Question

Asked Sep 17, 2019

Step 1

**Note:**

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Step 2

**Explanation:**

Part (1):

Here, the number of levels of Factor A is, *k*_{1} = 6.

The number of levels of Factor B is, *k*_{2} = 4.

The number of replications per cell is, *r* = 6.

Thus, the degrees of freedom (df) for Factor A is, *k*_{1} – 1 = 5; the df for Factor B is, *k*_{2} – 1 = 3.

Since Interaction = Factor A × Factor B, df for Interaction is 5 × 3 = 15.

Since there are 6 levels of Factor A and 4 levels of Factor B, there are a total of 6 × 4 = 24 cells. Each cell has 6 replicates. So, total number of observations, *n* = 24 × 6 = 144. Thus, total df is, *n* – 1 = 143.

As a result, error df is, 143 – 5 – 3 – 15 = 120.

The MS or mean of squares is the SS or sum of squares, divided by the corresponding df.

The F for a particular factor is the MS for that factor, divided by the error MS; same for the Interaction.

Using this information, the ANOVA table is completed as follows:

Step 3

**Parts 2 and 3:**

The *F*-test is used for testing the significant effects of Factor A, Factor B, and Interaction.

The numerator and denominator df’s for testing the effect of Factor A are, 5, 120.

The significance level is, *α* = 0.05.

Thus, the critical value is, *F*0.05; 5, 120.

Using Excel formula, **=F.INV.RT(0.05,5,120)**, the critical value for Factor A is **2.29**.

The numerator and denominator df’s for testing the effect of Factor B are, 3, 120.

Thus, the critical value is, *F*0.05; 3, 120.

Using Excel formula, **=F.INV.RT(0.05,3,120)**, the critical value for Factor A is **2.68**.

The numerator and denominator df&r...

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