Sum of Squares - Introduction to Statistics in Psychology - Lecture Slides, Slides of Statistics for Psychologists

Download Sum of Squares - Introduction to Statistics in Psychology - Lecture Slides and more Slides Statistics for Psychologists in PDF only on Docsity! Introduction to Statistics in Psychology PSY 201 Lecture 35 ANalysis Of VAriance How do changes in one variable correspond to changes in another variable? TESTING 4 STEPS 1. State the hypothesis. : H0 : µ1 = µ2 = ... = µK , Ha : µi != µj for some i, j. 2. Set the criterion. F(K"1,N"K) 3. Compute the test statistic. F = MSB/MSW 4. Interpret results. 2 SUM OF SQUARES calculating the sum of squares (within, between, total) can be messy can make it a bit easier for each group, k, let Tk = nk! i Xik and let T = ! k Tk then SSB = ! k T 2k nk " T 2 N SSW = ! k nk! i X2ik " ! k T 2k nk and SST = ! k nk! i X2ik " T 2 N 3 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 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 EXAMPLE 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 ! k nk! i X2ik = 200, 428 T = 2320 5 (1) HYPOTHESES for one-way ANOVA the hypotheses are H0 : µ1 = µ2 = µ3 Ha : µi != µk for some i, k Test with ! = 0.05 6 docsity.com (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] df = N " 1 = 27" 1 = 26 From F distribution table we find Fcv = 3.40 7 (3) STATISTIC from the data SSB = ! k T 2k nk " T 2 N SSB = (767)2 9 + (819)2 9 + (734)2 9 "(2320) 2 27 = 408.074 SSW = ! k nk! i X2ik " ! k T 2k nk SSW = 200428" " ##$ (767)2 9 + (819)2 9 + (734)2 9 % &&' = 671.778 and SST = ! k nk! i X2ik " T 2 N SST = 200428" (2320)2 27 = 1079.852 check SST = SSB+SSW = 408.074+671.778 = 1079.852 8 (3) STATISTIC MSB = SSB K " 1 = 408.074 2 = 204.037 MSW = SSW N "K = 671.778 24 = 27.991 so F = MSB MSW = 204.037 27.991 = 7.29 9 (4) INTERPRET since 7.29 = F > Fcv = 3.40 we reject H0. The methods of presentation are not equally e!ective. (Note, does not tell us which pair of means are di!erent!) 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. 10 ASSUMPTIONS to use ANOVA validly, the data must meet some restrictions 1. The observations are random and in- dependent samples from the popula- tions. 2. The distributions of the populations from which samples are selected are nor- mal. 3. The variances of the distributions in the populations are equal. Homogene- ity of variance. How do we check these assumptions? 1. Sample properly. 2. "2-test. 3. Bartlett’s test. 11 ANOVA TESTING 4 STEPS 1. State the hypothesis. : H0 : µ1 = µ2 = ... = µK , Ha : µi != µj for some i, j. 2. Set the criterion. F(K"1,N"K) 3. Compute the test statistic. F = MSB/MSW 4. Interpret results. 12 docsity.com