When the difference in performance is captured by the chi-square test, the test does not formally point out where the difference comes from, for example, from lowest, middle, or highest levels of performance. An indication of the sources for the differences can be found by examining the contributions from each cell into the chi-square test statistic given in Minitab's output. Examine the last row of each cell in the output to see that the largest values of the (Observed Expected )2 Expected proportions of males and females who scored at level 1 differ. correspond to mathematics level 1. We can do an additional test to determine if the Column 010 is labeled \"MTH_modied\". It has value 1 if a person scored at level 1 on the mathematics test, and value 0 otherwise. Repeat the chi-square analysis for this modied column by using Stat>Tables>Cross Tabulation and Chi-Square. Select \"Gender\" for rows, \"MTH_modied\" for columns. Click on the Display tab and check the appropriate boxes to display chi-square test for association and expected cell counts, and each cell's contribution to the chi-square statistic. Use Minitab's output to answer the questions: 14. The value of the test statistic (Pearson chi-square) is 15. Degrees of freedom is 1 6. P-value is When each categorical variable has 2 levels, and the resulting contingency table is 2 by 2, the comparison of the proportions of males and females who score at level 1 can be done using a 2-sample z-test for proportions (Section 12.3, text). In this case the hypotheses are H,J :pl p2 = 0 versus H, :pl p2 at 0. The Z test statistic squared equals the chi-square test statistics, and the p-values found using two approaches are the same because chi-square distribution with one degree of freedom (see question #15) is the same as Z (standard normal) squared. To see this, use the data in the last Minitab output to get: The sample proportion of females who scored at level 1: 1=63l415=.1518 17. The sample proportion of males who scored at level 1: 2= Please keep 4 decimals. 18. The pooled estimate of the common proportion of all students who scored at level 1 (see p. 487, text) f7: \"1131 +"2f'2 = X1+X2 = n1+n2 n1+n2 19. The 2 test statistic z = %= Please keep at least 2 decimals. m i+i '"1 "2