91Ó°ÊÓ

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

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

The significance level is $5\%$.

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

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

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" $H_0={\mathrm{μ}}_1â‰\yen {\mathrm{μ}}_2$

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

Hypotheses is left-tailed.

Determine the level of relevance

The significance level is$5\%$set at or $\mathrm{Î\pm }=0.05$.

Compute the value of test statistics

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

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

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

$⇒s_p=4.7718$

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

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

$⇒t_0=-4.058$

Identify critical values.

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

$⇒\text{df}=18$

By table IV we get

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

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