Consider a two-way ANOVA with three levels for factor A, five levels for factor
ID: 3369661 • Letter: C
Question
Consider a two-way ANOVA with three levels for factor A, five levels for factor B, and three replicates in each of the 15 cells, with SSA 48, SSB- 32, SSE 120, and SST 264 Complete parts (a) through (c). Degrees of Freedom Mean Square Sum of Squares 48 32 64 120 264 24 30 otal a. At the 0.025 level of significance, is there an effect due to factor A? Determine the hypotheses. Choose the correct answer below. O A. Ho: There is no effect due to factor A. H4: There is a positive effect due to factor A. O B. H: There is an effect due to factor A. H There is no effect due to factor A. C. Ho . There is no effect due to factor A. H: There is an effect due to factor A. O D. Ho: There is no effect due to factor A. H4: There is a negative effect due to factor A. Determine the value of the test statisticExplanation / Answer
a.
The correct hypotheses are -
C. H0: There is no effect due to factor A.
H1: There is an effect due to factor A.
From the table, F statistic for A is 6.
So, Fstat = 6
Numerator Degree of freedom = Degree of freedom of A = 2
Denominator Degree of freedom = Degree of freedom of error = 30
P-value for DF = 2, 30 is,
P[F > 6] = 0.006
As, p-value is less than the significance level of 0.025,
D. Reject H0 because there is sufficient evidence to conclude that there is an effect due to factor A.
b.
The correct hypotheses are -
A. H0: There is no effect due to factor B.
H1: There is an effect due to factor B.
From the table, F statistic for B is 2.
So, Fstat = 2
Numerator Degree of freedom = Degree of freedom of B = 4
Denominator Degree of freedom = Degree of freedom of error = 30
P-value for DF = 4, 30 is,
P[F > 2] = 0.120
As, p-value is greater than the significance level of 0.025,
B. Do not reject H0 because there is insufficient evidence to conclude that there is an effect due to factor B.
c.
The correct hypotheses are -
A. H0: There is no interaction effect.
H1: There is an interaction effect.
From the table, F statistic for AB is 2.
So, Fstat = 2
Numerator Degree of freedom = Degree of freedom of AB = 8
Denominator Degree of freedom = Degree of freedom of error = 30
P-value for DF = 8, 30 is,
P[F > 2] = 0.081
As, p-value is greater than the significance level of 0.025,
D. Do not reject H0 because there is insufficient evidence to conclude that there is an interaction effect.