Introduction to Statistics in Psychology PSY 201 Professor Greg Francis Lecture 34 ANalysis Of VAriance Measure twice cut once MULTIPLE t TESTS ANOVA VARIABLES if we have K 5 population means we might want to compare each mean to all the others independent variables variable that forms groupings requires K K 1 c 10 2 different t tests one way ANOVA one independent variable levels number of groups number of populations suppose each test is with 0 05 e g Method of teaching is an independent variable The probability of making a Type I error over the whole set is much larger you may teach in 17 different ways levels and have 17 different sample groups with sample means X 1 X 2 X 16 X 17 so that for your hypothesis test you would want to test whether all the population means of the different levels are the same 2 3 ANOVA VARIABLES HYPOTHESES INTUITION we need additional subscripts to keep track of variables for one way ANOVA the hypotheses are the basic approach of ANOVA is to make two calculations of variance xik H0 1 2 K is the score for the ith subject in the kth level group Ha i k for some i k nk the null hypothesis is that all population means are the same is the number of scores in the kth level i xik the alternative hypothesis is that at least one mean is different from another nk k i 2 We can also calculate the variance among all the group means relative to a grand mean between variance s2B these estimates will be the same if H0 is true is the sum of scores in the kth level 1 We can calculate variance of each group separately and combine them to estimate the variance of all scores within variance s2W these estimates will be different if H0 is not true xik is the sum of all scores 4 5 6 INTUITION SCORES SUM OF SQUARES we compare the estimates using the F ratio s2 F 2B sW what contributes to a particular score assume a linear model Xik k eik we want to estimate e2 variance of population if H0 is true need sum of squares k i Xik X 2 Xik is the ith score in the kth group is the grand mean for the population across all groups k k is the effect of belonging to group k eik is random error associated with the score if F 1 do not reject H0 if F 1 reject H0 how big depends on the sample sizes significance consider one score Xik X Xik X k X k X so Xik X 2 Xik X k X k X 2 or Xik X 2 Xik X k 2 2 X k X Xik X k X k X 2 eik changes because of random sampling normally distributed mean of zero e2 nk i 7 8 9 SUM OF SQUARES SUM OF SQUARES DEVIATIONS if we sum across all subjects in category k now we sum across the k groups to get the total sum of squares Xik X 2 nk i Xik X k 2 2 X k X nk i nk Xik X k X k X 2 i k i Xik X 2 n k k i Xik X k 2 nk X k X 2 which becomes since deviations from a mean equal zero this reduces to n k i Xik X 2 n k i n k Xik X k 2 X k X 2 i k i Xik X 2 n k k i Xik X k 2 nk X k X 2 k or in addition i X k X 2 nk X k X 2 so we get n k i 2 Xik X n k i Xik X k 2 nk X k X 2 10 SSW nk k i Xik X k 2 what causes this to be greater than zero since Xik k eik SST SSW SSB where n k SST is the total sum of squares SSW is the within sum of squares Deviation of scores from the group mean SSB is the between sum of squares Deviation of group means from the grand mean 11 k is fixed as i varies thus deviations from X k must be due to the eik term random error 12 DEVIATIONS ESTIMATE OF e2 since within each group deviations from the mean are due to the error terms eik so Xik k eik the mean of group k is X e X k i ik k i ik nk nk so Xik X k X X k 2 s2k i ik e2 nk 1 e k eik k i ik nk ieik eik nk so only the random variation across subjects is considered here to get a better estimate pool across all groups SSW M SW e2 N K here M SW stands for mean squares within N K is the degrees of freedom DEVIATIONS SSB k nk X k X 2 what causes this to be greater than zero since Xik k eik the mean of group k is X e X k i ik k i ik nk nk as k changes X stays the same so any deviations from X are due to changes in k changes between groups or to changes in nieik k random error 13 14 15 ESTIMATE OF e2 F statistic F critical if H0 is true then all k 0 and any deviations must be due only to the random error terms ieik nk so we do not know what e2 is but we have two estimates as always for inferential statistics we need to know if F is significantly greater than 1 0 so we can again estimate e2 as SSB n X X 2 M SB k k k e2 K 1 K 1 here K 1 is degrees of freedom on the other hand if H0 is not true then M SB includes deviations due to positive k so M SB e2 16 1 M SW always estimates e2 2 M SB estimates e2 if H0 is true Larger than e2 if H0 is false compare the estimates by computing M SB F M SW if H0 is true should get F 1 if H0 is not true should get F 1 17 depends on degrees of freedom look up Fcv from the F distribution table using K 1 d f in the numerator and N K d f in the denominator also written as F K 1 N K 18 TESTING CONCLUSIONS NEXT TIME 4 STEPS testing multiple means an example 1 State the hypothesis H0 1 2 K Ha i j for some i j two estimates of population variance assumptions underlying ANOVA 2 Set the criterion F K 1 N K 3 Compute the test statistic F M SB M SW 4 Interpret results 19 one estimate always estimates variance other estimate is true only if H0 is true measure of association repeated measures ANOVA The last topic lets us test H0 20 21