Question

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.

Short answer

We can be $90\%$certain that the average time served by inmates in the fraud category is between $4.96$ and $12.36$ months shorter than that of inmates in the firearm category.

Step-by-step solution

Given Information

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$5\%$.

Explanation

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,

localid="1651398653000" $$\begin{gathered} t_{\alpha /2}=t_{0.10/2} \\ =t_{0.05} \\ =1.734 \end{gathered}$$

Find the confidence interval's endpoints.

Pooled standard deviation,

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

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

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

$\Rightarrow s_p=4.6345$

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" $=-12.36to-4.96$