91Ó°ÊÓ

Given in the question that, the sample data is supplied from two male and female groups.

${\stackrel{¯}{x}}_1=37.6$

$s_1=38.5$

${\stackrel{¯}{x}}_2=55.8$

$s_2=48.3$

The significance level is $5\%$.

Let's consider the first population:

Male, ${\stackrel{¯}{x}}_1=37.6$

$s_1=38.5$and

$n_1=30$

Let's consider the second population:

Female, ${\stackrel{¯}{x}}_2=55.8$

$s_2=48.3$and

$n_2=30$

The main goal is to calculate a $98\%$confidence interval for the difference between the two population means, ${\mathrm{μ}}_1$and ${\mathrm{μ}}_2$

The null and alternate hypotheses should be stated.

Null hypotheses: $H_0={\mathrm{μ}}_1â‰\yen {\mathrm{μ}}_2$

Alternate hypotheses: $H_a={\mathrm{μ}}_1<{\mathrm{μ}}_2$

Hypotheses have a left-tailed structure.

Use Table IV to calculate $t_{\mathrm{Î\pm }/2}$for a confidence level of $1-a$with $\mathrm{df}=n_1+n_2-2$

for a $98\%$confidence level.

$$\begin{gathered} \text{df}=n_1+n_2-2 \\ =(30+30-2) \\ =58. \end{gathered}$$

When $\mathrm{df}=58$, using table IV for critical values,

Critical value, $$\begin{gathered} t_{\mathrm{Î\pm }/2}=t_{0.02/2} \\ =t_{0.01} \\ =2.392 \end{gathered}$$

Find the confidence interval's endpoints.

Pooled standard deviation,

$s_p=\sqrt{\frac{(30-1)(38.5{)}^2+(30-1\left)\right(48.3{)}^2}{30+30-2}}$

$s_p=\sqrt{\frac{29(1482.25)+29(2352.25)}{58}}$

$s_p=43.786$

Confidence interval $=\left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)Â\pm t_{\mathrm{Î\pm }/2}·s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

Confidence interval $=(37.6-55.8)Â\pm 2.392×43.786\sqrt{\frac{1}{30}+\frac{1}{30}}$

Confidence interval$=-18.2Â\pm 27.043$

Confidence interval $=-45.24to8.84$.