Chapter 7.1, Problem 19P

a.

To determine

Obtain the 90% confidence interval for the population mean annual number of reported larceny cases in such communities.

Find the margin of error.

Answer to Problem 19P

The 90% confidence interval for the population mean annual number of reported larceny cases in such communities is $\left(127.3862,149.6138\right)$.

The margin of error is 11.1138.

Explanation of Solution

Here, $\overline{x}=138.5,\sigma =42.6,n=30,\text{and }\alpha =0.1$.

From Table 5: Areas of a Standard Normal Distribution, the value corresponding to 0.0495 (which is approximately 0.05) is –1.65. That is $z_{\frac{\alpha }{2}=0.05}=-1.65\text{and}{z}_{1-\frac{\alpha }{2}=0.95}=-1.65$.

The 90% confidence interval for the population mean annual number of reported larceny cases in such communities is computed as follows:

$\begin{array}{c}\overline{x}\pm E=\overline{x}\pm z_{0.05}\frac{\sigma }{\sqrt{n}}\\ =138.5\pm 1.65\frac{42.6}{\sqrt{30}}\\ =\left(\left(138.5-11.1138\right),\left(138.5+11.1138\right)\right)\\ =\left(127.3862,149.6138\right)\end{array}$

Therefore, the 90% confidence interval for the population mean annual number of reported larceny cases in such communities is $\left(127.3862,149.6138\right)$.

The margin of error (E) is 11.1138.

b.

To determine

Obtain the 95% confidence interval for the population mean annual number of reported larceny cases in such communities.

Find the margin of error.

Answer to Problem 19P

The 95% confidence interval for the population mean annual number of reported larceny cases in such communities is $\left(125.2981,151.7019\right)$.

The margin of error is 13.2019.

Explanation of Solution

Here, $\alpha =0.05$.

From Table 5: Areas of a Standard Normal Distribution, the value corresponding to 0.0250 is –1.96. That is $z_{\frac{\alpha }{2}=0.025}=-1.96\text{and}{z}_{1-\frac{\alpha }{2}=0.975}=1.96$.

The 95% confidence interval for the population mean annual number of reported larceny cases in such communities is computed as follows:

$\begin{array}{c}\overline{x}\pm E=\overline{x}\pm z_{0.05}\frac{\sigma }{\sqrt{n}}\\ =138.5\pm 1.96\frac{42.6}{\sqrt{30}}\\ =\left(\left(138.5-13.2019\right),\left(138.5+13.2019\right)\right)\\ =\left(125.2981,151.7019\right)\end{array}$

Therefore, the 95% confidence interval for the population mean annual number of reported larceny cases in such communities is $\left(125.2981,151.7019\right)$.

The margin of error (E) is 13.2019.

c.

To determine

Obtain the 99% confidence interval for the population mean annual number of reported larceny cases in such communities.

Find the margin of error.

Answer to Problem 19P

The 99% confidence interval for the population mean annual number of reported larceny cases in such communities is $\left(121.122,155.878\right)$.

The margin of error is 17.378.

Explanation of Solution

Here, $\alpha =0.01$.

From Table 5: Areas of a Standard Normal Distribution, the value corresponding to 0.0049 (which is approximately 0.005) is –2.58. That is, $z_{\frac{\alpha }{2}=0.005}=-2.58\text{and}{z}_{1-\frac{\alpha }{2}=0.995}=2.58$.

The 99% confidence interval for the population mean annual number of reported larceny cases in such communities is computed as follows:

$\begin{array}{c}\overline{x}\pm E=\overline{x}\pm z_{0.05}\frac{\sigma }{\sqrt{n}}\\ =138.5\pm 2.58\frac{42.6}{\sqrt{30}}\\ =\left(\left(138.5-17.378\right),\left(138.5+17.378\right)\right)\\ =\left(121.122,155.878\right)\end{array}$

Therefore, the 99% confidence interval for the population mean annual number of reported larceny cases in such communities is $\left(121.122,155.878\right)$.

The margin of error (E) is 17.378.

d.

To determine

Compare the margin of errors and explain whether the margin of errors increases as the confidence levels increase.

Explanation of Solution

The margin of error obtained for 90% confidence level is 11.1138 (from Part (a)); the margin of error obtained for 95% confidence level is 13.2019 (from Part (b)); and the margin of error obtained for 99% confidence level is 17.378  (from Part (c)).

It is clear that the margin of error increases as the confidence levels increase.

e.

To determine

Compare the lengths of confidence intervals and explain whether the confidence intervals increase in length as the confidence levels increase.

Explanation of Solution

From Parts (a)–(c), it is clear that the length of the confidence interval for 95% confidence level is greater than the length of the confidence interval for 90% confidence level and the length of the confidence interval for 99% confidence level is greater than that for 95% and 90% confidence levels.

Therefore, the confidence interval increases in length as the confidence levels increase.