Chapter 11.5, Problem 11.5E

(a)

To determine

To complete: the ANOVA table.

Answer to Problem 11.5E

The Analysis of Variance is given by,

Source	$\begin{array}{l}df\hfill \end{array}$	$SS$	$MSS$	$F$
Treatments	$3$	$339.8$	$113.27$	$16.98$
Error	$20$	$133.4$	$6.67$
Total	$23$	$473.2$

Explanation of Solution

Given:

$\text{Total }SS=473.2$ and $SST=339.8$ .

Calculation:

We know that,

Source	$\begin{array}{l}df\hfill \end{array}$	$SS$
Treatments	$3$	$339.8$
Error	$20$
Total	$23$	$473.2$

Now, calculate,

  $\begin{array}{l}SSE=TotalSS-SST\\ =473.2-339.8\\ =133.4\end{array}$

  $\begin{array}{l}MST=SST/df\\ =\frac{339.8}{3}\\ =113.2667\end{array}$

  $\begin{array}{l}MSE=SSE/df\\ =\frac{133.4}{20}\\ =6.67\end{array}$

  $\begin{array}{l}F=\frac{MST}{MSE}\\ =\frac{113.27}{6.67}\\ =16.98\end{array}$

Therefore, the Analysis of Variance is given by,

Source	$\begin{array}{l}df\hfill \end{array}$	$SS$	$MSS$	$F$
Treatments	$3$	$339.8$	$113.27$	$16.98$
Error	$20$	$133.4$	$6.67$
Total	$23$	$473.2$

Conclusion: Therefore, the Analysis of Variance is given by,

Source	$\begin{array}{l}df\hfill \end{array}$	$SS$	$MSS$	$F$
Treatments	$3$	$339.8$	$113.27$	$16.98$
Error	$20$	$133.4$	$6.67$
Total	$23$	$473.2$

(b)

To determine

To find: the number of degrees of freedom associated with the F statistic.

Answer to Problem 11.5E

The degrees of freedom are associated with the F statistic for testing $H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4$ are $F_{(0.05;3,20)}=3.10$ .

Explanation of Solution

Given:

$\text{Total }SS=473.2$ and $SST=339.8$ .

  $H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4$ .

Calculation:

To test the null hypothesis,

  $H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4$ .

Versus the alternative hypothesis,

  $H_a:$ At least one of the means is different from the others.

We have,

  $\begin{array}{l}MSE=6.67\\ MST=113.27\end{array}$

And we know that,

The test statistic is,

  $\begin{array}{l}F=\frac{MST}{MSE}\\ =\frac{113.27}{6.67}\\ =16.982009\\ =16.98\end{array}$

And this test statistic has an F distribution with $d{f}_1=(k-1)=3$ and $d{f}_2=(n-k)=20$ degrees of freedom i.e., $F_{(0.05;3,20)}=3.10$ .

Conclusion: Thus, the degrees of freedom are associated with the F statistic for testing $H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4$ are $F_{(0.05;3,20)}=3.10$ .

(c)

To determine

To give: the rejection region for the test in part b for $\alpha =.05$ .

Answer to Problem 11.5E

There is sufficient evidence to indicate that at least one of the four treatment means is different from at least one of the others.

Explanation of Solution

Given:

$\text{Total }SS=473.2$ and $SST=339.8$ .

Calculation:

Rejection region:

Using the critical value approach with $\alpha =.05$ , we can reject $H_0$ .

Since, $F_{cal}=16.98>F_{(0.05;3,20)}=3.10$ .

The observed value, $F=16.98$ , exceeds the critical value, so we reject $H_0$ .

Therefore, we conclude that, there is sufficient evidence to indicate that at least one of the four treatment means is different from at least one of the others.

Conclusion: There is sufficient evidence to indicate that at least one of the four treatment means is different from at least one of the others.

(d)

To determine

To find: whether the given data is an evidence to indicate differences among the population means.

Answer to Problem 11.5E

There is sufficient evidence to support the claim that there is a difference in the population means.

Explanation of Solution

Given:

$\text{Total }SS=473.2$ and $SST=339.8$ .

Calculation:

If the value of the test statistic is within the rejection regions, then the null hypothesis is rejected.

From part c, rejection region contain all values greater than or equal to $3.10$ .

From part a, $F=16.98$ .

  $16.98>3.10\Rightarrow \text{reject }{H}_0$ .

Yes,

There is sufficient evidence to support the claim that there is a difference in the population means.

Conclusion: Therefore, there is sufficient evidence to support the claim that there is a difference in the population means.

(e)

To determine

To estimate: the p -value for the test and explain whether the value confirms part d conclusion.

Answer to Problem 11.5E

The required value is $P-value=0.000$ .

This value confirms the conclusions in part d.

Explanation of Solution

Given:

$\text{Total }SS=473.2$ and $SST=339.8$ .

Calculation:

The P-value is the number or interval in the row title of the F-distribution table in the appendix containing h F-value in the row $d{f}_2=20$ and $d{f}_1=3$ .

  $P<0.005$ .

If the P-value is less than the significance level, then reject the null hypothesis.

  $P<0.05\Rightarrow {\text{Reject H}}_0$ ,

Hence we can conclude that the data provide sufficient evidence to conclude that at least one of the four treatment means is different from at least one of the others.

Conclusion: Therefore, required value is $P-value=0.000$ . This value confirms the conclusions in part d.