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 of-fenders 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 is6months.

Consider the given question,

Population standard deviation is 6months.

Sample size, n is 100.

Sample mean is$\overline{x}$is17.8months.

Assume $\mu$to be the mean length of imprisonment for motor-vehicle-theft offenders.

We have to test the hypotheses,

$$\begin{gathered} H_0:\mu =16.7vs \\ {H}_a:\mu \ne 16.7 \end{gathered}$$

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

Test statistics $z=\frac{\overline{x}-{\mu }_0}{{\displaystyle \frac{\sigma }{\sqrt{n}}}}$

$$\begin{gathered} z=\frac{17.8-16.7}{{\displaystyle \frac{6}{\sqrt{100}}}} \\ =1.83 \end{gathered}$$

As the test is two tailed test with $\sigma$ is 0.05, the critical values are $\pm z_{0.025}=\pm 1.96$.

Here, the rejection region is $z<-z_{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}orz>z_{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$, i.e., $z<-1.96orz>1.96$

Here, $-1.96<z<1.96$.

Therefore, we do not reject $H_0$ at 5% level of significance as the value of the test statistic, z does not falls in the rejection region.

Hence, at 5% significance level, the data do not provide sufficient evidence to conclude that the mean length of imprisonment of motor-vehicle-theft offenders differs from the national mean.