商业统计学英文课件：ch10_ch11Two-Sample Tests and

1、Chap 10-1Chapter 10Two-Sample Tests and One-Way ANOVAStatistics For Managers Using Microsoft Excel6th EditionChap 10-2Learning ObjectivesIn this chapter, you learn hypothesis testing procedures to test: The means of two independent populations The means of two related populations The proportions of

2、two independent populations The variances of two independent populations The means of more than two populationsChap 10-3Chapter OverviewOne-Way Analysis of Variance (ANOVA)F-testTukey-Kramer testTwo-Sample TestsPopulation Means, Independent SamplesMeans, Related SamplesPopulationProportionsPopulatio

3、nVariancesChap 10-4Two-Sample TestsTwo-Sample TestsPopulation Means, Independent SamplesMeans, Related SamplesPopulation VariancesMean 1 vs. independent Mean 2Same population before vs. after treatment Variance 1 vs.Variance 2Examples:Population ProportionsProportion 1 vs. Proportion 2 Chap 10-5Diff

4、erence Between Two MeansPopulation means, independent samples1 and 2 knownGoal: Test hypothesis or form a confidence interval for the difference between two population means, 1 2 The point estimate for the difference isX1 X2*1 and 2 unknown, assumed equal 1 and 2 unknown, not assumed equal Chap 10-6

5、Independent SamplesPopulation means, independent samples Different data sources Unrelated Independent Sample selected from one population has no effect on the sample selected from the other population Use the difference between 2 sample means Use Z test, a pooled-variance t test, or a separate-varia

6、nce t test*1 and 2 known1 and 2 unknown, assumed equal 1 and 2 unknown, not assumed equal Chap 10-7Difference Between Two MeansPopulation means, independent samples1 and 2 known*Use a Z test statisticUse Sp to estimate unknown , use a t test statistic and pooled standard deviation1 and 2 unknown, as

7、sumed equal 1 and 2 unknown, not assumed equal Use S1 and S2 to estimate unknown 1 and 2, use a separate-variance t testChap 10-8Population means, independent samples1 and 2 known1 and 2 KnownAssumptions: Samples are randomly and independently drawn Population distributions are normal or both sample

8、 sizes are 30 Population standard deviations are known*1 and 2 unknown, assumed equal 1 and 2 unknown, not assumed equal Chap 10-9Population means, independent samples1 and 2 knownand the standard error of X1 X2 isWhen 1 and 2 are known and both populations are normal or both sample sizes are at lea

9、st 30, the test statistic is a Z-value222121XXnn21(continued)1 and 2 Known*1 and 2 unknown, assumed equal 1 and 2 unknown, not assumed equal Chap 10-10Population means, independent samples1 and 2 known2221212121nnXXZThe test statistic for 1 2 is:1 and 2 Known*(continued)1 and 2 unknown, assumed equa

10、l 1 and 2 unknown, not assumed equal Chap 10-11Hypothesis Tests forTwo Population MeansLower-tail test:H0: 1 2H1: 1 2i.e.,H0: 1 2 0H1: 1 2 2i.e.,H0: 1 2 0H1: 1 2 0Two-tail test:H0: 1 = 2H1: 1 2i.e.,H0: 1 2 = 0H1: 1 2 0Two Population Means, Independent SamplesChap 10-12Two Population Means, Independe

11、nt SamplesLower-tail test:H0: 1 2 0H1: 1 2 0Two-tail test:H0: 1 2 = 0H1: 1 2 0a aa a/2a a/2a a-za-za/2zaza/2Reject H0 if Z ZaReject H0 if Z Za/2 Hypothesis tests for 1 2 Chap 10-13Population means, independent samples1 and 2 known22212121nnZXXThe confidence interval for 1 2 is:Confidence Interval, 1

12、 and 2 Known*1 and 2 unknown, assumed equal 1 and 2 unknown, not assumed equal Chap 10-14Population means, independent samples1 and 2 known1 and 2 Unknown, Assumed EqualAssumptions: Samples are randomly and independently drawn Populations are normally distributed or both sample sizes are at least 30

