91Ó°ÊÓ

Let $\mathrm{μ}$be the average sentence duration for motor vehicle theft offenses.

For a reason,

$\mathrm{σ}=6$months populations standard deviation

The theories must be tested.

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

$H_a:\mathrm{μ}≠16.7$months

We'll run the test at a significance threshold of $5\%$, or $\mathrm{Î\pm }=0.05$.

$n=100$is the sample size.

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

Statistical test,$z=\frac{\stackrel{¯}{x}-{\mathrm{μ}}_0}{\frac{\mathrm{σ}}{\sqrt{n}}}$

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

$=1.83$

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

$=Â\pm 1.96$

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

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

Perhaps $z<-1.96$or $z≯1.96$(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.

$∴$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.