Question

In each of Exercises 10.39-10.44, we have provided summary statistics for independent simple random samples from two populations. In each case, use the pooled $t$-test and the pooled $t$-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.
10.40 ${\overline{x}}_1=10,s_1=4,n_1=15,\overline{x}2=12,s_2=5,n_2=15$
a. Two-tailed test, $\alpha =0.05$
b. $95\%$confidence level.

Short answer

(a) The given data do not provide sufficient evidence to reject null hypotheses at a significance level of $5\%$.

(b) The difference between the means of two populations is somewhere between $-2.7478$and $-1.2522$, with a $95\%$confidence interval.

Step-by-step solution

Part(a) Step 1: Given information

To conduct the two-tailed test for ${\overline{x}}_1=10,s_1=4,n_1=15,$and${\overline{x}}_2=12,s_2=5,n_2=15$then obtain the specified confidence interval.

Part (a) Step 2: Explanation

Let the hypothesis test is two-tailed and the significance level is $5\%$
Population $1:{\overline{x}}_1=10,s_1=4,n_1=15$
Population $2:{\overline{x}}_2=12,s_2=5,n_2=15$.
The main goal is to calculate a $95\%$confidence interval for the difference between two population mean ${\mu }_1$and ${\mu }_2$.
Null hypotheses:$H_0:{\mu }_1={\mu }_2$
Alternate hypotheses:$H_a:{\mu }_1\ne {\mu }_2$
Hypotheses is two-tailed.

Part(a) Step 3: Explanation

Determine the significance level:
Significance level is $5\%$. which is $\alpha =0.05$.
Calculate the value of test statistics as:
Pooled standard deviation,

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

$\Rightarrow s_p=\sqrt{\frac{(15-1)(4{)}^2+(15-1\left)\right(5{)}^2}{15+15-2}}$

$\Rightarrow s_p=\sqrt{\frac{14\left(16\right)+14\left(25\right)}{28}}$

$\begin{array}{r}\Rightarrow s_p=4.5277\end{array}$

Then, the test statistic as:

$t_0=\frac{{\overline{x}}_1-{\overline{x}}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$\Rightarrow t_0=\frac{10-12}{4.5277\sqrt{\frac{1}{15}+\frac{1}{15}}}$$\Rightarrow t_0=\frac{10-12}{4.5277\sqrt{\frac{1}{15}+\frac{1}{15}}}$

$\Rightarrow t_0=-1.2097$

Part (a) Step 4: Explanation

Identify the critical values as:

$df=n_1+n_2-2$

$=15+15-2$

$=28$

$\Rightarrow df=28$

When $df=28$, use table $IV$for important values:

$\pm t_{a/2}=\pm t_{0.05/2}$

$=\pm t_{0.025}$

$=\pm 2.048$is the critical value.

Then,$t_0=-1.2097$, in other words the test statistic does not fall into the two-tailed hypotheses test rejection zone.
As a result, null hypotheses are not ruled out.

Part (b) Step 1: Given information

To obtain the specified confidence interval for $95\%$of the given data.

Part (b) Step 2: Explanation

Let, Population $1:{\overline{x}}_1=10,s_1=4,n_1=15$

And population $2:{\overline{x}}_2=12,s_2=5,n_2=15$

The main goal is to determine $95\%$confidence interval for the difference between two population mean ${\mu }_1$and${\mu }_2$.
Null hypotheses is $H_0:{\mu }_1={\mu }_2$
Alternate hypotheses is $H_a:{\mu }_1\ne {\mu }_2$
Hypotheses is two-tailed.
Table $IV$may be used to determine $t_{\alpha /2}$with a confidence level of$1-\alpha$using$df=n_1+n_2-2$.
For $95\%$confidence level,$\alpha =0.05$.
$df=n_1+n_2-2$

$=(15+15-2)$

$=28$

When $df=28$, use table $IV$for important values.

Critical value is $t_{\alpha /2}=t_{0.05/2}$

$=t_{0.025}$

$=2.048$

Part (b) Step 3: Explanation

Determine the endpoints of the confidence interval as:
$\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}\times \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

Confidence interval $=(10-12)\pm 2.048\sqrt{\frac{1}{15}+\frac{1}{15}}$

Confidence interval $=-2\pm 0.7478$

Confidence interval $=-2.7478$to $-1.2522$