13、 Population variances are unknown but assumed equal*1 and 2 unknown, assumed equal 1 and 2 unknown, not assumed equal Chap 10-15Population means, independent samples1 and 2 known(continued)*Forming interval estimates: The population variances are assumed equal, so use the two sample variances and po

14、ol them to estimate the common 2 the test statistic is a t value with (n1 + n2 2) degrees of freedom1 and 2 unknown, assumed equal 1 and 2 unknown, not assumed equal 1 and 2 Unknown, Assumed EqualChap 10-16Population means, independent samples1 and 2 knownThe pooled variance is(continued)1)n(nS1nS1n

15、S212222112p() 1*1 and 2 unknown, assumed equal 1 and 2 unknown, not assumed equal 1 and 2 Unknown, Assumed EqualChap 10-17Population means, independent samples1 and 2 knownWhere t has (n1 + n2 2) d.f.,and212p2121n1n1SXXtThe test statistic for 1 2 is:*1)n() 1(nS1nS1nS212222112p(continued)1 and 2 unkn

16、own, assumed equal 1 and 2 unknown, not assumed equal 1 and 2 Unknown, Assumed EqualChap 10-18Population means, independent samples1 and 2 known212p2-nn21n1n1StXX21The confidence interval for 1 2 is:Where1)n() 1(nS1nS1nS212222112p*Confidence Interval, 1 and 2 Unknown1 and 2 unknown, assumed equal 1

17、and 2 unknown, not assumed equal Chap 10-19Pooled-Variance t Test: ExampleYou 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 NASDAQNumber 21 25Sample mean 3.27 2.53Sample std d

18、ev 1.30 1.16Assuming both populations are approximately normal with equal variances, isthere a difference in average yield (a = 0.05)?Chap 10-20Calculating the Test Statistic1.50211)25(1)-(211.161251.301211)n() 1(nS1nS1nS22212222112p2.0402512115021.102.533.27n1n1SXXt212p2121The test statistic is:Cha

19、p 10-21SolutionH0: 1 - 2 = 0 i.e. (1 = 2)H1: 1 - 2 0 i.e. (1 2)a a = 0.05df = 21 + 25 - 2 = 44Critical Values: t = 2.0154Test Statistic:Decision:Conclusion:Reject H0 at a a = 0.05There is evidence of a difference in means.t0 2.0154-2.0154.025Reject H0Reject H0.0252.0402.0402512115021.12.533.27tChap

20、10-22Population means, independent samples1 and 2 known1 and 2 Unknown, Not Assumed EqualAssumptions: Samples are randomly and independently drawn Populations are normally distributed or both sample sizes are at least 30 Population variances are unknown but cannot be assumed to be equal*1 and 2 unkn

21、own, assumed equal 1 and 2 unknown, not assumed equal Chap 10-23Population means, independent samples1 and 2 known(continued)*Forming the test statistic: The population variances are not assumed equal, so include the two sample variances in the computation of the t-test statistic the test statistic

22、is a t value (statistical software is generally used to do the necessary computations)1 and 2 unknown, assumed equal 1 and 2 unknown, not assumed equal 1 and 2 Unknown, Not Assumed EqualChap 10-24Population means, independent samples1 and 2 known2221212121nSnSXXtThe test statistic for 1 2 is:*(conti

23、nued)1 and 2 unknown, assumed equal 1 and 2 unknown, not assumed equal 1 and 2 Unknown, Not Assumed EqualChap 10-25Related Populations Tests Means of 2 Related PopulationsPaired or matched samples Repeated measures (before/after) Use difference between paired values: Eliminates Variation Among Subje

24、cts Assumptions: Both Populations Are Normally Distributed Or, if not Normal, use large samplesRelated samplesDi = X1i - X2iChap 10-26Mean Difference, D KnownThe ith paired difference is Di , whereRelated samplesDi = X1i - X2i The point estimate for the population mean paired difference is D :nDDn1i

