Chapter 8.1, Problem 19P

FBI Report: Larceny Thirty small communities in Connecticut (population near 10.000 each) gave an average of $\bar{x}=138.5$ reported cases of larceny per year. Assume that $\sigma$ is known to be 42.6 cases per year (Reference: Crime in the United States, Federal Bureau of Investigation).

(a) Find a 90% confidence interval for the population mean annual number of reported larceny cases in such communities What is the margin of error? (b) Find a 95% confidence interval for the population mean annual number of reported larceny cases in such communities What is the margin of error?

(c) Find a 99% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error?

(d) Compare the margins of error for parts (a) through (c). As the confidence levels increase, do the margins of error increase?

(c) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the confidence levels increase, do the confidence intervals increase in length?

To determine

To find: The 90% confidence interval for the population mean annual number of reported larceny cases in such communities. Also find the margin of error.

Answer to Problem 19P

Solution:

The 90% confidence interval for $\mu$ is (125.71, 151.29) and margin of error $E\text{is}12.79\text{ larceny cases per year}.$

Explanation of Solution

Calculation:

Given that x be a random variable that represents the communities in Connecticut. The average of $\overline{x}=138.5$ reported cases of larceny per year and it is assumed that $\sigma$ is known to be $\text{42}\text{.6}$ cases per year.

We have to find 90% confidence interval,

$\begin{array}{l}c=0.90\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.90}=1.645\\ \sigma =42.6\\ n=30\\ E=z_c\frac{\sigma }{\sqrt{n}}\\ E=1.645\frac{42.6}{\sqrt{30}}\\ E=12.79425\\ E\approx 12.79\end{array}$

90% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 138.5-12.79<\mu <138.5+12.79\\ 125.71<\mu <151.29\end{array}$

The 90% confidence interval for $\mu$ is (125.71, 151.29).

Margin of error is $E=12.79\text{ larceny cases per year}\text{.}$

To determine

To find: A 95% confidence interval for the population mean annual number of reported larceny cases in such communities. Also find the margin of error.

Answer to Problem 19P

Solution:

The 95% confidence interval for $\mu$ is (123.26, 153.74) and margin of error $E\text{is}15.24\text{ larceny cases per year}.$

Explanation of Solution

Calculation:

Given that x be a random variable that represents the communities in Connecticut. The average of $\overline{x}=138.5$ reported cases of larceny per year and it is assumed that $\sigma$ is known to be $\text{42}\text{.6}$ cases per year.

We have to find 95% confidence interval,

$\begin{array}{l}c=0.95\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.95}=1.96\\ \sigma =42.6\\ n=30\\ E=z_c\frac{\sigma }{\sqrt{n}}\\ E=1.96\frac{42.6}{\sqrt{30}}\\ E=15.24421\\ E\approx 15.24\end{array}$

95% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 138.5-15.24<\mu <138.5+15.24\\ 123.26<\mu <153.74\end{array}$

The 90% confidence interval for $\mu$ is (123.26, 153.74).

Margin of error is $E=15.24\text{ larceny cases per year}\text{.}$

To determine

To find: A 99% confidence interval for the population mean annual number of reported larceny cases in such communities. Also find the margin of error.

Answer to Problem 19P

Solution:

The 99% confidence interval for $\mu$ is (118.43, 158.57) and margin of error $E\text{is}20.07\text{ larceny cases per year}.$

Explanation of Solution

Calculation:

Given that x be a random variable that represents the communities in Connecticut. The average of $\overline{x}=138.5$ reported cases of larceny per year and it is assumed that $\sigma$ is known to be $\text{42}\text{.6}$ cases per year.

We have to find 99% confidence interval,

$\begin{array}{l}c=0.99\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.99}=2.58\\ \sigma =42.6\\ n=30\\ E=z_c\frac{\sigma }{\sqrt{n}}\\ E=2.58\frac{42.6}{\sqrt{30}}\\ E=20.06636\\ E\approx 20.07\end{array}$

99% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 138.5-20.07<\mu <138.5+20.07\\ 118.43<\mu <158.57\end{array}$

The 99% confidence interval for $\mu$ is (118.43, 158.57).

Margin of error is $E=20.07\text{ larceny cases per year}\text{.}$

To determine

The comparison of the margins of error for part (a) through (c). Also determine whether the margins of error increases or decreases when confidence level increases.

Answer to Problem 19P

Solution:

The margins of errors increases as the confidence level increases.

Explanation of Solution

Calculation:

For 90% confidence interval, the margin of error is $E\text{is}12.79\text{ larceny cases per year}.$

For 95% confidence interval, the margin of error is $E=15.24\text{ larceny cases per year}\text{.}$

For 99% confidence interval, the margin of error is $E=20.07\text{ larceny cases per year}\text{.}$

As the confidence level increases the margins of errors also increases.

To determine

The comparison of confidence interval for part (a) through (c). Also determine whether the confidence intervals increases in length as confidence level increases.

Answer to Problem 19P

Solution:

The length of confidence interval increases as the confidence level increases.

Explanation of Solution

For 90%confidence interval, the length of confidence interval is $151.29-125.71=\mathrm{25.58.}$

For 95%confidence interval, the length of confidence interval is $153.74-123.26=\mathrm{30.48.}$

For 99%confidence interval, the length of confidence interval is $158.57-118.43=\mathrm{40.14.}$

As the confidence level increases the length of confidence interval also increases.