商业统计学英文课件：ch12 Chi-Square Tests And Nonparametric Tests

1、Chap 11-1Chapter 12Chi-Square Tests And Nonparametric TestsStatistics For Managers Using Microsoft Excel6th EditionChap 11-2Learning ObjectivesIn this chapter, you learn:How and when to use the chi-square test for contingency tablesHow to use the Marascuilo procedure for determining pairwise differe

2、nces when evaluating more than two proportionsChap 11-3Contingency TablesContingency TablesnUseful in situations involving multiple population proportionsnUsed to classify sample observations according to two or more characteristicsnAlso called a cross-classification table.Chap 11-4Contingency Table

3、 ExampleLeft-Handed vs. Gender Dominant Hand: Left vs. Right Gender: Male vs. Female 2 categories for each variable, so called a 2 x 2 table Suppose we examine a sample of size 300Chap 11-5Contingency Table ExampleSample results organized in a contingency table:(continued)GenderHand PreferenceLeftRi

4、ghtFemale12108120Male2415618036264300120 Females, 12 were left handed180 Males, 24 were left handedsample size = n = 300:Chap 11-62 Test for the Difference Between Two ProportionsnIf H0 is true, then the proportion of left-handed females should be the same as the proportion of left-handed malesnThe

5、two proportions above should be the same as the proportion of left-handed people overallH0: 1 = 2 (Proportion of females who are left handed is equal to the proportion of males who are left handed) H1: 1 2 (The two proportions are not the same Hand preference is not independent of gender)Chap 11-7Th

6、e Chi-Square Test Statisticnwhere:fo = observed frequency in a particular cellfe = expected frequency in a particular cell if H0 is true 2 for the 2 x 2 case has 1 degree of freedom(Assumed: each cell in the contingency table has expected frequency of at least 5)cells alle2eo2f)ff (The Chi-square te

7、st statistic is:Chap 11-8Decision Rule 2 22UDecision Rule:If 2 2U, reject H0, otherwise, do not reject H0The 2 test statistic approximately follows a chi-square distribution with one degree of freedom0 Reject H0Do not reject H0Chap 11-9Computing the Average ProportionHere: 120 Females, 12 were left

8、handed180 Males, 24 were left handedi.e., the proportion of left handers overall is 0.12, that is, 12%nXnnXXp212112.0300361801202412pThe average proportion is: Chap 11-10Finding Expected FrequenciesnTo obtain the expected frequency for left handed females, multiply the average proportion left handed

9、 (p) by the total number of femalesnTo obtain the expected frequency for left handed males, multiply the average proportion left handed (p) by the total number of malesIf the two proportions are equal, then P(Left Handed | Female) = P(Left Handed | Male) = .12i.e., we would expect (.12)(120) = 14.4

10、females to be left handed(.12)(180) = 21.6 males to be left handedChap 11-11Observed vs. Expected FrequenciesGenderHand PreferenceLeftRightFemaleObserved = 12Expected = 14.4Observed = 108Expected = 105.6120MaleObserved = 24Expected = 21.6Observed = 156Expected = 158.418036264300Chap 11-12GenderHand

11、PreferenceLeftRightFemaleObserved = 12Expected = 14.4Observed = 108Expected = 105.6120MaleObserved = 24Expected = 21.6Observed = 156Expected = 158.4180362643000.7576158.4158.4)(15621.621.6)(24105.6105.6)(10814.414.4)(12f)f(f2222cells alle2eo2The Chi-Square Test StatisticThe test statistic is:Chap 11

12、-13Decision RuleDecision Rule:If 2 3.841, reject H0, otherwise, do not reject H03.841 d.f. 1 with , 0.7576 isstatistic test The2U2Here, 2 = 0.7576 2U, reject H0, otherwise, do not reject H0Where 2U is from the chi-square distribution with c 1 degrees of freedomChap 11-17The Marascuilo ProcedurenUsed

13、 when the null hypothesis of equal proportions is rejectednEnables you to make comparisons between all pairsnStart with the observed differences, pj pj, for all pairs (for j j) . . .n. . .then compare the absolute difference to a calculated critical rangeChap 11-18The Marascuilo ProcedurenCritical R

14、ange for the Marascuilo Procedure:n(Note: the critical range is different for each pairwise comparison)nA particular pair of proportions is significantly different if| pj pj| critical range for j and j(continued) j j jjjj2Un)p(1pn)p(1prange CriticalChap 11-19Marascuilo Procedure ExampleA University

15、is thinking of switching to a trimester academic calendar. A random sample of 100 administrators, 50 students, and 50 faculty members were surveyedOpinion Administrators Students FacultyFavor 63 20 37Oppose 37 30 13Totals 100 5050At the 0.01 level of significance, is there evidence that the groups d

16、iffer in attitude? If so, how?Business Statistics, A First Course (4e) 2006 Prentice-Hall, Inc.Chap 11-20Marascuilo Procedure ExampleExpected Cell Frequencies:Opinion Admin. Students FacultyFavor 60 30 30Oppose 40 20 20Totals 100 5050DataLevel of Significance0.01Number of Rows2Number of Columns3Degr

17、ees of Freedom2ResultsCritical Value9.210351Chi-Square Test Statistic12.79167p-Value0.001668Reject the null hypothesisH0: 1 = 2 = = cH1: Not all of the j are equal (j = 1, 2, , c)The test statistic is greater than the critical value, so H0 is rejected.Chap 11-21Marascuilo Procedure: SolutionExcel Ou

18、tput:compareAt the 0.01 level of significance, there is evidence of a difference in attitude between students and facultyChap 11-222 Test of IndependencenSimilar to the 2 test for equality of more than two proportions, but extends the concept to contingency tables with r rows and c columnsH0: The tw

19、o categorical variables are independent(i.e., there is no relationship between them)H1: The two categorical variables are dependent(i.e., there is a relationship between them)Chap 11-232 Test of Independencenwhere:fo = observed frequency in a particular cell of the r x c tablefe = expected frequency

20、 in a particular cell if H0 is true 2 for the r x c case has (r-1)(c-1) degrees of freedom(Assumed: each cell in the contingency table has expected frequency of at least 1)cells alle2eo2f)ff (The Chi-square test statistic is:(continued)Chap 11-24Expected Cell FrequenciesnExpected cell frequencies:nt

21、otal columntotalrow feWhere: row total = sum of all frequencies in the rowcolumn total = sum of all frequencies in the column n = overall sample sizeChap 11-25Decision RulenThe decision rule isIf 2 2U, reject H0, otherwise, do not reject H0Where 2U is from the chi-square distribution with (r 1)(c 1)

22、 degrees of freedomChap 11-26ExamplenThe meal plan selected by 200 students is shown below:ClassStandingNumber of meals per weekTotal20/week10/weeknoneFresh.24321470Soph.22261260Junior1014630Senior14161040Total 708842200Chap 11-27ExamplenThe hypothesis to be tested is:(continued)H0: Meal plan and cl

23、ass standing are independent(i.e., there is no relationship between them)H1: Meal plan and class standing are dependent(i.e., there is a relationship between them)Chap 11-28ClassStandingNumber of meals per weekTotal20/wk10/wknoneFresh.24321470Soph.22261260Junior1014630Senior14161040Total 708842200Cl

24、assStandingNumber of meals per weekTotal20/wk10/wknoneFresh.24.530.814.770Soph.21.026.412.660Junior10.513.26.330Senior14.017.68.440Total 708842200Observed:Expected cell frequencies if H0 is true:5.102007030ntotal columntotalrow feExample for one cell:Example: Expected Cell Frequencies(continued)Chap 11-29