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

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

Last week

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

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, 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?

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.

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)

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)

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)

Example Three groups of preschoolers and their language scores, whether they are overall different? Group 1 ScoresGroup 2 ScoresGroup 3 Scores

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

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

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

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

F test steps Group 1 Scoresx squareGroup 2 Scoresx squareGroup 3 Scoresx square n10 N30 ∑x ∑∑X

F-test Between sum of squares within sum of squares total sum of squares

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 groups Within gruops Total

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

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

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

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

ANOVA: a single factor Anova: Single Factor SUMMARY GroupsCountSumAverageVariance Group 1 Scores Group 2 Scores Group 3 Scores ANOVA Source of VariationSSdfMSFP-valueF crit Between Groups Within Groups Total

Exercise S-p