91Ó°ÊÓ

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

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

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

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

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

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

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

Then, the test statistic as:

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

$⇒t_0=\frac{10−12}{4.5277\sqrt{\frac{1}{15}+\frac{1}{15}}}$$⇒t_0=\frac{10−12}{4.5277\sqrt{\frac{1}{15}+\frac{1}{15}}}$

$⇒t_0=−1.2097$

Identify the critical values as:

$df=n_1+n_2-2$

$=15+15-2$

$=28$

$⇒df=28$

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

$Â\pm t_{a/2}=Â\pm t_{0.05/2}$

$=Â\pm t_{0.025}$

$=Â\pm 2.048$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.

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

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

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

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

$=(15+15-2)$

$=28$

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

Critical value is $t_{\mathrm{Î\pm }/2}=t_{0.05/2}$

$=t_{0.025}$

$=2.048$

Determine the endpoints of the confidence interval as:
$\left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)Â\pm t_{\mathrm{Î\pm }/2}×\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

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

Confidence interval $=-2Â\pm 0.7478$

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