A two-factor factorial design has 5 degree of freedom for factor A 4 degrees of
ID: 3176188 • Letter: A
Question
A two-factor factorial design has 5 degree of freedom for factor A 4 degrees of for factor B of 20 degrees of freedom for interaction variatio 20 degrees of freedom for random variation, and 149 degrees for total variation. Suppose also th ssE sST MSA 40, "60 MSAB 60, and SSA 240, a 2,400, 6,040. MSB 20. Complete parts (a) through (d) a. What is the value of the F test statistic for the interaction effect? STAT (Simplify your answer.) STAT b. What is the value of the F test statistic for the factor A effect? STAT (simplify your answer.) STAT c. What is the value of the F test statistic for the factor B effect? STAT (Simplify your answer. STAT d. Form the ANOVA summary table and fill in all values in the body of the table. Mean Square Sum of Squares of Freedom Degrees Sum of Mean Square 7. a two-way ANOVA with two levels for of Freedom Squares (Variance factor A, four levels for factor B, and four replicates Source in each of the 8 cells, with SSA 5, SSB 20, SSE 120, and 40 SST 335. Complete parts (a) through (c) 30 335 a. At the 0.01 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. H1: There is a negative effect due to factor A. O B. Ho: There is no effect due to factor A. H1: There is an effect due to factor A. O c. Ho: There is an effect due to factor A. H1: There is no effect due to factor A. O D. Ho: There is no effect due to factor A. H1: There is a positive effect due to factor A. Determine the value of the test statistic. Simplify your answer. STAT Determine the p-value.Explanation / Answer
a.The null and alternative hypothesis are
B. H0: There is no effect due to factor A
H1: There is an effect due to factor A
The test statistic FSTAT=1 (from row of factor A in table above.)
Degree of freedom of main effect test statistics F = (1,24)
p-value= P(F(1,24)>1) = 0.32729 using excel function =FDIST(1,1,24)
As p-value>0.01, we do not reject the null hypothesis , so answer is option-D (Do not rejectH0,because there is insufficients evidence to conclude that there is an effect due to factor A.
b. The null and alternative hypothesis are
A. H0: There is no effect due to factor B
H1: There is an effect due to factor B
The test statistic FSTAT=8 (from row of factor B in table above.)
Degree of freedom of main effect test statistics F = (3,24)
p-value= P(F(3,24)>8) = 0.0007 using excel function =FDIST(8,3,24)
As p-value<0.01, we reject the null hypothesis , so answer is option-B (Rejct H0,because there is sufficient evidence to conclude that there is an effect due to factor B.
c. The null and alternative hypothesis are
C. H0: There is no interaction effect
H1: There is interaction effect
The test statistic FSTAT=6 (from row of factor AB in table above.)
Degree of freedom of main effect test statistics F = (3,24)
p-value= P(F(3,24)>6) = 0.0034 using excel function =FDIST(6,3,24)
As p-value<0.01, we reject the null hypothesis , so answer is option-C (Reject H0,because there is sufficient evidence to conclude that there is an interaction effect.
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