91Ó°ÊÓ

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

${\stackrel{¯}{x}}_1=10.12$

$s_1=4.90$

${\stackrel{¯}{x}}_2=18.78$

and

$s_1=4.64$.

The significance level is$5\%$.

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

$s_1=4.90$and

$n_1=10$

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

$s_1=4.64$and

$n_1=10$.

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

Specify the null and alternative hypotheses.

Hypotheses with no support: $H_0={\mathrm{μ}}_1â‰\yen {\mathrm{μ}}_2$

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

Hypotheses have a left tail.

Table IV may be used to find $t_{\mathrm{Î\pm }/2}$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,

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

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

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

$⇒s_p=4.6345$

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

Confidence interval$=-8.660Â\pm 3.7$

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