We know from STAT.1030 that the relevant test statistic for equality of proportions is:

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1 2. Chi 2 -tests for equality of proportions Introduction: Two Samples Consider comparing the sample proportions p 1 and p 2 in independent random samples of size n 1 and n 2 out of two populations which have a certain characteristic with respective probabilities π 1 and π 2. We know from STAT.1030 that the relevant test statistic for equality of proportions is: z = p 1 p 2 ( ) N(0, 1) under H 0 : π 1 = π 2, 1 p(1 p) n n 2 where p = n 1p 1 + n 2 p 2 n 1 + n 2 denotes the combined proportion of both samples. Recall that this z-statistic is based upon the normal approximation of the binomial distribution and requires therefore sufficiently large sample sizes (n 30). SAS EG does not offer this z-test. But the same test may be equivalently formulated as a χ 2 independence test as follows. 71

2 Consider the contingency table below: success failure sum pop.1 n 1 p 1 n 1 (1 p 1 ) n 1 = f 1 pop.2 n 2 p 2 n 2 (1 p 2 ) n 2 = f 2 sum: np = f 1 n(1 p) = f 2 n Since p 1 and p 2 are unbiased estimators of their population counterparts π 1 and π 2, H 0 : π 1 = π 2 is equivalent to independence of the success and population variables. Applying the formula for two-way tables, χ 2 = n(f 11f 22 f 12 f 21 ) 2 f 1 f 2 f 1 f 2, yields after some manipulation: χ 2 = (p 1 p 2 ) 2 p(1 p) ( 1 n1 + 1 n 2 ) χ 2 (1) under H 0. It should not surprise us that this is just the square of our familiar z-statistic, since we know from STAT.1030 that the square of a N(0, 1)-distributed random variable is χ 2 (1)- distributed. 72

3 Homogeneity tests in SAS EG Since SAS does not offer the familiar z-test, we resort to the equivalent χ 2 -independence test instead. The advantage of the χ 2 -test is that we may use it to compare arbitrarily many sample proportions and/or variables with arbitrarily many outcomes (rather than just success/failure). For example, our introductory example (satisfaction with the companies management) of section may be regarded as a comparison of the conditional distribution of satisfaction on a 3-level scale for employees both with and without vocational education. The disadvantage of the χ 2 -test as compared to the z-test is that the alternative is always two-sided (H 1 : π 1 π 2 ). To perform onesided tests such as H 1 : π 1 < π 2, check first that p 1 p 2 has the sign as suggested by H 1, and reject H 0 in favour of H 1 if α > p 2, where p denotes the p-value of the χ 2 -independence test. 73

4 Example. Consider the following table about consumption of alcohol for men and women: Consumption: yes no female (pop.1) male (pop.2) As is evident from the output on the next slide, we may not reject H 0 : π 1 = π 2 against the two-sided alternative H 1 : π 1 π 2 (p = ), but we may reject H 0 against the one-sided alternative H 1 : π 2 > π 1 (p = ). Using Fishers exact test, the p-value of H 0 : π 1 = π 2 against the two-sided alternative H 1 : π 1 π 2 is p = , leading to the same conclusion as above. But against the onesided alternative H 1 : π 2 > π 1 it is p = , which means that we cannot even reject H 0 in a one-sided test at α = 5%. If we have access to software, we prefer using the output from Fishers exact test. It is however too hard to calculate by hand, and in larger tables even software might fail to calculate it. 74

5 Table Analysis Results 1 Table of suku by käyttä suku(suku) käyttä(käyttä) Frequency Row Pct 1 0 Total nainen mies Total Statistics for Table of suku by käyttä Statistic DF Value Prob Chi-Square Likelihood Ratio Chi-Square Continuity Adj. Chi-Square Mantel-Haenszel Chi-Square Phi Coefficient Contingency Coefficient Cramer's V Fisher's Exact Test Cell (1,1) Frequency (F) 68 Left-sided Pr <= F Right-sided Pr >= F Table Probability (P) Two-sided Pr <= P Sample Size = 214

6 General test for homogeneity of proportions The χ 2 -test for equality of proportions may be generalized to more than two independent samples as follows. Assemble the frequencies of success np i and failure n i (1 p i ) for the respective populations i in a (2 c) contingeny table, where n i and p i denote the size and proportion of success in population i, and c denotes the number of populations. The homogeneity of the populations (with respect to the success variable) H 0 : π 1 = π 2 = = π c versus H 1 : Not all π i, i = 1,..., c are equal is then assessed by a χ 2 independence test of the population and success variables. 76

