S519: Evaluation of Information Systems

Social Statistics

Inferential Statistics

Chapter 11: ANOVA

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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, 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?

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)

iablegroupwithin

iablegroupbetween

MS

MSF

within

between

var

var

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 Scores Group 2 Scores Group 3 Scores87 87 8986 85 9176 99 9656 85 8778 79 8998 81 9077 82 8966 78 9675 85 9667 91 93

F test steps

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

ANOVA)

3210 : H

3211 : XXXH

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 Scores x square Group 2 Scores x square Group 3 Scores x square87 7569 87 7569 89 792186 7396 85 7225 91 828176 5776 99 9801 96 921656 3136 85 7225 87 756978 6084 79 6241 89 792198 9604 81 6561 90 810077 5929 82 6724 89 792166 4356 78 6084 96 921675 5625 85 7225 96 921667 4489 91 8281 93 8649

n 10 10 10 N 30∑x 766 852 916 ∑∑X 2534

76.6 85.2 91.6 214038.5333

59964 72936 84010

216910 58675.6 72590.4 83905.6 215171.6

X)( 2 X

nX /)( 2

NX /)( 2 2)(X

nX /)( 2

F-test

Between sum of squares

215171.6-214038.53 1133.07

within sum of squares 216910-215171.60 1738.40total sum of squares 216910-214038.53 2871.47

NXnX /)(/)( 22

nXX /)()( 22

NXX /)()( 22

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

sourcesums of squares df

mean sums of squares F

Between groups 1133.07 2 566.53 8.799Within gruops 1738.40 27 64.39 Total 2871.47 29

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

SUMMARYGroups Count Sum Average Variance

Group 1 Scores 10 766 76.6 143.1556Group 2 Scores 10 852 85.2 38.4Group 3 Scores 10 916 91.6 11.6

ANOVASource of Variation SS df MS F P-value F crit

Between Groups 1133.067 2 566.5333 8.799126 0.001142 3.354131Within Groups 1738.4 27 64.38519

Total 2871.467 29

Exercise

S-p241 1 2 3