Chapter 12, Problem 11E

a.

To determine

Obtain the null and the alternative hypotheses.

Explanation of Solution

The null and alternative hypotheses are given below:

Null Hypothesis

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

That is, the mean of all treatments are equal.

Alternative Hypothesis

$H_1:$ At least one treatment mean is different from others.

b.

To determine

Give the decision rule.

Explanation of Solution

The treatment and error degrees of freedom are given below:

Treatment degrees of freedom:

$\begin{array}{c}\text{Degrees}\text{of}\text{freedom = }k-1\\ =3-1\\ =2\end{array}$

Error degrees of freedom:

$\begin{array}{c}\text{Degrees}\text{of}\text{freedom = }n-k\\ =12-3\\ =9\end{array}$

Here, the level of significance $\left(\alpha \right)$ is 0.05.

Step-by-step procedure to obtain the critical F value using Excel-MegaStat:

• In EXCEL, Select Add-Ins > MegaStat > Probability.
• Choose probability> F-distribution> calculate F given probability.
• Enter P as 0.05.
• Enter df1 as 2.
• Enter df2 as 9.
• Click Ok.

Output using the Excel-MegaStat software is given below:

From the output, the critical F value is 4.26.

Decision rule:

If $F>4.26$, then reject the null hypothesis.

Therefore, the decision rule is to reject $H_0$ if the computed value of F exceeds 4.26.

c.

To determine

Find the values of SST, SSE and SS total.

Answer to Problem 11E

The value of SST is 107.20.

The value of SSE is 9.47.

The value of SS total is 116.67.

Explanation of Solution

Here, the level of significance $\left(\alpha \right)$ is 0.05.

Step-by-step procedure to obtain the sum of square total, sum of square treatment and sum of square error using Excel-MegaStat:

• Choose MegStat > Analysis of Variance > One-Factor ANOVA.
• Select the column of Treatment 1, Treatment 2 and Treatment 3 in Input range.
• Click OK.

Output using the Excel-MegaStat software is given below:

From the output, the values of SST is 107.20, SSE is 9.47 and SS total is 116.67.

d.

To determine

Find an ANOVA table.

Explanation of Solution

From the output in Part (c), the ANOVA table is obtained.

The ANOVA table is given below:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatments	107.2	2	53.6	50.96
Error	9.47	9	1.05
Total	116.67	11

e.

To determine

Find the decision regarding the null hypothesis.

Explanation of Solution

Conclusion:

The F value is 50.96 and the F critical value is 4.26.

Here, F value is greater than F critical value. That is, 50.96 > 4.26.

Using rejection rule, reject the null hypothesis.

Therefore, there is sufficient evidence that at least one mean of all treatment is differ from others.

f.

To determine

Check whether there is significant difference between treatment 1 and treatment 2, if null hypothesis is rejected by using the 95% level of confidence.

Explanation of Solution

A 95% confidence interval is as follows:

$\left(\begin{array}{l}\text{Confidence interval}\text{for}\\ \text{the}\text{difference}\text{in}\text{treatment}\\ \text{means}\end{array}\right)=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t\sqrt{\text{MSE}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}$

Where,

${\overline{x}}_1$ is the mean of treatment 1, ${\overline{x}}_2$ is the mean of treatment 2, $n_1$ is the number of observations in the treatment 1, $n_2$ is the number of observations in the treatment 2, t is the critical value with $\left(n-k=9\right)$ degrees of freedom and MSE is mean square term obtained from the ANOVA table.

From the output in Part (c), the mean of treatment 1 is 9.7, mean of treatment 2 is 2.2, and MSE is 1.052.

Step-by-step procedure to obtain t-critical value using Excel-MegaStat:

• In EXCEL, Select Add-Ins > MegaStat > Probability > t-Distribution.
• Select calculate t given P.
• Enter probability as 0.05.
• Enter df as 9.
• Under Shading, choose two-tail.
• Click Ok.

Output using the Excel-MegaStat software is given below:

From the output, the t is $\pm 2.26$. Now the 95% confidence interval is calculated as follows:

$\begin{array}{c}\left(\begin{array}{l}\text{Confidence interval}\text{for}\\ \text{the}\text{difference}\text{in}\text{treatment}\\ \text{means}\end{array}\right)=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t\sqrt{\text{MSE}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ =\left(9.7-2.2\right)\pm 2.26\sqrt{1.052\left(\frac{1}{3}+\frac{1}{5}\right)}\\ =7.5\pm \left(2.26\times 0.75\right)\\ =7.5\pm 1.70\end{array}$

                                                   $\begin{array}{c}=\left[7.5-1.70,7.5+1.70\right]\\ =\left[5.8,9.2\right]\end{array}$

Therefore, a 95% confidence interval for that difference is 5.8 and 9.2. Here, 0 does not include in the confidence interval.

It means that there is a significant difference between the means of treatment 1 and treatment 2 because the endpoints have same sign or does not include zero.