Question

The Federal Bureau of Prisons publishes data in Prison Statistics on the times served by prisoners released from federal institutions for the first time. Independent random samples of released prisoners in the fraud and firearms offense categories yielded the following information on time served, in months.

At the $5\%$ significance level, do the data provide sufficient evidence to conclude that the meantime served for fraud is less than that for firearms offenses? (Note: ${\overline{x}}_1=10.12,s_1=4.90,{\overline{x}}_2=18.78$, and $s_2=4.64$.)

Short answer

The presented statistics provide adequate evidence to infer that, on average, the meantime served by fraud offenders is less than that spent by firearms offenders at a significance level of $5\%$.

Step-by-step solution

Given Information

A sample of data from two populations of fraud and firearm offenses is provided.

${\overline{x}}_1=10.12,s_1=4.90,{\overline{x}}_2=18.78,\text{and}{s}_1=4.64$

The significance level is $5\%$.

Explanation

Population $1$: Fraud offences, ${\overline{x}}_1=10.12,s_1=4.90,\text{and}{n}_1=10$.

Population localid="1651157783900" $2$: Firearms offences, ${\overline{x}}_2=18.78,s_1=4.64,\text{and}{n}_2=10$.

The main goal is to come to the conclusion that the average time spent for fraud is less than that for firearms offences.

Define null and alternate hypotheses.

Null hypotheses: localid="1651157858528" $H_0={\mu }_1\ge {\mu }_2$

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

Hypotheses is left-tailed.

Determine the level of relevance

The significance level is$5\%$set at or $\alpha =0.05$.

Calculation

Compute the value of test statistics

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

$\Rightarrow s_p=\sqrt{\frac{(10-1)(4.90{)}^2+(10-1\left)\right(4.64{)}^2}{10+10-2}}$

$\Rightarrow s_p=\sqrt{\frac{9(24.01)+9(21.5296)}{18}}$

$\Rightarrow s_p=4.7718$

Test statistic,$t_0=\frac{{\overline{x}}_1-{\overline{x}}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$\Rightarrow t_0=\frac{10.12-18.78}{4.7718\sqrt{\frac{1}{10}+\frac{1}{10}}}$

$\Rightarrow t_0=-4.058$

Identify critical values.

Here, localid="1651300680058" role="math" $$\begin{gathered} df=n_1+n_2-2 \\ =10+10-2 \\ =18 \end{gathered}$$

$\Rightarrow \text{df}=18$

By table IV we get

localid="1651300689725" role="math" $$\begin{gathered} \text{Critical value,}{t}_{\alpha }=t_{0.05} \\ =-1.734 \end{gathered}$$

Since $t_0<t_{\alpha }$, i.e. the test statistic falls in the left-tailed hypotheses test rejection zone. As a result, null hypotheses are ruled out.