Q 8.22RP. - Elementary Statistics

Question

Prison Sentences. Refer to Problem $21$

a. Find the margin of error, $E$

b. Explain the meaning of $E$as far as the accuracy of the estimate is concerned.

c. Determine the sample size required to have a margin of error of \(0.5\) year and a $90\%$ confidence level.

d. Find a $90\%$ confidence interval for $\mu$ if a sample of the size determined in part (c) yields a mean of $10.1$ years.

Step-by-Step Solution

Answer

Part (a) The margin of error is $1.060$ years.

Part (b) For $90\%$ confidence level, the population means $\left(\mu \right)$ can be estimated to be within $1.060$ years.

Part (c) The sample size required for a $0.5$ year margin of error and a $90\%$ confidence level is $3,203$

Part (d) The $90\%$ confidence interval for $\mu$ is $(9.60,10.60)$

Part (a) Step 1: Given information

$712$ federally sentenced adult male convicts received an average sentence of $9.15$ years. Assume that the median sentence term for federally sentenced adult male offenders is $17.2$ years.

Part (a) Step 2: Concept

The formula used: The margin of error $E=z_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)$ and $\overline{x}\pm z_{\frac{a}{2}}\left(\frac{\sigma }{\sqrt{n}}\right)$

Part (a) Step 3: Calculation

Find the margin of error, $E$

Consider $\overline{x}=9.15, \sigma =17.2$ and $n=712$, and the confidence level is $90\%$

The needed value of $z_{\frac{\alpha }{2}}$ with a $90\%$ confidence level is $1.645$, according to "Table II Areas under the standard normal curve."

The margin of error is,

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right) \\ =1.645\left(\frac{17.2}{\sqrt{712}}\right) \\ =1.645(0.6446) \\ =1.060 \end{gathered}$$

Thus, the margin of error is $1.060$ years.

Part (b) Step 1: Explanation

Meaning of $E$ :

The population mean $\left(\mu \right)$ can be predicted to be within $1.060$ years at a $90\%$ confidence level.

Part (c) Step 1: Calculation

Obtain the sample size required to have a $0.5$ year margin of error and a $90\%$ confidence level.

The sample size formula is as follows: 

$$\begin{gathered} n={\left(\frac{z_a}{2}\sigma \right)}^2 \\ ={\left(\frac{(1.645)(17.2)}{0.5}\right)}^2 \\ =3,202.2 \\ \approx 3,203 \end{gathered}$$

As a result, the sample size required for a $0.5$ year margin of error and a $90\%$ confidence level is $3,203$

Part (d) Step 1: Calculation

If a sample of the size determined in component (c) returns a mean of $10.1$ years, find a $90\%$ confidence interval for the population mean.

The $90\%$ confidence interval is,

$$\begin{gathered} \overline{x}\pm z_{\frac{a}{2}}\left(\frac{\sigma }{\sqrt{n}}\right)=10.1\pm 1.645\left(\frac{17.2}{\sqrt{3,203}}\right) \\ =10.1\pm 0.4999 \\ =(10.1-0.4999,10.1+0.4999) \\ =(9.60,10.60) \end{gathered}$$

Therefore, the $90\%$ confidence interval for $\mu$ is $(9.60,10.60)$

If a sample of the size determined in component (c) provides a mean of $10.1$ years, the $90\%$ confidence interval for population mean is $(9.60,10.60)$ years.