25、i Suppose the population standard deviation of the difference scores, D, is knownn is the number of pairs in the paired sampleChap 10-27The test statistic for the mean difference is a Z value:Paired samplesnDZDDMean Difference, D Known(continued)WhereD = hypothesized mean differenceD = population st

26、andard dev. of differencesn = the sample size (number of pairs)Chap 10-28Confidence Interval, D KnownThe confidence interval for D isPaired samplesnZDDWhere n = the sample size(number of pairs in the paired sample)Chap 10-29If D is unknown, we can estimate the unknown population standard deviation w

27、ith a sample standard deviation:Related samples1n)D(DSn1i2iD The sample standard deviation isMean Difference, D UnknownChap 10-30 Use a paired t test, the test statistic for D is now a t statistic, with n-1 d.f.:Paired samples1n)D(DSn1i2iDnSDtDDWhere t has n - 1 d.f. and SD is:Mean Difference, D Unk

28、nown(continued)Chap 10-31The confidence interval for D isPaired samples1n)D(DSn1i2iDnStDD1nwhereConfidence Interval, D UnknownChap 10-32Lower-tail test:H0: D 0H1: D 0Two-tail test:H0: D = 0H1: D 0Paired SamplesHypothesis Testing for Mean Difference, D Unknowna aa a/2a a/2a a-ta-ta/2tata/2Reject H0 i

29、f t taReject H0 if t ta/2 Where t has n - 1 d.f.Chap 10-33 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 t Test Example Number of Complaints: (2) - (1)Salesperson Befor

30、e (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 -21D =Din5.671n)D(DS2iD = -4.2Chap 10-34 Has the training made a difference in the number of complaints (at the 0.01 level)? - 4.2D =1.6655.67/04.2n/SDtDDH0: D = 0H1: D 0Test Statistic:Critical Value = 4.6

31、04 d.f. = n - 1 = 4Rejecta a/2 - 4.604 4.604Decision: Do not reject H0(t stat is not in the reject region)Conclusion: There is not a significant change in the number of complaints.Paired t Test: Solution Rejecta a/2 - 1.66a a = .01Chap 10-35Two Population ProportionsGoal: test a hypothesis or form a

32、 confidence interval for the difference between two population proportions, 1 2 The point estimate for the difference isPopulation proportionsAssumptions: n1 1 5 , n1(1- 1) 5n2 2 5 , n2(1- 2) 5 21pp Chap 10-36Two Population ProportionsPopulation proportions2121nnXXpThe pooled estimate for the overal

33、l proportion is:where X1 and X2 are the numbers from samples 1 and 2 with the characteristic of interestSince we begin by assuming the null hypothesis is true, we assume 1 = 2 and pool the two sample estimatesChap 10-37Two Population ProportionsPopulation proportions 212121n1n1)p(1pppZThe test stati

34、stic for p1 p2 is a Z statistic:(continued)2221112121nXp , nXp , nnXXpwhereChap 10-38Confidence Interval forTwo Population ProportionsPopulation proportions22211121n)p(1pn)p(1pZppThe confidence interval for 1 2 is:Chap 10-39Hypothesis Tests forTwo Population ProportionsPopulation proportionsLower-ta

35、il test:H0: 1 2H1: 1 2i.e.,H0: 1 2 0H1: 1 2 2i.e.,H0: 1 2 0H1: 1 2 0Two-tail test:H0: 1 = 2H1: 1 2i.e.,H0: 1 2 = 0H1: 1 2 0Chap 10-40Hypothesis Tests forTwo Population ProportionsPopulation proportionsLower-tail test:H0: 1 2 0H1: 1 2 0Two-tail test:H0: 1 2 = 0H1: 1 2 0a aa a/2a a/2a a-za-za/2zaza/2R

36、eject H0 if Z ZaReject H0 if Z Za/2 (continued)Chap 10-41Example: 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 Y

