Chapter 12, Problem 46CE

a.

To determine

Compute SSE.

Answer to Problem 46CE

The sum of squares due to errors is 603.99.

Explanation of Solution

It is given that the total sum of square is 1,099.61.

$SStotal=1,099.61$

The sum of squares due to errors can be obtained as follows:

$\begin{array}{c}SSE=126.29+233.34+166.79+77.57\\ =603.99\end{array}$

Thus, the error sum of squares is 603.99.

b.

To determine

Determine SST.

Answer to Problem 46CE

The sum of square due to treatments is 495.62.

Explanation of Solution

The sum of square due to treatments is given below:

$\begin{array}{c}SST=SStotal-SSE\\ =1,099.61-603.99\\ =495.62\end{array}$

The sum of square due to treatments is 495.62.

c.

To determine

Construct an ANOVA table.

Answer to Problem 46CE

The complete ANOVA table is as follows:

Source of variation	Sum of Squares	Degrees of freedom	Mean Squares	F-statistic
Treatments	495.62	3	165.207	9.847
Error	603.99	36	16.778
Total	1,099.61	39

Explanation of Solution

There are four radio stations in Midland, and a sample of 10 randomly hours from each station is selected.

Degrees of freedom:

The degrees of freedom for treatments $\left(k-1\right)$ is 3.

The total degrees of freedom $\left(n-1\right)$ is 39.

The degrees of freedom for error $\left(n-k\right)$ is 36.

Mean sum of squares:

The mean sum of squares can be obtained by dividing each sum of squares by its respective degrees of freedom.

The mean sum of squares for the treatments can be calculated as shown below:

$\begin{array}{c}MST=\frac{SST}{df}\\ =\frac{495.62}{3}\\ =165.2067\end{array}$

The mean sum of squares for error can be calculated as follows:

$\begin{array}{c}MSE=\frac{SSE}{df}\\ =\frac{603.99}{36}\\ =16.778\end{array}$

F-Statistic:

F-statistic is the ratio of mean square for the treatments to the mean square error.

$\begin{array}{c}F=\frac{MST}{MSE}\\ =\frac{165.2067}{16.778}\\ =9.847\end{array}$

The ANOVA table and computed F value are given below:

Source of variation	Sum of Squares	Degrees of freedom	Mean Squares	F-statistic
Treatments	495.62	3	165.207	9.847
Error	603.99	36	16.778
Total	1,099.61	39

d.

To determine

Find whether there is a difference in the treatment means.

Answer to Problem 46CE

There is a difference in the treatment means.

Explanation of Solution

The null and alternative hypotheses are stated below:

$H_0:$ There is no difference in the treatment means.

 $H_1:$ There is a difference in the treatment means.

There are four treatments; thus, the numerator degrees of freedom $\left(k-1\right)$ is 3.

There are 40 observations; thus, the denominator degrees of freedom $\left(n-k\right)$ is 36.

From appendix B.6A, at the 0.05 significance level, the F-test critical value is 2.88.

Decision rule:

Reject the null hypothesis if the computed F test statistic value is greater than the critical value. Otherwise, fail to reject the null hypothesis.

Conclusion:

The F-test statistic value is greater than the critical value at the 0.05 significance level. One can reject the null hypothesis at the 0.05 significance level.

Therefore, there is a difference in the treatment means.

e.

To determine

Find whether there is a difference in the means of hard rock station and country/western station at the 0.05 significance level.

Answer to Problem 46CE

There is no difference in the means of hard rock station and country/western station.

Explanation of Solution

The null and alternative hypotheses are stated below:

$H_0:$ There is no difference in the means of hard rock station and country/western station.

$H_1:$ There is a difference in the means of means of hard rock station and country/western station.

It is given that the mean for the hard rock station is 51.32 and the mean for the country/western station is 50.85. The sample size of each station is 10. The mean square error is 16.778.

That is, ${\overline{x}}_1=51.32$, ${\overline{x}}_2=50.85$, $n_1=10$, $n_2=10$, and $MSE=16.778$.

From Appendix B.5, the critical value of t with 18 degrees of freedom is 2.101.

The 95% confidence interval for the difference in the means is given below:

$\begin{array}{c}CI=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t\sqrt{MSE\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ =\left(51.32-50.85\right)\pm 2.101\sqrt{\left(16.778\right)\left(\frac{1}{10}+\frac{1}{10}\right)}\\ =0.47\pm 2.101\left(1.83183\right)\\ =0.47\pm 3.85\\ =\left(-3.38,\text{ 4}\text{.32}\right)\end{array}$

Conclusion:

The confidence interval for the difference in the means of hard rock station and country/western station includes zero. Thus, the two means are not significantly different.

There is no difference in the means of hard rock station and country/western station.