Question

A partial result of the computation for a two-factor factorial experiment is given as follows (assume that the number of experimental units in each group is fixed):

Source of Variation	SS	df	MS	F
Factor A	48	4	12	1
Factor B		6		5
Interaction (AB)		24	15
Error		105		-
Total		139	-	-

A) Fill in all the missing results in the ANOVA table.

B) At the 0.05 level of significance, is there an interaction effect? (Show your work: your hypotheses, test statistic, critical value, and decision).

C) At the 0.05 level of significance, is there evidence of an effect due to factor A? (Show your work: your hypotheses, test statistic, critical value, and decision).

Answer 1

Part A

MS A / MS error = F

12/a = 1

a = 12

MS error = 12

SS/df = MS

MS interaction = 15

15 = SS interaction / 24

SS interaction = 15*24 = 360

MS B / MS error = 5

MS B /12 = 5

MS B = 60

Required ANOVA table is given as below:

Source of Variation	SS	df	MS	F
Factor A	48	4	12	1
Factor B	360	6	60	5
Interaction (AB)	360	24	15	1.25
Error	1260	105	12	-
Total	2028	139	-	-

Part b

H₀: There is no interaction.

H_a: There is an interaction.

α = 0.05

Test statistic = F = MS interaction / MS error = 60/12 = 5

Critical value = 1.621485

Test statistic > Critical value

So, we reject the null hypothesis

There is sufficient evidence to conclude that there is a statistically significant interaction.

Part c

H₀: There is no significant effect due to factor A.

H_a: There is a significant effect due to factor A.

α = 0.05

F = 1

Critical value = 2.45821

Test statistic F is less than critical value

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that there is significant effect due to factor A.