37、es Test at the .05 level of significanceChap 10-42 The hypothesis test is:H0: 1 2 = 0 (the two proportions are equal)H1: 1 2 0 (there is a significant difference between proportions) The sample proportions are: Men: p1 = 36/72 = .50 Women: p2 = 31/50 = .62.5491226750723136nnXXp2121 The pooled estima

38、te for the overall proportion is:Example: Two population Proportions(continued)Chap 10-43The test statistic for 1 2 is:Example: Two population Proportions(continued).025-1.961.96.025-1.31Decision: Do not reject H0Conclusion: There is not significant evidence of a difference in proportions who will v

39、ote yes between men and women. 1.31501721.549)(1.5490.62.50n1n1p(1pppz212121)Reject H0Reject H0Critical Values = 1.96For a a = .05Chap 10-44Hypothesis Tests for VariancesTests for TwoPopulation VariancesF test statisticH0: 12 = 22H1: 12 22Two-tail testLower-tail testUpper-tail testH0: 12 22H1: 12 22

40、*Chap 10-45Hypothesis Tests for VariancesTests for TwoPopulation VariancesF test statistic2221SSF The F test statistic is: = Variance of Sample 1 n1 - 1 = numerator degrees of freedom n2 - 1 = denominator degrees of freedom = Variance of Sample 221S22S*(continued)Chap 10-46 The F critical value is f

41、ound from the F table There are two appropriate degrees of freedom: numerator and denominator In the F table, numerator degrees of freedom determine the column denominator degrees of freedom determine the rowThe F Distributionwhere df1 = n1 1 ; df2 = n2 12221SSF Chap 10-47F 0 Finding the Rejection R

42、egionL2221U2221FSSFFSSFn rejection region for a two-tail test is:aFL Reject H0Do not reject H0F 0 aFU Reject H0Do not reject H0F 0 a/2Reject H0Do not reject H0FU H0: 12 = 22H1: 12 22H0: 12 22H1: 12 22FL a/2Reject H0Reject H0 if F FUChap 10-48Finding the Rejection RegionF 0 a/2Reject H0Do not reject

43、H0FU H0: 12 = 22H1: 12 22FL a/2Reject H0(continued)2. Find FL using the formula:Where FU* is from the F table with n2 1 numerator and n1 1 denominator degrees of freedom (i.e., switch the d.f. from FU)*ULF1F 1. Find FU from the F table for n1 1 numerator and n2 1 denominator degrees of freedomTo fin

44、d the critical F values:Chap 10-49F Test: An ExampleYou 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 NASDAQNumber 2125Mean3.272.53Std dev1.301.16Is there a difference in the vari

45、ances between the NYSE & NASDAQ at the a a = 0.05 level?Chap 10-50F Test: Example Solution Form the hypothesis test:H0: 21 22 = 0 (there is no difference between variances)H1: 21 22 0 (there is a difference between variances) Numerator: n1 1 = 21 1 = 20 d.f. Denominator: n2 1 = 25 1 = 24 d.f. FU

46、 = F.025, 20, 24 = 2.33 Find the F critical values for a a = 0.05: Numerator: n2 1 = 25 1 = 24 d.f. Denominator: n1 1 = 21 1 = 20 d.f. FL = 1/F.025, 24, 20 = 1/2.41 = 0.415FU:FL:Chap 10-51 The test statistic is:0 256.116.130.1SSF222221a/2 = .025FU=2.33Reject H0Do not reject H0H0: 12 = 22H1: 12 22F T

47、est: Example Solution F = 1.256 is not in the rejection region, so we do not reject H0(continued) Conclusion: There is not sufficient evidence of a difference in variances at a = .05FL=0.43a/2 = .025Reject H0F Chap 10-52Two-Sample Tests in EXCELFor independent samples:Independent sample Z test with

