Question

The following table summarizes the results of a two-factor ANOVA evaluating an independent-measures experiment with 2 levels of factor A, 3 levels of factor B, and n = 6 participants in each treatment condition.

A. Fill in all missing values in the table. Show your work (i.e., all computational steps for finding the missing values). Hint: start with the df values.

B.  Do these data indicate any significant effects (assume p < .05 for hypothesis testing of all three effects)?

           Note: in order to receive full credit, your answer for each effect should include:

• the null hypothesis H₀ & the alternative hypothesis H₁
• the critical F value used for the decision about H₀
• if the difference is statistically significant, the computed effect size, η²
• the conclusion in APA style format

                      Source                 SS          df            MS                

          Between Treatments 75 ____

               Factor A                    ____       ____      ____     F_A = ___   

               Factor B                    ____       ____ 15 F_B = ___

               A x B    ____     ____      ____      F_AxB = 6.00

          Within Treatments           ____       ____             

          Total    165          ____

Answer 1

A.

Total no. of observations=n=6*3*2=36

Source	SS	df	MS=SS/df	F= MS due effect/MS due error
Between Treatments	75	1+2+2=5	75/5=15
Factor A	75-30-36=9	2-1=1	9	FA= 9/3=3
Factor B	15*2=30	3-1=2	15	FB=15/3=5
AxB	18*3=36	(2-1)(3-1)=2	6.00*3=18	FAxB=6.00
Within Treatments	165-75=90	30	90/30=3
Total	165	36-1=35

B.

Null hypothesis, H_0A: Main effect A is absent

Alternative hypothesis, H_aA: Main effect A is present.

p-value=P(F>3|F~F_1,30)=0.0935>0.05

so we fail to reject H_0A at 5% level of significance and conclude that effect of factor A is insignificant.

Null hypothesis, H_0B: Main effect B is absent

Alternative hypothesis, H_aB: Main effect B is present.

p-value=P(F>5|F~F_2,30)=0.0134<0.05

so we reject H_0B at 5% level of significance and conclude that effect of factor B is significant.

Effect size= η² =SSB/SST=30/165=0.1818

Null hypothesis, H_0AxB: Interaction effect AxB is absent

Alternative hypothesis, H_aAxB: Interaction effect AxB is present.

p-value=P(F>6|F~F_2,30)=0.0064<0.05

so we reject H_0AxB at 5% level of significance and conclude that effect of interaction AxB is significant.

Effect size= η² =SS(AxB)/SST=36/165=0.2182