Question

Complete the following two-way ANOVA table. Use α=0.05α=0.05.

Source	Degrees of Freedom	Sum of Squares	Mean Square	F	P−P−value
Row	3	126.98
Column	4	37.49
Interaction		380.82
Error	60
Total		1278.94

Answer 1

SOLUTION:

df(interaction)=3*4=12

MS=SS/df

F=MS(effect)/MS(error)

p-value=FDIST(F,df(effect),df(error))

	df	SS	MS	F	P-Value
Row	3	126.98	42.3267	3.4616	0.0217
Column	4	37.49	9.3725	0.7665	0.5512
Interaction	12	380.82	31.735	2.5954	0.0076
Error	60	733.65	12.2275
total	79	1278.94

a) As,p-value for interactions effect is less than 0.05,we reject the null hypothesis and can conclude that interaction effect are significant.

b) As,p-value for row effects is less than 0.05,we reject the null hypothesis and can conclude that row effects are significant.

c) As,p-value for column effects is greater than 0.05,we fail to reject the null hypothesis and can conclude that column effects are not significant.

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