Doing Time. Refer to Exercise 10.45 and obtain a 90% confidence interval for the difference between the mean times served by prisoners in the fraud and firearms offense categories.

Data from two populations of fraud and firearms offenses are provided as samples.

${\overline{x}}_1=10.12$

$s_1=4.90$

${\overline{x}}_2=18.78$

and

$s_1=4.64$.

The significance level is$5\%$.

Population 1: Fraud offenses,${\overline{x}}_1=10.12$,

$s_1=4.90$and

$n_1=10$

Population 2: Firearms offenses, ${\overline{x}}_2=18.78$,

$s_1=4.64$and

$n_1=10$.

The primary goal is to determine the 90% confidence interval for the difference between two population means ${\mu }_1$and ${\mu }_2$.

Specify the null and alternative hypotheses.

Hypotheses with no support: $H_0={\mu }_1\ge {\mu }_2$

Alternative hypotheses: $H_0={\mu }_1<{\mu }_2$

Hypotheses have a left tail.

Table IV may be used to find $t_{\alpha /2}$with $df=n\_\left\{1\right\}+n\_\left\{2\right\}-2$with a confidence level of $1-a$.

$a=0.10$for a $90\%$confidence level.

Let's find the degree of freedom

localid="1651398590402" $$\begin{gathered} \text{df}=n_1+n_2-2 \\ =(10+10-2) \\ =18. \end{gathered}$$

when $\mathrm{df}=18$use table IV for critical values.

Critical value,

Find the confidence interval's endpoints.

Pooled standard deviation,

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

$\Rightarrow s_p=\sqrt{\frac{(10-1)(4{)}^2+(15-1\left)\right(5{)}^2}{10+15-2}}$

$\Rightarrow s_p=\sqrt{\frac{9\left(16\right)+14\left(25\right)}{23}}$

$\Rightarrow s_p=4.6345$

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 $=(10.12-18.78)\pm 1.734\times 4.6345\sqrt{\frac{1}{10}+\frac{1}{10}}$

Confidence interval$=-8.660\pm 3.7$

Confidence interval localid="1651146771778" $=-12.36to-4.96$