesis

Chap 10-37 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Two Proportion Test In Minitab Shows The Same Conclusions Chap 10-37 DCOVA Test and CI for Two Proportions Sample X N Sample p 1 36 72 0.500000 2 35 50 0.700000 Difference = p (1) - p (2) Estimate for difference: -0.2 95% CI for difference: (-0.371676, -0.0283244) Test for difference = 0 (vs not = 0): Z = -2.28 P-Value = 0.022

Chap 10-38 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-38 Confidence Interval for Two Population Proportions Population proportions The confidence interval for π 1 – π 2 is: DCOVA

Chap 10-39 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-39 Testing for the Ratio Of Two Population Variances Tests for Two Population Variances F test statistic H 0 : σ 1 2 = σ 2 2 H 1 : σ 1 2 σ 2 2 H 0 : σ 1 2 σ 2 2 H 1 : σ 1 2 > σ 2 2 * HypothesesF STAT = Variance of sample 1 (the larger sample variance) n 1 = sample size of sample 1 = Variance of sample 2 (the smaller sample variance) n 2 = sample size of sample 2 n 1 –1 = numerator degrees of freedom n 2 – 1 = denominator degrees of freedom Where: DCOVA

Chap 10-40 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-40 The F critical value is found from the F table There are two degrees of freedom required: numerator and denominator The larger sample variance is always the numerator When In the F table, numerator degrees of freedom determine the column denominator degrees of freedom determine the row The F Distribution df 1 = n 1 – 1 ; df 2 = n 2 – 1 DCOVA

Chap 10-41 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-41 Finding the Rejection Region H 0 : σ 1 2 = σ 2 2 H 1 : σ 1 2 σ 2 2 H 0 : σ 1 2 σ 2 2 H 1 : σ 1 2 > σ 2 2 F 0 FαFα Reject H 0 Do not reject H 0 Reject H 0 if F STAT > F α F 0 /2 Reject H 0 Do not reject H 0 F α/2 Reject H 0 if F STAT > F α/2 DCOVA

Chap 10-42 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-42 F Test: An Example You are a financial analyst for a brokerage firm. You want to compare dividend yields between stocks listed on the NYSE & NASDAQ. You collect the following data : NYSE NASDAQ Number 2125 Mean3.272.53 Std dev1.301.16 Is there a difference in the variances between the NYSE & NASDAQ at the = 0.05 level? DCOVA

Chap 10-43 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-43 F Test: Example Solution Form the hypothesis test: ( there is no difference between variances) ( there is a difference between variances) Find the F critical value for = 0.05: Numerator d.f. = n 1 – 1 = 21 –1 = 20 Denominator d.f. = n2 – 1 = 25 –1 = 24 F α/2 = F.025, 20, 24 = 2.33 DCOVA

Chap 10-44 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-44 The test statistic is: 0 /2 =.025 F 0.025 =2.33 Reject H 0 Do not reject H 0 H 0 : σ 1 2 = σ 2 2 H 1 : σ 1 2 σ 2 2 F Test: Example Solution F STAT = 1.256 is not in the rejection region, so we do not reject H 0 (continued) Conclusion: There is insufficient evidence of a difference in variances at =.05 F DCOVA

Chap 10-45 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Two Variance F Test In Excel Chap 10-45 DCOVA F Test for Differences in Two Variables Data Level of Significance0.05 Larger-Variance Sample Sample Size21 Sample Variance1.6900=1.3^2 Smaller-Variance Sample Sample Size25 Sample Variance1.3456=1.16^2 Intermediate Calculations F Test Statistic1.2561.255945 Population 1 Sample Degrees of Freedom20=B6 - 1 Population 2 Sample Degrees of Freedom24=B9 - 1 Two-Tail Test Upper Critical Value2.327=FINV(B4/2,B14,B15) p-value0.589=2*FDIST(B13,B14,B15) Do not reject the null hypothesis=IF(B19 0.05 = α.

Chap 10-46 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Two Variance F Test In Minitab Yields The Same Conclusion Chap 10-46 DCOVA Test and CI for Two Variances Null hypothesis Sigma(1) / Sigma(2) = 1 Alternative hypothesis Sigma(1) / Sigma(2) not = 1 Significance level Alpha = 0.05 Statistics Sample N StDev Variance 1 21 1.300 1.690 2 25 1.160 1.346 Ratio of standard deviations = 1.121 Ratio of variances = 1.256 95% Confidence Intervals CI for Distribution CI for StDev Variance of Data Ratio Ratio Normal (0.735, 1.739) (0.540, 3.024) Tests Test Method DF1 DF2 Statistic P-Value F Test (normal) 20 24 1.26 0.589

Chap 10-47 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-47 Chapter Summary Compared two independent samples Performed pooled-variance t test for the difference in two means Performed separate-variance t test for difference in two means Formed confidence intervals for the difference between two means Compared two related samples (paired samples) Performed paired t test for the mean difference Formed confidence intervals for the mean difference

Chap 10-48 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-48 Chapter Summary Compared two population proportions Formed confidence intervals for the difference between two population proportions Performed Z-test for two population proportions Performed F test for the ratio of two population variances (continued)