Political Prisoners. In Exercise $8.73$, you found a $95\%$ confidence interval of $18.8$months to $48.0$months for the mean duration of imprisonment $\mu$of all East German political prisoners with chronic PTSD.

a. Determine the margin of error, $E$

b. Explain the meaning of $E$ in this context in terms of the accuracy of the estimate.

c. Find the sample size required to have a margin of error of $12$ months and a $99\%$ confidence level. (Recall that $\sigma =42$ months.)

d. Find a $99\%$ confidence interval for the mean duration of imprisoned $36.2$ months.

The population mean duration $\mu$ has a $95\%$ confidence interval of $18.8$ months to $48.0$ months.

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

$\therefore$ The length of the confidence interval

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

$\therefore$ 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 $\mu$ indicates that we are $95\%$ certain that the error in estimating the population mean $\mu$ by the sample mean $\overline{x}$ 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 $\mu$ we may anticipate around $95\%$ of the sample means to depart from $\mu$ by at most $14.6$ months.

The sample size necessary for $100(1-\alpha )\%$is $n$$C.I.$for $\mu$with a defined margin of error$E$ is calculated as follows:

$n={\left(\frac{Z_{a/2}\times \sigma }{E}\right)}^2$

Where $n$is rounded up to the nearest integer

Here margin of error $E=12$for months

Population S.D $\sigma =42$months

Confidence level $=99\%$

$$\begin{gathered} =100\times 0.99\% \\ \therefore 1-\alpha =0.99 \\ \Rightarrow \alpha =0.01 \end{gathered}$$

$\Rightarrow \frac{\alpha }{2}=0.005$

$\therefore Z_{\frac{a}{2}}=Z_{0,005}=2.57583$

$$\begin{gathered} \therefore n={\left\{\frac{Z_{0.005}\times \sigma }{E}\right\}}^2 \\ ={\left\{\frac{2.57583\times 42}{12}\right\}}^2 \end{gathered}$$

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

$\therefore$ The required sample size is $82$

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

Given that sample mean $\overline{x}=36.2$

Sample mean $n=82$

$99\%$C.I of population mean $\mu$is given by,

$\left(\overline{x}-Z_{0.005}\times \frac{\sigma }{\sqrt{n}},\overline{x}+Z_{0.005}\times \frac{\sigma }{\sqrt{n}}\right)=(\overline{x}-E,\overline{x}+E)$

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

$$\begin{gathered} \therefore E=Z_{0.005}\frac{\sigma }{\sqrt{n}} \\ =\frac{2.57583\times 42}{\sqrt{82}} \\ =11.95 \end{gathered}$$

$\therefore 99\%$C.l of the population mean $\mu$is

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

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