Chapter 12.3, Problem 15E

a.

To determine

To state: The hypothesis.

Answer to Problem 15E

The hypothesis for interaction is,

$H_0:$ There is no interaction between the ages of the sales people and the products they sell on the monthly sales.

$H_1:$ There is an interaction between the ages of the sales people and the products they sell on the monthly sales.

The hypothesis for age of salespersonis,

$H_0:$ There is no difference in the means of monthly sales of the two age groups.

$H_1:$ There is adifference in the means of monthly sales of the two age groups.

The hypothesis for product is,

$H_0:$ There is no difference in the means of monthly sales for the different products.

$H_1:$ There is a difference in the means of monthly sales for the different products.

Explanation of Solution

Given info:

The table shows age of the sales representative and type of item affect monthly sales.

Calculation:

The hypothesis for interaction is,

Null hypothesis:

$H_0:$ There is no interaction between the ages of the sales people and the products they sell on the monthly sales.

Alternative hypothesis:

$H_1:$ There is an interaction between the ages of the sales people and the products they sell on the monthly sales.

The hypothesis for age of salesperson is,

Null hypothesis:

$H_0:$ There is no difference in the means of monthly sales of the two age groups.

Alternative hypothesis:

$H_1:$ There is a difference in the means of monthly sales of the two age groups.

The hypothesis for productis,

Null hypothesis:

$H_0:$ There is no difference in the means of monthly sales for the different products.

Alternative hypothesis:

$H_1:$ There is a difference in the means of monthly sales for the different products.

b.

To determine

To find: The critical value for each F test.

Answer to Problem 15E

The critical F-value of A is 4.26.

The critical F-value of B and $A\times B$ is 3.40.

Explanation of Solution

Given info:

The level of significance is 0.05.

Calculation:

Consider factors A represent the type of paintand B represents the geographic location.

Here, $a=2\text{, }b=3\text{ and }n=5$

The formula to find the degrees of freedom for factor A is,

$\begin{array}{c}\text{d}\text{.f}\text{.N}=a-1\\ =2-1\\ =1\end{array}$

Thus, the degrees of freedom for factor A is1.

Here, $b=3$

The formula to find the degrees of freedom for factor B is,

$\begin{array}{c}\text{d}\text{.f}\text{.N}=b-1\\ =3-1\\ =2\end{array}$

Thus, the degrees of freedom for factor B is3.

The formula to find the degrees of freedom for factor $A\times B$ is,

$\begin{array}{c}\text{d}\text{.f}\text{.N}=\left(a-1\right)\left(b-1\right)\\ =\left(2-1\right)\left(3-1\right)\\ =1\times 2\\ =2\end{array}$

Thus, the degrees of freedom for factor $A\times B$ is2.

The degrees of freedom for the within (error) factor

$\begin{array}{c}\text{d}\text{.f}\text{.D}=ab\left(n-1\right)\\ =2\times 3\times \left(5-1\right)\\ =6\times 4\\ =24\end{array}$

The degrees of freedom for the within (error) factor is 24.

Critical value:

Here A having 1 degrees of freedom and B and $A\times B$ having 2 degrees of freedom.

For A:

The critical F-value of A is obtained using the Table H: The F-Distribution with the level of significance $\alpha =0.05$.

Procedure:

• Locate 24 in the degrees of freedom, denominator row of the Table H.
• Obtain the value in the corresponding degrees of freedom, numerator column below 1.

That is,

$\text{Critical value of }A=4.26$

Thus, the critical F-value of A is 4.26.

For B and $A\times B$:

The critical F-value of B and $A\times B$ is obtained using the Table H: The F-Distribution with the level of significance $\alpha =0.05$.

Procedure:

• Locate 24 in the degrees of freedom, denominator row of the Table H.
• Obtain the value in the corresponding degrees of freedom, numerator column below 2.

That is,

$\text{Critical value of }B\text{ and }A\times B=3.40$

Thus, the critical F-value of B and $A\times B$ is 3.40.

Decision criteria:

If $F_A\text{-value > Critical value}$ then reject the null hypothesis $\left(H_0\right)$.

If $F_B\text{-value > Critical value}$ then fail to reject the null hypothesis $\left(H_0\right)$.

If $F_{AB}\text{-value > Critical value}$ then fail to reject the null hypothesis $\left(H_0\right)$.

c.

To determine

To compute: The summary table.

To find: The test value.

Answer to Problem 15E

The summary table is,

The test valuesare 1.57, 8.21 and 37.09.

Explanation of Solution

Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistics and P-value using the MINITAB software:

• Choose Stat > ANOVA > Two-Way.
• In Response, enter the column of Sales.
• In Row Factor, enter the column of Ages of sales person.
• In Column Factor, enter the column of product.
• Click OK.

Output using the MINITAB software is given below:

From Minitab output, the test valuesare 1.57, 8.21 and 37.09.

d.

To determine

To make: The decision.

Answer to Problem 15E

The null hypothesis is rejected for $F_{AB}$ and no need to test the main effects.

Explanation of Solution

Conclusion:

From the result of part (c), the $F_A$, $F_B$ and $F_{AB}$ are1.57, 8.21 and 37.09.

Decision for $F_A$:

Here, the test value of $F_A$ is lesser than the critical value.

That is, $1.57<4.26$.

Thus, it can be concluding that, the null hypothesis for $F_A$ is not rejected.

Decision for $F_B$:

Here, the test value of $F_B$ is greaterthan the critical value.

That is, $8.21>3.40$.

Thus, it can be concluding that, the null hypothesis for $F_B$ is rejected.

Decision for $F_{AB}$:

Here, the test value of $F_{AB}$ is greater than the critical value.

That is, $37.09>3.40$.

Thus, it can be concluding that, the null hypothesis for $F_{AB}$ is rejected.

e.

To determine

To explain: The results.

Answer to Problem 15E

The result concludes that, there is an interception effect between the ages of salespeople and the type of products sold.

Explanation of Solution

Calculation:

From the results, it can be observed there is a significant between the ages of the sales people and the products they sell on the monthly sales.

Here, the integration effect is between the ages of the sales person and the type of product sold. So, the procedure for drawing the graph for the interaction plot is given below.

Minitab Procedure:

• Choose Stat > ANOVA >Interaction Plot.
• In Response, enter the column of Sales.
• In Factors, enter the column of Ages of sales person and product.
• Click OK.

Output using the MINITAB software is given below:

Interpretation:

From the graph, it can be observed that the lines cross. So, the interaction is disordinal interaction. Hence there is an interception effect between the ages of salespeople and the type of products sold.