91Ó°ÊÓ

The population mean duration $\mathrm{μ}$ has a $95\%$ confidence interval of $18.8$ months to $48.0$ months.

The formula used:$n={\left(\frac{Z_{a/2}×\mathrm{σ}}{E}\right)}^2$

$∴$ The length of the confidence interval

$∴$ Margin of error $E=\frac{1}{2}×$ length of the C.I

$∴$ Margin of error at$95\%$ confidence level is $14.6$ months

The margin of error $E=14.6$ months for a $95\%$confidence interval of the population mean $\mathrm{μ}$ indicates that we are $95\%$ certain that the error in estimating the population mean $\mathrm{μ}$ by the sample mean $\stackrel{¯}{x}$ is at most $14.6$ months.

In other words, if we take a large number of simple random samples of size $n$ from a population with a mean of $\mathrm{μ}$ we may anticipate around $95\%$ of the sample means to depart from $\mathrm{μ}$ by at most $14.6$ months.

The sample size necessary for $100(1-\mathrm{Î\pm })\%$is $n$$C.I.$for $\mathrm{μ}$with a defined margin of error$E$ is calculated as follows:

$n={\left(\frac{Z_{a/2}×\mathrm{σ}}{E}\right)}^2$

Where $n$is rounded up to the nearest integer

Here margin of error $E=12$for months

Population S.D $\mathrm{σ}=42$months

Confidence level $=99\%$

$$\begin{gathered} =100×0.99\% \\ ∴1-\mathrm{Î\pm }=0.99 \\ ⇒\mathrm{Î\pm }=0.01 \end{gathered}$$

$⇒\frac{\mathrm{Î\pm }}{2}=0.005$

$∴Z_{\frac{a}{2}}=Z_{0,005}=2.57583$

$$\begin{gathered} ∴n={\left\{\frac{Z_{0.005}×\mathrm{σ}}{E}\right\}}^2 \\ ={\left\{\frac{2.57583×42}{12}\right\}}^2 \end{gathered}$$

$$\begin{gathered} =81.27 \\ =82 \end{gathered}$$

$∴$ The required sample size is $82$

We have to obtain a $99\%\mathrm{Cl}$of population mean $\mathrm{μ}$

Given that sample mean $\stackrel{¯}{x}=36.2$

Sample mean $n=82$

$99\%$C.I of population mean $\mathrm{μ}$is given by,

$\left(\stackrel{¯}{x}-Z_{0.005}×\frac{\mathrm{σ}}{\sqrt{n}},\stackrel{¯}{x}+Z_{0.005}×\frac{\mathrm{σ}}{\sqrt{n}}\right)=(\stackrel{¯}{x}-E,\stackrel{¯}{x}+E)$

Where $E=Z_{0.005}\frac{\mathrm{σ}}{\sqrt{n}}$

$$\begin{gathered} ∴E=Z_{0.005}\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{2.57583×42}{\sqrt{82}} \\ =11.95 \end{gathered}$$

$∴99\%$C.l of the population mean $\mathrm{μ}$is

$$\begin{gathered} (\stackrel{¯}{x}-E,\stackrel{¯}{x}+E) \\ =(36.2-11.95,36.2+11.95) \\ =(24.25,48.15) \end{gathered}$$

i.e., we are $99\%$certain that the average length of incarceration$\mathrm{μ}$ is between $24.25$ and $48.15$ months.