Chapter 11, Problem 8E

For Exercises 7 and 8: (a) state the decision rule, (b) compute the pooled estimate of the population variance, (c) compute the test statistic, (d) state your decision about the null hypothesis, and (e) estimate the p-value.

8. The null and alternate hypotheses are:

$\begin{array}{c}{H}_0:{\text{μ}}_{\text{1}}={\text{μ}}_{\text{2}}\\ H_1:{\text{μ}}_{\text{1}}\ne {\text{μ}}_{\text{2}}\end{array}$

A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .10 significance level, is there a difference in the population means?

To determine

Determine the decision rule.

Explanation of Solution

Calculation:

Degrees of freedom:

The degrees of freedom is as follows:

$\begin{array}{c}df=n_1+n_2-2\\ =17+15-2\\ =30\end{array}$

Step-by-step procedure to obtain the critical value using MINITAB software:

1. 1. Choose Graph > Probability Distribution Plot choose View Probability > OK.
2. 2. From Distribution, choose ‘t’ distribution.
3. 3. In Degrees of freedom, enter 30.
4. 4. Click the Shaded Area tab.
5. 5. Choose P Value and Two Tail for the region of the curve to shade.
6. 6. Enter the probability value as 0.05.
7. 7. Click OK.

Output obtained using MINITAB software is given below:

From the MINITAB output, the critical value is $\pm 1.697$

The decision rule is,

If $t\text{ <}-1.697$, then reject the null hypothesis $H_0$.

If $t\text{>}1.697$, then reject the null hypothesis $H_0$.

To determine

Find the value of the pooled estimate of the population variance.

Answer to Problem 8E

The pooled estimate of the population variance is 187.20.

Explanation of Solution

Calculation:

Pooled estimate:

The pooled estimate of the population variance is as follows:

$=\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}$

Substitute $s_1$ as 12, $s_2$ as 15, $n_1$ as 15, and $n_2$ as 17 in the above formula as follows:

$\begin{array}{c}{s}_p^2=\frac{\left(15-1\right){\left(12\right)}^2+\left(17-1\right){\left(15\right)}^2}{15+17-2}\\ =\frac{\left(14\right)\left(144\right)+\left(16\right)\left(225\right)}{30}\\ =\frac{5,616}{30}\\ =187.20\end{array}$

Thus, the pooled estimate of the population variance is 187.20.

To determine

Find the value of test statistic.

Answer to Problem 8E

The value of the test statistic is 1.651.

Explanation of Solution

Test statistic:

The test statistic for the hypothesis test of ${\mu }_1-{\mu }_2$ when ${\sigma }_1$ and ${\sigma }_2$ are unknown is as follows:

$t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

Substitute ${\overline{x}}_1$ as 350, ${\overline{x}}_2$ as 342, $s_p^2$ as 187.20, $n_1$ as 15, and $n_2$ as 17 in the above formula as follows:

$\begin{array}{c}t=\frac{350-342}{\sqrt{187.20\left(\frac{1}{15}+\frac{1}{17}\right)}}\\ =\frac{8}{\sqrt{23.4918}}\\ =\frac{8}{4.8468}\\ =1.651\end{array}$

Thus, the test statistic is 1.651.

To determine

Determine the decision regarding $H_0$

Answer to Problem 8E

The decision is fail to reject the null hypothesis.

Explanation of Solution

The critical value is 1.697 and the value of test statistic is 1.651.

The value of test statistic is less than the critical value.

That is, $\text{Test statistic}\left(1.651\right)<\text{Critical}\text{value}\left(1.697\right)$.

From the decision rule, fail to reject the null hypothesis.

To determine

Find the p-value.

Answer to Problem 8E

The p-value is 0.109.

Explanation of Solution

The p-value is obtained below:

Step-by-step procedure to obtain the p-value using MINITAB software:

1. 1. Choose Graph > Probability Distribution Plot choose View Probability > OK.
2. 2. From Distribution, choose ‘t’ distribution.
3. 3. In Degrees of freedom, enter 30.
4. 4. Click the Shaded Area tab.
5. 5. Choose X Value and Two Tail for the region of the curve to shade.
6. 6. Enter the X value as 1.651.
7. 7. Click OK.

Output obtained using MINITAB software is given below:

From the MINITAB output, the p-value for one side is 0.05459.

$\begin{array}{c}p\text{-value}=2\left(0.05459\right)\\ =0.109\end{array}$

Thus, the p-value is 0.109.