1 252mvar 3 31 04 6 Tests for Equality of Variances a Introduction We have now covered tests for Equality of Multiple Means ANOVA Equality of Multiple Medians Kruskal Wallis and Friedman Equality of Multiple Proportions Chi squared So for completeness we need Tests for Equality of Multiple Variances b The Bartlett Test This test seems to require that the underlying distribution be Normal It should not be used to compare two columns since a simple F test described earlier is more appropriate Recall that in the test for comparing 2 means with equal variances we used a pooled variance Assume that we have 2 be s p n1 1 s12 or more rows is 2 d 1 1 3 c 1 c columns representing c s 2 p n1 1 s12 n2 1 s22 n1 n2 2 independent samples Then the pooled variance would n 2 1 n 3 1 n c 1 s c2 The test statistic used when there are 6 n1 n 2 n 3 n c c s 22 c 1 n s 32 n 2 30259 d 1 j 1 j n 1 log s 2p j 1 log s 2j where n j c 1 For smaller examples less than 6 rows a special table is required and the instructions that I have found are very confusing Use the computer Bartlett Test Example Kanji heavily edited Test the following data for equal variances s12 5 47 s 22 4 64 s 32 11 47 s 42 11 29 n1 31 n 2 15 n 3 20 n 4 42 n 1 s12 n 2 1 s 22 n3 1 s 32 n c 1 s c2 s p2 1 n1 n 2 n 3 n c c 30 5 47 14 4 64 19 11 47 41 11 29 164 1 64 96 217 93 462 89 104 31 15 20 42 4 909 88 n j 1 n j c The test 8 7488 Note that the denominator can be written as 104 c 1 2 30259 n j 1 log s 2p n j 1 log s 2j where statistic used is 2 d d 1 1 3 c 1 n 1 j 1 1 1 1 1 1 1 1 nj c 3 5 30 14 19 41 104 1 2 1 0 033333 0 071429 0 052632 0 024902 0096153 15 1 1 1726807 1 0115 15 c 1 2 30259 2 n j 1 log s 2p n j 1 log s 2j d 2 30259 104log 8 7488 30log 5 47 14log 4 64 19log 11 47 41 log 11 29 1 0115 2 30259 104 0 941948 30 0 737987 14 0 666518 19 1 059563 41 1 052694 1 0115 2 30259 97 962592 22 139610 9 331252 20 13164 43 160454 1 0115 2 30259 3 199636 7 2837 This has c 1 4 1 3 degrees of freedom and the chi 1 0115 1 squared table says that 3 2 05 7 8147 Since our computed chi squared is less than the table chi square do not reject the null hypothesis c The Levene Test This test is quite simple It can be used for non Normal data and can be used to compare two columns as well as more than two columns i Find the median of each column This is the middle number or the average of the two middle numbers ii Subtract the median of each column from the column from which it comes and take the absolute value of the result iii Do a 1 way ANOVA on the result If the results would lead you to reject the null hypothesis because the computed F is above the table F or the p value is below your significance level reject the null hypothesis of equal variances Levene Test Example Test the following data for equality of variances Method 1 Method 2 Method 3 Method 4 1 31 1 08 0 85 1 31 1 27 1 10 1 02 1 27 1 28 1 05 0 78 1 28 1 22 1 02 0 87 1 22 1 19 0 99 0 80 1 19 1 30 0 95 0 96 1 30 If we write out the numbers in Method 1 in order we get 1 19 1 22 1 27 1 28 1 30 1 31 The median for Method 2 is the halfway point between 1 27 and 1 28 or 1 275 The other medians are 1 035 0 86 and 1 275 If we subtract the medians we get the following Method 1 Method 2 Method 3 Method 4 0 035 0 005 0 005 0 055 0 085 0 025 0 045 0 065 0 015 0 015 0 045 0 085 0 01 0 16 0 08 0 01 0 06 0 10 0 035 0 005 0 005 0 055 0 085 0 025 3 If we take absolute values we get the following Method 1 Method 2 Method 3 0 035 0 005 0 005 0 055 0 085 0 025 0 045 0 065 0 015 0 015 0 045 0 085 0 01 0 16 0 08 0 01 0 06 0 10 Method 4 0 035 0 005 0 005 0 055 0 085 0 025 If we subject these results to a one way ANOVA we get the table below and conclude that we cannot reject the hypothesis of equal variances F Source SS DF MS F 05 H0 Between 0 00491 3 0 00164 1 10ns F 3 20 3 10 Variances equal Within 0 02980 Total 0 03471 For computer examples see 252mvarex 20 23 0 00149
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