Chapter 7.2, Problem 20P

(a)

To determine

Check the mean by using the calculator.

Check the standard deviation by using the calculator.

Explanation of Solution

Calculation:

Let variable x denotes crime rates.

Mean:

Use Ti 83 calculator tofind the mean as follows:

• Select STAT>Edit> Enter the values of Rate as L1.
• Click $\boxed{2^{\text{nd}}}$ button >STAT; take the arrow to the MATH menu, and then ‘3’ numbered key.
• Click $\boxed{2^{\text{nd}}}$ button; then ‘1’ numbered key to get L1, and close the ‘)’ bracket.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the mean value is approximately 64.2 crimes per 1000 population.

Standard deviation:

Use Ti 83 calculator tofind the standard deviation as follows:

• Select STAT>Edit> Enter the values of Rate as L1.
• Click $\boxed{2^{\text{nd}}}$ button >STAT; take the arrow to the MATH menu, and then ‘7’ numbered key.
• Click $\boxed{2^{\text{nd}}}$ button; then ‘1’ numbered key to get L1, and close the ‘)’ bracket.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the standard deviation value is approximately 27.9 crimes per 1000 population.

Hence, the sample mean and the sample standard deviation by using the calculator is verified.

(b)

To determine

Find the 80% confidence intervalfor the population mean crime rate for all Denver neighborhoods.

Answer to Problem 20P

The 80% confidence intervalfor the population mean crime rate for all Denver neighborhoodsis $58.8<\mu <69.6$.

Explanation of Solution

Calculation:

Confidence interval:

The confidence interval formula is,

$\overline{x}-E<\mu <\overline{x}+E$

In the formula $E=z_c\frac{s}{\sqrt{n}}$, $\overline{x}$ is sample mean, s is sample standard deviation, n is the sample size, and $t_c$ is critical value for confidence level c and degrees of freedom $d.f.=n-1$.

Let $\mu$ denotes the population mean crime rate for all Denver neighborhoods. There are 46 samples in the data and the confidence level is 80%.

Critical value:

The degrees of freedom is,

$\begin{array}{c}d.f.=46-1\\ =45\end{array}$

Use the Appendix II: Tables, Table 6: Critical Values for Student’s t Distribution:

• In d.f. column locate the value 45.
• In c row of locate the value 0.800.
• The intersecting value of row and columns is 1.301.

The critical value is $\pm 1.301$.

The error is,

$\begin{array}{c}E=\left(1.301\right)\frac{27.9}{\sqrt{46}}\\ =1.301\times 4.1136\\ =5.3518\\ \approx 5.4\end{array}$

Confidence interval:

Substitute $\overline{x}$ as 64.2, E as 5.4 in the confidence interval formula

$\begin{array}{c}64.2-5.4<\mu <64.2+5.4\\ 58.8<\mu <69.6\end{array}$

Hence, the 80% confidence intervalfor the population mean crime rate for all Denver neighborhoods is $58.8<\mu <69.6$ crimes per 1,000 population.

(c)

To determine

Explain whether the crime rate 57 is belowthe average population crime rate or not.

Explain whether fewer patrols can be assigned neighbourhood or not.

Explanation of Solution

From part (b), the 80% confidence intervalcrime rate is $\left(58.8,69.6\right)$ crimes per 1,000 population.

The rate of crime in neighbourhood is 57 crimes per 1,000 population. The value 57 is slightly less than the lower bound of 80% confidence interval. May be fewer patrols could be assigned to neighbourhood, but cannot determine how far it could be safe as 57 is closer to lower bound.

(d)

To determine

Explain whether the crime rate 75 is above the average population crime rate or not.

Explain whether more patrols should be assigned neighbourhood or not.

Explanation of Solution

From part (b), the 80% confidence intervalcrime rate is $\left(58.8,69.6\right)$ crimes per 1,000 population.

The rate of crime in neighbourhood is 75 crimes per 1,000 population. The value 75 is larger than the upper bound of 80% confidence interval. This shows that, more patrols could be assigned to neighbourhood because of higher crime rate.

(e)

To determine

Find the 80% confidence intervalfor the population mean crime rate for all Denver neighborhoods.

Explain whether the crime rate 57 is belowthe average population crime rate or not.

Explain whether fewer patrols can be assigned neighbourhood having crime rate of 57 or not.

Explain whether the crime rate 75 is above the average population crime rate or not.

Explain whether more patrols should be assigned neighbourhood having crime rate of 75 or not.

Answer to Problem 20P

The 95% confidence intervalfor the population mean crime rate for all Denver neighborhoodsis $55.9<\mu <72.5$.

Explanation of Solution

Calculation:

There are 46 samples in the data and the confidence level is 95%.

Critical value:

The degrees of freedom is,

$\begin{array}{c}d.f.=46-1\\ =45\end{array}$

Use the Appendix II: Tables, Table 6: Critical Values for Student’s t Distribution:

• In d.f. column locate the value 45.
• In c row of locate the value 0.950.
• The intersecting value of row and columns is 2.014.

The critical value is $\pm 2.014$.

The error is,

$\begin{array}{c}E=\left(2.014\right)\frac{27.9}{\sqrt{46}}\\ =2.014\times 4.1136\\ =8.2848\\ \approx 8.3\end{array}$

Confidence interval:

Substitute $\overline{x}$ as 64.2, E as 8.3 in the confidence interval formula

$\begin{array}{c}64.2-8.3<\mu <64.2+8.3\\ 55.9<\mu <72.5\end{array}$

Hence, the 95% confidence intervalfor the population mean crime rate for all Denver neighborhoods is $55.9<\mu <72.5$ crimes per 1,000 population.

The 95% confidence interval for crime rate is $\left(55.9,72.5\right)$ crimes per 1,000 populations.

The rate of crime in neighbourhood is 57 crimes per 1,000 populations. The value 57 is within 95% confidence interval. This shows that, it is not necessary to assign patrols to neighbourhood.

The rate of crime in another neighbourhood is 75 crimes per 1,000 population. The value 75 is larger than the upper bound of 95% confidence interval. This shows that, more patrols could be assigned to neighbourhood because of higher crime rate.

(f)

To determine

Explain whether it is necessary to make any assumptions for the problem.

Answer to Problem 20P

No, it is not necessary to make any assumptions for the problem.

Explanation of Solution

A random sample of 46 Denver neighbourhoods is taken from the population.

Based on central limit theorem, the sample size n must be greater than or equal to 30 for making the distribution of sample means $\overline{x}$ to be approximately normal, even if the population is not normally distributed.

In the scenario, the sample size is greater than 30. This shows that, $\overline{x}$ is normally distributed and it is not necessary to make any assumptions for the problem.