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 localid="1653225892125" $17.8$months. At the localid="1653225896253" $5\%$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 localid="1653225901113" $6.0$months.

Let $\mu$be the average sentence duration for motor vehicle theft offenses.

For a reason,

$\sigma =6$months populations standard deviation

The theories must be tested.

$H_0:\mu =16.7$months $\mathrm{Vs}$

$H_a:\mu \ne 16.7$months

We'll run the test at a significance threshold of $5\%$, or $\alpha =0.05$.

$n=100$is the sample size.

$\overline{x}=17.8$months, arithmetic mean.

Statistical test,$z=\frac{\overline{x}-{\mu }_0}{\frac{\sigma }{\sqrt{n}}}$

$=\frac{17.8-16.7}{\frac{6}{\sqrt{100}}}$

$=1.83$

The crucial values are $\pm z_{\alpha /2}=\pm z_{0.025}$because the test is a two-tailed test with $\alpha =0.05$.

$=\pm 1.96$

A regression line is $z<-z_{\alpha /2}$or $z>z_{\alpha /2}$in this case.

i.e., $z<-1.96$ or $z>1.96$

Perhaps $z<-1.96$or $z\ngtr 1.96$(Actually in fact $-1.96<z<1.96$)

As that of the amount of a statistical test,$z$, doesn't really fall as in rejection zone, we need not reject $H_0$at the $5\%$significance level.

$\therefore$The records may not provide credible evidence to infer that its average duration of sentence for motor vehicle theft offenses varies first from nation mean in $.$just at $5\%$level of significance.