week 8.Analysis of Variance

Question 1

ANOVA

Answer 1

(Analysis of Variance) allows researchers to test the significance of the differences between 2 or more means.

X_ij=µ + r_j + €_ij

where i and j are groups, r=treatment effect for group and €=error.

ie Basic concept of predicting a value, given group mean, grand group mean, and error.

Question 2

Assumptions of Anova

Answer 2

1. Assume variances of each group are the same (if 2 populations have the same variance, we assume sample variances will be alike)
2. Assumes that the variables are normally distributed
3. assumes that the scores of one group are entirely independent of scores from another group.Also, each individual score within a group should be entirely independent of any other particiapnts in group.

Question 3

Logic of Anova

Answer 3

1. If assume the variances of groups are the same, and assume normally distributed, then groups will only differ in their mean.
2. Total variability in the data has 2 components:
   a) variability associated with the treatment
   b) error variance

Question 4

Error Variance

Answer 4

Provides an index of variability between people who have been treated the same. ie Within one group, there will be chance variations of motivation, personality, IQ, and experimental error etc. Also called Within-Group Variation.

Question 5

Between-treatments Variance

Answer 5

Also called Treatment variance.Differences between the means of the groups may be due to the different treatment, or chance.

Question 6

F-statistic

Answer 6

For ANOVA, we use and F-statistic. We need to determine if the differences between treatment groups are larger than would be expected by chance alone.

F=variance between treatments/error variance

F=treatment effect+differences due to chance/differences due to chance

ie, if there is no treatment effect (H₀ is true), then the F ratio will equal 1. If there is a treatment effect, F>1.

F=MS_treatment/MS_error

Question 7

ANOVA calculations

Answer 7

SS=Sum of squares=Σ(X-X^-)²

For Anova, we have SS_treatment

SS_error and

SS_total .

SS_treatment=nΣ(X^-_j -X^-…)²

where n=group sample size,

X^-_j=group mean

and X^-…=grand mean.

SS_error=Σ(X_ij-X^-_j)²

where X_ij=score of person i in group j

and X^-_jis the mean of group j.

SS_total=SS_treatment + SS_error

Then if we divide an SS by it’s respective degrees of freedom, we get Mean Squares (MS).

SS/df=MS

df_total=N-1=df_treatment + df_error

df_treatment=k-1 where k=number of treatments

df_error=k(n-1)

Question 8

Evaluating the significance of the F-statistic

Answer 8

Use the F_critical tables.

the degrees of freedom for treatment are the numerator degrees of freedom

and the error degrees of freedom are the denominator degrees of freedom.

If F_{(dftreatment,dferror)} >F_critical, then we reject the null hypothesis.

Question 9

Effect size

Answer 9

To evaluate the effect size of the Anova finding, we use partial Eta squared (n²). This can be interpreted as a squared correlation coefficient. eg if n²=0.23, then 23% of variation in dependent variable can be attributed to differences in the variable which was manipulated across different groups.

In general. n²=0.01=small effect

n²=0.06=medium effect

n²=0.14=large effect.

Question 10

Comparison between Anova and t-test

Answer 10

Anova allows comparisons between >2 group means. For comparison of group A, B and C,

would need 3 seperate t-tests (A vs B, Avs C, and Bvs C) whereas can do 1 Anova. If do multiple t-tests, increase the liklihood of a type 1 error, simply because each single test has a 5% chance of error.

Question 11

Levene’s test

Answer 11

Levene’s test us is to test the assumption that different groups are of homogenous variance. (one of the assumptions of ANOVA.)

If the results of Levene’s (also denoted by F) has an accompanying p value and p < 0.05 then it is concluded that the assumption of homogenity has been violated. Some might then recommend using different statistics to anova.