CRIM 320 Lecture Notes - Lecture 8: Multiple Comparisons Problem, Descriptive Statistics, Chi-Squared Distribution
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21 Mar 2016
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Lec 8 analysis of variance i (anova: one-way anova. T-test can only compare 2 groups at the same ime. With 7 groups, 21 separate t-tests are required. Given an alpha of 0. 05, 1 in 20 chances of incorrectly rejecing the null hypothesis. Therefore, with 21 t-tests, one should be signiicant by chance alone. F = ms(between)/ ms(within: on f distribuion, homogeneity of within group variance and heterogeneity of between-group variance. The essence is the sum of squares: hand calculaion. Sum of squares (x x ) : sum of the squared deviaion from the mean, iniial step for measuring total variaion as well as variaion between and within groups, three types. When you add up all the (x x ) you get total sum of squares. Then you have to do for each groups that"s in there: ss(within) = (x x (group)) , ss(between) = ngroup (xgroup x total) . Ms(between) = ss(between) / df(between: df = k-1.