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### L36

Course: PSY 201, Fall 2010
School: Purdue
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Word Count: 753

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to Introduction Statistics in Psychology PSY 201 Professor Greg Francis Lecture 36 ANalysis Of VAriance The last topic! REPEATED MEASURES SUM OF SQUARES one way ANOVA deals with independent samples scores for an individual are dependent scores for dierent individuals are independent we want to consider one situation with a dependency suppose we repeat our measurements from the same individuals at dierent...

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to Introduction Statistics in Psychology PSY 201 Professor Greg Francis Lecture 36 ANalysis Of VAriance The last topic! REPEATED MEASURES SUM OF SQUARES one way ANOVA deals with independent samples scores for an individual are dependent scores for dierent individuals are independent we want to consider one situation with a dependency suppose we repeat our measurements from the same individuals at dierent times e.g., tracking health patterns across years, grades throughout school,... SST = SSI + SSO + SSRes where SST is the total sum of square SSI is the variation among individuals SSO is the variation among test occasions SSRes is any other type of variation 2 3 INDIVIDUALS OBSERVATIONS RESIDUAL the combined variation among individuals is the combined variation across observations is we need a term that correspond to SSW SSI = iK (X i X )2 SSO = k n (X k X )2 where X X i = k ik K is the average for the ith individual across all observations where X X k = i ik n is the average for the k th observation across all subjects SSI deviation of individual means from overall mean SSO deviation of observation mean from overall mean does not correspond to SSW or SSB in the normal ANOVA similar to SSB in the normal ANOVA 4 5 we can directly calculate the total sum of squares SST = ki (Xik X )2 if there is variation beyond SSI and SSO , we can calculate it as SSRes = SST SSI SSO this is similar to SSW factors out variation due to individuals and variation due to observations 6 VARIANCE ESTIMATES VARIANCE ESTIMATES F RATIO SSRes > 0 due to random sampling (choice of individuals) SSRes M SRes = (K 1)(n 1) estimates the variance of the population distribution SSO can vary due to random sampling, or due to dierences across observations as before we compare these estimates with the F statistic M SO F= M SRes if H0 is true H0 : 1 = 2 = ... = K then there are no dierences across observations, so all variation must be due to random sampling. So, SSO M SO = K 1 estimates the variance of the population distribution if H0 is true the degrees of freedom associated with this estimate is (K 1)(n 1) if H0 is true F 1.0 if H0 is not true F > 1.0 look up Fcv for (K-1), (K-1)(n-1) degrees of freedom everything else is the same as before otherwise it overestimates it 7 8 9 EXAMPLE (1) HYPOTHESIS (2) CRITICAL VALUE H 0 : 1 = 2 = 3 for the numerator (observation sum of squares) we have A school principal traces reading comprehension scores on a standardized test for a random sample of dyslexic students across three The years. data are given below. Complete the ANOVA using = 0.05. Student Third Grade Fourth Grade Fifth Grade 1 2.8 3.2 4.5 2 2.6 4.0 5.1 3 3.1 4.3 5.0 4 3.8 4.9 5.7 5 2.5 3.1 4.4 6 2.4 3.1 3.9 7 3.2 3.8 4.3 8 3.0 3.6 4.4 Tk 23.40 30.00 37.30 ki Ha : i = k for some i and k test with = 0.05 df = K 1 = 3 1 = 2 for the denominator (residual sum of squares) we have df = (K 1)(n1) = (31)(81) = 14 Ti 10.50 11.70 12.40 14.40 10.00 9.40 11.30 11.00 90.70 so from the F distribution table we nd the F critical value to be Fcv = 3.74 2 Xik = 361.43 10 11 12 (3) TEST STATISTIC (3) TEST STATISTIC there are computational formulas that help manage the calculations 2 T N where T is the sum of all scores SST = k i 2 Xik SSO = so (90.70)2 = 361.43 342.77 24 SST = 18.66 SST = 361.43 and T2 T2 SSI = i K N i k 2 Tk T 2 n N (23.40)2 (30.0)2 (37.30)2 + + 342.77 8 8 8 SSO = 354.86 342.77 = 12.09 SSO = so any remaining variation is residual SSRes = SST SSI SSO SSRes = 18.66 5.67 12.09 = 0.90 again, these cannot be negative! (3) TEST STATISTIC now calculate SSO 12.09 M SO = = = 6.05 K 1 2 and SSRes 0.90 M SRes = = = 0.06 (K 1)(n 1) 14 and get the F statistic M SO 6.05 F= = = 100.83 M SRes 0.06 where Ti is the sum of observations for the ith individual so SSI = 348.44 342.77 = 5.67 Note, this cannot be negative! 13 14 15 (4) INTERPRET SUMMARY TABLE ASSUMPTIONS since F = 100.83 > 3.74 = Fcv we reject H0. there is evidence that the reading scores for these subjects are dierent across the years Source of Sum of Degrees of Variance variation squares freedom estimate F ratio Occasions 12.09 2 6.05 100.83 Individuals 5.67 7 Residual 0.90 14 0.06 Total 18.66 23 With = 0.05, Fcv = 3.74. Therefore H0 is rejected. ANOVA for repeated measures depends on four assumptions 1. The sample was randomly selected for a population. 2. The dependent variable (e.g., reading scores) is normally distributed in the population. 3. The population variances for the test occasions are equal. (homogeneity of variance) 4. The population correlation coecients between pairs of test occasion scores are equal. deviation from (2) tends to not cause serious problems deviations from (3) and (4) can be compensated for sometimes 16 17 18 CONCLUSIONS NEXT TIME restrictions on use of one-way ANOVA nish up details of class strength of association ANOVA repeated measures Study guide for nal exam How does it feel to not be a schmuck? partitioning the variance accounts for dependence among scores for same subjects 19 20
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