Chapter 13, Problem 18E

(a)

To determine

To graph: The interaction plot displaying systolic blood pressure on the y-axis and training level on the x-axis.

Explanation of Solution

Graph: The problem compares two factors systolic blood pressure on y-axis with training level on x-axis. The factor systolic blood pressure is further classified for men and women and the other factor training level is classified to endurance trained and sedentary men and women. Thus, the table is created for the means of two factor as shown below;

The marginal means for Endurance is calculated by using the function $=\text{SUM}\left(C3:D3\right)$

The marginal means for Sedentary is calculated by using the function $=\text{SUM}\left(C3:D3\right)$

The marginal means for Systolic blood pressure for men is calculated by using the function $=\text{SUM}\left(C3:D3\right)$

The marginal means for Systolic blood pressure for women is calculated by using the function $=\text{SUM}\left(C3:D3\right)$

The table is obtained as:

The interaction plot is drawn by following these steps:

Step 1: Open Excel sheet and write the data value. The screenshot is shown below:

Step 2: INSERT>Recommended Charts>All Charts>Line Chart. The screenshot is shown below:

The interaction plot is obtained as shown below:

Interpretation: From the chart above, the two lines are not parallel, hence, there is an interaction present between two factors. There is a main effect of training level, training sedentary takes much lower value for women when compared to training endurance for women.

(b)

To determine

To find: The ANOVA table.

Answer to Problem 18E

Solution: The ANOVA table is obtained as:

Explanation of Solution

Given: The following values of the Sum of Squares is provided,

$\begin{array}{c}\text{SSA}=677.12\\ \text{SSB}=0.72\\ \text{SSAB}=147.92\\ \text{SSE}=2478\end{array}$,

where A is the sex effect and B is the training level.

Calculation: The ANOVA table is obtained by following these steps:

Step 1: The degree of freedom for “A” is obtained as follows:

$\begin{array}{c}\text{DFA}=I-1\\ =2-1\\ =1\end{array}$

The degree of freedom for “B” is obtained as follows;

$\begin{array}{c}\text{DFB}=J-1\\ =2-1\\ =1\end{array}$

The degree of freedom for “AB” is obtained as follows:

$\begin{array}{c}\text{DFAB}=\left(I-1\right)\left(J-1\right)\\ =\left(2-1\right)\times \left(2-1\right)\\ =1\end{array}$

The degree of freedom for “E” is obtained as follows:

$\begin{array}{c}\text{DFE}=N-IJ\\ =32-\left(2\times 2\right)\\ =28\end{array}$

Step 2: The mean squares for “A” is obtained as follows:

$\begin{array}{c}\text{MSA}=\frac{\text{SSA}}{\text{DFA}}\\ =\frac{677.12}{1}\\ =677.12\end{array}$

The mean squares for “B” is obtained as follows:

$\begin{array}{c}\text{MSB}=\frac{\text{SSB}}{\text{DFB}}\\ =\frac{0.72}{1}\\ =0.72\end{array}$

The mean squares for “AB” is obtained as follows:

$\begin{array}{c}\text{MSAB}=\frac{\text{SSAB}}{\text{DFAB}}\\ =\frac{147.92}{1}\\ =147.92\end{array}$

The mean squares for “E” is obtained as follows:

$\begin{array}{c}\text{MSE}=\frac{\text{SSE}}{\text{DFE}}\\ =\frac{2478}{28}\\ =88.5\end{array}$

Step 3: The F-value for “A” is obtained as follows:

$\begin{array}{c}F=\frac{\text{MSA}}{\text{MSE}}\\ =\frac{677.12}{88.5}\\ =7.6511\end{array}$

The F-value for “B” is obtained as follows:

$\begin{array}{c}F=\frac{\text{MSB}}{\text{MSE}}\\ =\frac{0.72}{88.5}\\ =0.0081\end{array}$

The F-value for “AB” is obtained as follows:

$\begin{array}{c}F=\frac{\text{MSAB}}{\text{MSE}}\\ =\frac{147.92}{88.5}\\ =1.6714\end{array}$

To test the hypothesis for the obtained F-values, the F-critical value is obtained through the F-distribution table as $F\left(1,28\right)=4.19$

Interpretation: The comparisons of F-values with F-critical are as follows:

$\begin{array}{l}{F}_A>F_{critical}\\ F_B<F_{critical}\\ F_{AB}<F_{critical}\end{array}$

The value of F_A is greater than F-critical; hence, the null hypothesis can be rejected significantly, which states that there is a main effect of training level. While the other factor systolic blood pressure and the interaction of these two factors have F-value less than F-critical, which means that they do not show the significant difference in means.

(c)

To determine

The benefit of considering pretest measurement.

Answer to Problem 18E

Solution: It increases the power of the test statistic.

Explanation of Solution

The benefit of pre/post measurement or repeated measure improves the test statistic by reducing the within group variability when comparing the mean differences. The power of the test statistic can be understood as the probability of detecting the significant difference when comparing means, increases as a result of reduction in variance of means.