Gender and Direction. Refer to Exercise 10.46 and obtain a 98% confidence interval for the difference between the mean absolute pointing errors for males and females.

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

${\overline{x}}_1=37.6$

$s_1=38.5$

${\overline{x}}_2=55.8$

$s_2=48.3$

The significance level is $5\%$.

Let's consider the first population:

Male, ${\overline{x}}_1=37.6$

$s_1=38.5$and

$n_1=30$

Let's consider the second population:

Female, ${\overline{x}}_2=55.8$

$s_2=48.3$and

$n_2=30$

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

The null and alternate hypotheses should be stated.

Null hypotheses: $H_0={\mu }_1\ge {\mu }_2$

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

Hypotheses have a left-tailed structure.

Use Table IV to calculate $t_{\alpha /2}$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, $$\begin{gathered} t_{\alpha /2}=t_{0.02/2} \\ =t_{0.01} \\ =2.392 \end{gathered}$$

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 $=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}·s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

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

Confidence interval$=-18.2\pm 27.043$

Confidence interval $=-45.24to8.84$.