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.
10.39 $x_1=10,s_1=2.1,n_1=15,x_2=12,s_2=2.3,n_2=15$
a. Two-tailed test, $\alpha =0.05$
b.$95\%$confidence interval

To conduct the two-tailed test for $x_1=10,s_1=2.1,n_1=15$and $x_2=12,s_2=2.3,n_2=15$then obtain the specified confidence interval.

Let the hypothesis test is two-tailed and the significance level is $5\%$.
Population $1:{\overline{x}}_1=10,s_1=2.1,n_1=15$
Population $2:{\overline{x}}_2=12,s_2=2.3,n_2=15$.
The most important goal is to perform a two-tailed hypothesis test.
The null and alternate hypotheses should be stated as:
Null hypotheses:
$H_0:{\mu }_1={\mu }_2$
Alternate hypotheses:$H_a:{\mu }_1\ne {\mu }_2$
Hypotheses is two-tailed.

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

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

$\begin{array}{r}\Rightarrow s_p=\sqrt{\frac{(15-1)(2.1{)}^2+(15-1\left)\right(2.3{)}^2}{15+15-2}}\end{array}$

$\Rightarrow s_p=\sqrt{\frac{14\left(4.41\right)+14\left(5.29\right)}{28}}$

$\begin{array}{r}\Rightarrow s_p=2.20\end{array}$

Then, the test statistic is

$t_0=\frac{{\overline{x}}_1-{\overline{x}}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{m_2}}}$

$\Rightarrow t_0=\frac{10-12}{2.20\sqrt{\frac{1}{15}+\frac{1}{15}}}$

$\Rightarrow t_0=-2.4896$

Identify the critical values as:

$\mathrm{df}=n_1+n_2-2$

$=15+15-2$

$=28$

$\Rightarrow \mathrm{df}=28$

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

Critical value:

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

$=\pm t_{0.025}$

$=\pm 2.048$

Comparison:$t_0=-2.4896$, indicating that the test statistic is in the rejection zone of the two-tailed hypotheses test.

As a result, null hypotheses are ruled out.

To obtain the specified confidence interval of $95\%$of the given data$x_1=10,s_1=2.1,n_1=15,$and $x_2=12,s_2=2.3,n_2=15$.

Let, the hypotheses test is two-tailed and significance level is $5\%$.
Population $1:{\overline{x}}_1=10,s_1=2.1,n_1=15$
Population $2:{\overline{x}}_2=12,s_2=2.3,n_2=15$.
The main goal is to calculate a $95\%$confidence interval for the difference between two population means (${\mu }_1$and ${\mu }_2$).
Null hypotheses: $H_0:{\mu }_1={\mu }_2$
Alternate hypotheses: $H_a:{\mu }_1\ne {\mu }_2$
Hypotheses is two-tailed.

Table$IV$may be used to find $t_{\alpha /2}$with $df=n_1+n_2-2$for a confidence level of $1-\alpha$.

Let, $\alpha =0.05$for a$95\%$ confidence level.
$df=n_1+n_2-2$

$=(15+15-2)$

$=28$

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

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

$=t_{0.025}$

$=2.048$

Determine the confidence interval's endpoints 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 $=-2.7478to-1.2522$

One can be $95\%$confident that the difference between the means of two population is somewhere between $-2.7478$to $-1.2522$.