1. S519: Evaluation of Information Systems Social Statistics Inferential Statistics Chapter 11: ANOVA

2. Last week

3. This week When to use F statstic How to compute and interpret Using FTEST and FDIST functions How to use the ANOVA

4. Analysis of variance Goudas, M.; Theodorakis, Y.; and Karamousalidis, G. (1998). Psychological skills in basketball: Preliminary study for development of a Greek form of the Athletic Coping Skills Inventory. Perceptual and Motor Skills, 86, 59-65 Group 1: athletes with 6 years of experience or less Group 2: athletes with 7 to 10 years of experience Group 3: athletes with more than 10 years of experience The athletes are not being tested more than once. One factor: psychological skills (experiences) Which statistic test should we use?

5. Simple analysis of variance There is one factor or one treatment variable being explored and there are more than two levels within this factor. Simple ANOVA: one-way analysis of variance or single factor It tests the difference between the means of more than two groups on one factor or dimension.

6. Simple ANOVA Any analysis where There is only one dimension or treatment or one variable There are more than two levels of the grouping factor, and One is looking at differences across groups in average scores Using simple ANOVA (F test)

7. F value Logic: if there are absolutely no variability within each group (all the scores were the same), then any difference between groups would be meaningful. ANOVA: compares the amount of variability between groups (which is due to the grouping factor) to the amount of variability within groups (which is due to chance)

8. F value F = 1 The amount of variability due to within-group differences is equal to the amount of variability due to between-group differences  any difference between groups would not be significant F increase The average different between-group gets larger  the difference between groups is more likely due to something else (the grouping factor) than chance (the within-group variation) F decrease The average different between-group gets smaller  the difference between groups is more likely due to chance (the within-group variation) rather than due to other reasons (the grouping factor)

9. Example Three groups of preschoolers and their language scores, whether they are overall different? Group 1 ScoresGroup 2 ScoresGroup 3 Scores 87 89 868591 769996 568587 787989 988190 778289 667896 758596 679193

10. F test steps Step1: a statement of the null and research hypothesis One-tailed or two-tailed (there is no such thing in ANOVA)

11. F test steps Step2: Setting the level of risk (or the level of significance or Type I error) associated with the null hypothesis 0.05

12. F test steps Step3: Selection of the appropriate test statistics See Figure 11.1 (S-p227) Simple ANOVA

13. F test steps Step4: Computation of the test statistic value the between-group sum of squares = the sum of the differences between the mean of all scores and the mean of each group score, then squared The within-group sum of squares = the sum of the differences between each individual score in a group and the mean of each group, then squared The total sum of square = the sum of the between-group and within-group sum of squares

14. F test steps Group 1 Scoresx squareGroup 2 Scoresx squareGroup 3 Scoresx square 877569877569897921 867396857225918281 765776999801969216 563136857225877569 786084796241897921 989604816561908100 775929826724897921 664356786084969216 755625857225969216 674489918281938649 n10 N30 ∑x 766 852 916 ∑∑X2534 76.6 85.2 91.6 214038.5333 59964 72936 84010 216910 58675.6 72590.4 83905.6 215171.6

15. F-test Between sum of squares 215171.6-214038.531133.07 within sum of squares 216910-215171.601738.40 total sum of squares 216910-214038.532871.47

16. F test steps Between-group degree of freedom=k-1 k: number of groups Within-group degree of freedom=N-k N: total sample size source sums of squaresdf mean sums of squaresF Between groups1133.072566.538.799 Within gruops1738.402764.39 Total2871.4729

17. F test steps Step5: determination of the value needed for rejection of the null hypothesis using the appropriate table of critical values for the particular statistic Table B3 (S-p363) df for the denominator = n-k=30-3=27 df for the numerator = k-1=3-1=2 3.36

18. F test steps Step6: comparison of the obtained value and the critical value If obtained value > the critical value, reject the null hypothesis If obtained value < the critical value, accept the null hypothesis 8.80 and 3.36

19. F test steps Step7 and 8: decision time What is your conclusion? Why? How do you interpret F (2, 27) =8.80, p<0.05

20. Excel: ANOVA Three different ANOVA: Anova: single factor Anova: two factors with replication Anova: two factors without replication

21. ANOVA: a single factor Anova: Single Factor SUMMARY GroupsCountSumAverageVariance Group 1 Scores1076676.6143.1556 Group 2 Scores1085285.238.4 Group 3 Scores1091691.611.6 ANOVA Source of VariationSSdfMSFP-valueF crit Between Groups1133.0672566.53338.7991260.0011423.354131 Within Groups1738.42764.38519 Total2871.46729

22. Exercise S-p241 1 2 3