Question

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.

Short answer

We can be $98\%$ sure that a male's mean absolute pointing inaccuracy is between $45.24$and $8.84$ degrees less than a female's.

Step-by-step solution

Given Information

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 $5\%$.

Explanation

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,

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

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

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

$s_p=43.786$

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 $=-45.24to8.84$.