TESTING SUM OF SQUARES 4 STEPS calculating the sum of squares within between total can be messy can make it a bit easier Lecture 35 1 State the hypothesis H0 1 2 K Ha i j for some i j ANalysis Of VAriance 2 Set the criterion F K 1 N K Introduction to Statistics in Psychology PSY 201 Professor Greg Francis How do changes in one variable correspond to changes in another variable 3 Compute the test statistic F M SB M SW 4 Interpret results for each group k let Tk n k i Xik and let T k Tk then 2 T2 Tk SSB N k nk SSW n k k i 2 Xik k Tk2 nk and SST EXAMPLE A college professor wants to determine the best way to present an important lecture topic to his class He has the following three choices 1 he can lecture 2 he can lecture plus assign supplementary reading or 3 he can show a film and assign supplementary reading He decides to do an experiment to evaluate the three options He solicits 27 volunteers from his class and randomly assigns 9 to each of three conditions In condition 1 he lectures to the students In condition 2 he lectures plus assigns supplementary reading In condition 3 the students see a film on the topic plus n k k i T2 N 2 3 EXAMPLE 1 HYPOTHESES Lecture Lecture Reading Film Reading Condition 1 Condition 2 Condition 3 92 86 81 86 93 80 87 97 72 76 81 82 80 94 83 87 89 89 92 98 76 83 90 88 84 91 83 T1 767 T2 819 T3 734 nk k i for one way ANOVA the hypotheses are H 0 1 2 3 Ha i k for some i k Test with 0 05 2 Xik 200 428 T 2320 receive the same supplementary reading as the students in condition 2 The students are subsequently tested on the material The following scores percentage correct were obtained 4 2 Xik 5 6 2 CRITICAL VALUE between degrees of freedom numerator number of groups minus 1 df K 1 3 1 2 within degrees of freedom denominator number of subjects minus number of groups df N K 27 3 24 Total degrees of freedom number of subjects minus 1 3 STATISTIC from the data 2 T2 Tk SSB N k nk 767 2 819 2 734 2 2320 2 SSB 408 074 9 9 9 27 2 nk T k 2 SSW Xik k i k nk 2 819 2 734 2 767 SSW 200428 671 778 9 9 9 and nk T2 2 Xik SST N k i 2320 2 SST 200428 1079 852 27 df N 1 27 1 26 From F distribution table we find 3 STATISTIC M SB SSB 408 074 204 037 K 1 2 M SW 671 778 SSW 27 991 N K 24 so F M SB 204 037 7 29 M SW 27 991 check SST SSB SSW 408 074 671 778 1079 852 Fcv 3 40 7 8 9 4 INTERPRET ASSUMPTIONS ANOVA TESTING since to use ANOVA validly the data must meet some restrictions 4 STEPS 1 The observations are random and independent samples from the populations 1 State the hypothesis H0 1 2 K Ha i j for some i j 7 29 F Fcv 3 40 we reject H0 The methods of presentation are not equally effective Note does not tell us which pair of means are different SUMMARY TABLE Source of Sum of Degrees of Variance variation squares freedom estimate F ratio Between 408 074 2 204 037 7 29 Within 671 778 24 27 991 Total 1079 852 26 With 0 05 Fcv 3 40 Therefore H0 is rejected 2 The distributions of the populations from which samples are selected are normal 3 The variances of the distributions in the populations are equal Homogeneity of variance 2 Set the criterion F K 1 N K 3 Compute the test statistic F M SB M SW 4 Interpret results How do we check these assumptions 1 Sample properly 2 2 test 3 Bartlett s test 10 11 12 ASSUMPTIONS SIGNIFICANCE SIGNIFICANCE it turns out that ANOVA tells us if an effect difference of means is statistically significant what percentage of the variation in the data is accounted for by the independent variable changes in groups SS B K 1 M SW 2 SST M SW in our example problem last time we plug in the numbers 408 074 2 27 991 2 0 3178 1079 852 27 991 violations of normality have small effects on Type I error rates violations of homogeneity of variance have a big effect if the population sizes are different means that ANOVA is robust as long as the sample sizes are the same across populations Note though the restrictions to apply ANOVA are more stringent than for t tests no restrictions on population distributions it does not tell us if it is important one measure that helps us determine importance is the strength of association 2 measures association between independent and dependent variables similar to correlation coefficient of determination r2 the different lecture styles account for approximately 32 of the variance suggests that the results are not really that strong 13 14 15 t tests t tests EXAMPLE if we have only two groups Ha 1 2 we can use either ANOVA or the t tests discussed previously it turns out that the F distribution for K 1 N K 1 N 2 degrees of freedom is simply the t distribution for N 2 df squared A sociologist wants to determine whether sorority or dormitory women date more often He randomly samples 12 women who live in sororities and 12 women who live in dormitories and determines the number of dates they each have during the ensuing month The following are the results they give identical results t2cv Fcv H 0 1 2 t2 F so using either technique produces the same results reject or not reject 16 17 Sorority Dormitory Women X1 Women X2 8 9 5 7 6 3 4 4 12 4 7 8 9 7 10 5 5 8 3 6 7 3 5 5 T1 81 T2 69 X 1 6 750 X 2 5 750 18 t TEST t TEST test with 0 05 two tailed so standard error is 1 1 2 sX 1 X 2 s n1 n2 sX 1 X 2 0 963 and X 1 X 2 1 2 t sX 1 X 2 H 0 1 2 0 Ha 1 2 0 we have equal numbers of subjects so we do not need to test for homogeneity of variance df n1 n2 2 12 12 2 22 so from the t distribution table we find 1 0 t 1 038 2 074 tcv 0 963 tcv 2 074 which is not in the region of rejection so do not reject H0 from data we calculate the pooled estimate of population variance no evidence for a difference in number of dates s2 5 570 ANOVA H 0 1 2 Ha 1 …