Question

Serving Time. According to the Bureau of Crime Statistics and Research of Australia, as reported on Lawlink, the mean length of imprisonment for motor-vehicle theft offenders in Australia is 16.7 months. One hundred randomly selected motor-vehicle-theft offenders in Sydney, Australia, had a mean length of imprisonment of 17.8 months. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle theft offenders in Sydney differs from the national mean in Australia? Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in Sydney is 6.0 months.

Short answer

Ans: Since $z$does not fall in the rejection region.

Thus, role="math" localid="1652215192012" $H_0$is not rejected at a role="math" localid="1652215197979" $5\%$level of significance of the test value statistic.

At the $1\%$significance level, the data provided is sufficient evidence for concluding that the mean post-work heart rate for easting workers exceeds the normal resting heart rate of role="math" localid="1652215186430" $72bpm$.

Step-by-step solution

Step 1. Given information.

given,

Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in Sydney is 6.0 months.

Step 2. Let's assume the mean post-work heart rate for casting workers to be μ.

Given that,

Population standard deviation is,

$\sigma =6.8$years

Now, test the hypotheses,

$\begin{array}{r}{H}_0:\mu =72\text{b.p.m}\\ H_a:\mu >72\text{b.p.m}\end{array}$

Perform the test at a $5\%$level of significance i.e., $\alpha =0.05$

The sample size was $n=29$

The sample mean is,

$\overline{x}=78.3bpm$

Step 3. Now,

$\begin{array}{r}\text{Test statistic,}z=\frac{\overline{x}-{\mu }_0}{\frac{\sigma }{\sqrt{n}}}\\ =\frac{78.3-72}{\frac{11.2}{\sqrt{29}}}\\ =3.03\end{array}$

Since the test is left the tailed test with $\alpha =0.05$,

and the critical value is,

localid="1651233032648" $z_{\alpha }=z_{0.05}=1.645$

Here, the region is,

$z>z_{0.05}$

i.e., $z>1.645$

Step 4. Then,

Step 5. Here,

$z=3.03>z_{0.05}=1.645$

Since $z$ does not fall in the rejection region.

Thus, $H_0$is not rejected at a $5\%$level of significance of the test value statistic.

At the $1\%$significance level, the data provided is sufficient evidence for concluding that the mean post-work heart rate for easting workers exceeds the normal resting heart rate of $72bpm$.