48、variances known:Tools | data analysis | z-test: two sample for meansPooled variance t test:Tools | data analysis | t-test: two sample assuming equal variancesSeparate-variance t test:Tools | data analysis | t-test: two sample assuming unequal variancesFor paired samples (t test):Tools | data analysi

49、s | t-test: paired two sample for meansFor variances:F test for two variances:Tools | data analysis | F-test: two sample for variancesChap 10-53One-Way Analysis of VarianceOne-Way Analysis of Variance (ANOVA)F-testTukey-Kramer testChap 10-54General ANOVA Setting Investigator controls one or more ind

50、ependent variables Called factors (or treatment variables) Each factor contains two or more levels (or groups or categories/classifications) Observe effects on the dependent variable Response to levels of independent variable Experimental design: the plan used to collect the dataChap 10-55One-Way An

51、alysis of Variance Evaluate the difference among the means of three or more groupsExamples: Accident rates for 1st, 2nd, and 3rd shift Expected mileage for five brands of tires Assumptions Populations are normally distributed Populations have equal variances Samples are randomly and independently dr

52、awnChap 10-56Hypotheses of One-Way ANOVA All population means are equal i.e., no treatment effect (no variation in means among groups) At least one population mean is different i.e., there is a treatment effect Does not mean that all population means are different (some pairs may be the same) c3210:

53、H1:Not all of the population means are the sameHChap 10-57One-Way ANOVA All Means are the same:The Null Hypothesis is True (No Treatment Effect)c3210:Hsame the are all Not:Hj1321Chap 10-58One-Way ANOVA At least one mean is different:The Null Hypothesis is NOT true (Treatment Effect is present)c3210:

54、Hsame the are all Not:Hj1321321or(continued)Chap 10-59Partitioning the Variation Total variation can be split into two parts:SST = Total Sum of Squares (Total variation)SSA = Sum of Squares Among Groups (Among-group variation)SSW = Sum of Squares Within Groups (Within-group variation)SST = SSA + SSW

55、Chap 10-60Partitioning the VariationTotal Variation = the aggregate dispersion of the individual data values across the various factor levels (SST)Within-Group Variation = dispersion that exists among the data values within a particular factor level (SSW)Among-Group Variation = dispersion between th

56、e factor sample means (SSA)SST = SSA + SSW(continued)Chap 10-61Partition of Total VariationVariation Due to Factor (SSA)Variation Due to Random Sampling (SSW)Total Variation (SST)Commonly referred to as: Sum of Squares Within Sum of Squares Error Sum of Squares Unexplained Within-Group VariationComm

57、only referred to as: Sum of Squares Between Sum of Squares Among Sum of Squares Explained Among Groups Variation=+d.f. = n 1d.f. = c 1d.f. = n cChap 10-62Total Sum of Squaresc1jn1i2ijj)XX(SSTWhere:SST = Total sum of squaresc = number of groups (levels or treatments)nj = number of observations in gro

58、up jXij = ith observation from group j X = grand mean (mean of all data values)SST = SSA + SSWChap 10-63Total VariationGroup 1Group 2Group 3Response, XX2cn212211)XX(.)XX()XX(SSTc(continued)Chap 10-64Among-Group VariationWhere:SSA = Sum of squares among groupsc = number of groupsnj = sample size from

59、 group j Xj = sample mean from group j X = grand mean (mean of all data values)2jc1jj)XX(nSSASST = SSA + SSWChap 10-65Among-Group VariationVariation Due to Differences Among Groupsij2jc1jj)XX(nSSA1cSSAMSAMean Square Among = SSA/degrees of freedom(continued)Chap 10-66Among-Group VariationGroup 1Group

60、 2Group 3Response, XX1X2X3X2222211)xx(n.)xx(n)xx(nSSAcc(continued)Chap 10-67Within-Group VariationWhere:SSW = Sum of squares within groupsc = number of groupsnj = sample size from group j Xj = sample mean from group jXij = ith observation in group j2jijn1ic1j)XX(SSWjSST = SSA + SSWChap 10-68Within-Group Variat