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Consider the following partially completed two-way ANOVA table. Suppose there ar

ID: 2908948 • Letter: C

Question

Consider the following partially completed two-way ANOVA table. Suppose there are 4 levels of Factor A and 2 levels of Factor B. The number of replications per cell is 4. Use the 0.01 significance level. (Hint: estimate the values from the F table.)

Complete an ANOVA table. (Round MS and F to 2 decimal places.)

Sources: Factor A, Factor B, Interaction, Error, total.

SS: 70, 50, 210, 400, 730

Find the critical values to test for equal means. (Round your answers to 2 decimal places.)

Determine if there is a significant difference in Factor A means, Factor B means.

Determine if there is a significant difference in interaction means.

Explanation / Answer

To complete the table above, we should first obtain the table of formulas needed.

n= 4 * 2 *4 = 32

ANOVA

Source

SS

df

MS

F

Factor A

Factor A

k-1

MSA= SSA/(k-1)

MSA/MSE

Factor B

Factor B

b-1

MSB = SSB/(b-1)

MSB/MSE

Interaction

SSI

(k-1)(b-1)

MSI=SSI/(k-1)(b-1)

MSI/MSE

Error

SSE

n-kb

MSE = SSE/(n-kb)

Total

SS Total

n-1

Therefore, the completed ANOVA table is as follows:

ANOVA

Source

SS

df

MS

F

Factor A

70

3

23.33

1.339

Factor B

50

1

50

2.99

Interaction

210   

70

4.199

Error

400

24

16.667

Total

730

31

To test whether there is a significant difference in factor A means, we will use critical value from F table with 0.05 significance level, df numerator (k-1) = 3, and df denominator (n-kb) = 24. The critical value is 3.01. The computed value of factor A is smaller than the critical value, therefore, we fail to reject the null hypothesis. We can conclude that the factor A means are equal.

To test whether there is a significant difference in factor B means, we will use critical value from F table with 0.05 significance level, df numerator (b-1) = 1, and df denominator (n-kb) = 24. The critical value is 4.26. The computed value of factor B is smaller than the critical value, therefore, we fail to reject the null hypothesis. We can conclude that the factor B means are equal.

To test whether there is a significant effect from the interaction between factor A and factor B, we will use critical value from F table with 0.05 significance level, df numerator (k-1)(b-1) = 3, and df denominator (n-kb) = 24. The critical value is 3.01. The computed value of interaction is higher than the critical value, therefore, we reject the null hypothesis of no interaction. We can conclude that the combination between factor A and factor B has a significant effect on the response factor.

ANOVA

Source

SS

df

MS

F

Factor A

Factor A

k-1

MSA= SSA/(k-1)

MSA/MSE

Factor B

Factor B

b-1

MSB = SSB/(b-1)

MSB/MSE

Interaction

SSI

(k-1)(b-1)

MSI=SSI/(k-1)(b-1)

MSI/MSE

Error

SSE

n-kb

MSE = SSE/(n-kb)

Total

SS Total

n-1