7 Example. An insurance company wants to test whether the proportion of people who submit claims for automobile accidents is about the same for the three age groups 25 and under, over 25 and under 50, and 50 and over: H 0 : The age goups are homogeneous wrt. claims, H 1 : The age goups are not homogeneous wrt. claims. The (2 3) contingency table for this example is: age< age< age Total Claim No claim Total The expected frequencies are: age< age< age Total Claim No claim Total The χ 2 statistic is: χ 2 (40 45)2 (35 45)2 (60 45)2 = (60 55) 2 (66 55)2 (40 55)2 + + = and the degrees of freedom are (2-1)(3-1)=2. Looking up in a table or calling CHIDIST(14.14;2) in Excel establishes that the p-value in this case is less than 0.1%, so we reject homogeneity with respect to claims in favour of inhomogeneity at all conventional significance levels. 77

8 The median test The hypotheses for the median test are H 0 : The c populations have the same median, H 1 : Not all c populations have the same median. This is in fact a special case of the homogeneity test for independent samples which we just discussed, with the indicator: value median as the success variable. Median means here grand median, which is the median of all observations regardless of the population. Example: An economist wants to test the null hypothesis that the median family income in three rural areas are approximately equal. Random samples of family incomes (in \$1000/year) in three regions (A/B/C) are given below: A B C

9 Example: (continued.) Arranging all observations in order reveals that the grand median is The observed and expected counts for each region are displayed below: Region A Region B Region C Total median (expected) (5) (5) (5) >median (expected) (5) (5) (5) Total Note that here: e ij = f i f j n = n/2 n i n = n i 2. The χ 2 statistic is: χ 2 (4 5)2 (5 5)2 (6 5)2 = (6 5) 2 (5 5)2 (4 5)2 + + = and the degrees of freedom are (2-1)(3-1)=2. Comparing this value with critical points χ 2 α(2) of the χ 2 -distribution with 2 degrees of freedom, we conclude that there is no evidence to reject the null hypothesis. The p-value of the test is CHIDIST(0.8;2)=

10 The median test in SAS EG You find the median test in SAS Enterprise Guide under the menu point Analyze/ANOVA/ Nonparametric One-Way ANOVA. In the Task Roles window chose the variable of interest as Dependent variable and the classification variable as Independent Variable. Then check the Median box in the Test Scores pane of the Analysis Window and click Run. SAS makes a finite-sample correction by multiplying χ 2 by (N 1)/N, where N denotes the number of observations. Nonparametric One-Way ANOVA Region Median Scores (Number of Points Above Median) for Variable Income Classified by Variable Region N Sum of Scores Expected Under H0 Std Dev Under H0 Mean Score Average scores were used for ties. Median One-Way Analysis Chi-Square DF 2 Asymptotic Pr > Chi-Square Exact Pr >= Chi-Square

11 McNemar test on paired proportions Occasionally there is a need to compare two proportions when the samples drawn are not independent. This is always the case when the same statistical unit generates a pair of observations, for example in studies with before and after design. The test requires the same n statistical units to be included in the before- and after measurements (matched pairs). The observed counts may be assembled in a two-way table of the form: Sample II: 1 0 sum Sample I: 1 a = np 11 b = np 10 a + b 0 c = np 01 d = np 00 c + d sum a + c b + d n The null hypothesis that the probability of success is the same in both samples reduces then to testing b = c = (b + c)/2, that is p 10 = p 01. The approximate sample statistic including a continuity correction is χ 2 = ( b c 1)2 b + c χ 2 (1) under H 0. 81

12 Example: (Bland 2000) 1391 schoolchildren were questioned on the prevalence of symptoms of severe cold at the age of 12 and again at the age of 14 years: Severe colds Severe colds at age 14 Total at age 12 Yes No Yes No Total In order to assess whether the increase is significant, calculate the test statistic χ 2 = ( ) = , which is far beyond the critical value χ (1) = 10.83, so the increase in the prevalence of symptoms of severe cold is statistically significant. Note: SAS skips the continuity correction and calculates the test statistics as χ 2 = = However, you may ask SAS for exact p-values which is even better than the continuity correction and works also in small samples. To apply the test in SAS check Measures in the Table Statistics/Agreement window under Describe/Table Analysis. 82

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