Chapter 7.1, Problem 23P

a.

To determine

Verify that $\overline{x}=36.0$ for the given data.

Explanation of Solution

Step-by-step procedure to obtain the average using Ti-83 calculator:

• Click on Stat.
• From EDIT, choose 1: Edit..
• In column L1, enter the data.
• Click on Stat.
• From CALC, choose 1: 1-Var Stats.
• Select 2nd > 1.
• Click Enter.

Output obtained is as follows:

From the output, the value of average is obtained as 35.9952 $\left(\approx 36\right)$.

It is verified that $\overline{x}=36.0$ for the given data.

b.

To determine

Obtain the 75% confidence interval for $\mu$, the average annual profit per employee for all successful banks.

Answer to Problem 23P

The 75% confidence interval for $\mu$, the average annual profit per employee for all successful banks, is $\left(34.19,37.81\right)$.

Explanation of Solution

Here, $\overline{x}=36,\sigma =10.2,n=42,\text{and}\alpha =0.25$.

From Table 5: Areas of a Standard Normal Distribution, the value corresponding to 0.1251 is –1.15. That is, $z_{\frac{\alpha }{2}=0.125}=1.15\text{and}{z}_{1-\frac{\alpha }{2}=0.875}=-1.15$.

The 75% confidence interval for $\mu$, the average annual profit per employee for all successful banks, is calculated as follows:

$\begin{array}{c}\overline{x}\pm E=\overline{x}\pm z_c\frac{\sigma }{\sqrt{n}}\\ =36\pm 1.15\frac{10.2}{\sqrt{42}}\\ =\left(36-1.81,36+1.81\right)\\ =\left(34.19,37.81\right)\end{array}$

Therefore, the 75% confidence interval for $\mu$, the average annual profit per employee for all successful banks, is $\left(34.19,37.81\right)$.

c.

To determine

Explain whether the annual profit of less than 30 thousand dollars is low when compared with other successful financial institutions.

Explanation of Solution

From Part (b), the 75% confidence interval for $\mu$, the average annual profit per employee for all successful banks, is $\left(34.19,37.81\right)$.

Since all the values in the interval are greater than 30, it is clear that the annual profit of less than 30 thousand dollars is low when compared with other successful financial institutions.

d.

To determine

Explain whether the annual profit of more than 40 thousand dollars is better when compared with other successful financial institutions.

Explanation of Solution

From Part (b), the 75% confidence interval for $\mu$, the average annual profit per employee for all successful banks, is $\left(34.19,37.81\right)$.

Here, all the values in the interval are less than 40. That is, 40 thousand dollars is greater than the upper bound of the interval. Therefore, the annual profit of more than 40 thousand dollars is better when compared with other successful financial institutions.

e.

To determine

Find the 90% confidence interval for $\mu$, the average annual profit per employee for the profit values 30 thousand and 40 thousand dollars.

Answer to Problem 23P

The 90% confidence interval for $\mu$, the average annual profit per employee for the profit value 30 thousand dollars, is $\left(27.4031,32.5969\right)$.

The 90% confidence interval for $\mu$, the average annual profit per employee for the profit value 40 thousand dollars, is $\left(37.4031,42.5969\right)$.

Explanation of Solution

Here, $\overline{x}=30,\sigma =10.2,n=42,\text{and}\alpha =0.25$.

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 $\mu$, the average annual profit per employee for the profit value 30 thousand dollars, is calculated as follows:

$\begin{array}{c}\overline{x}\pm E=\overline{x}\pm z_c\frac{\sigma }{\sqrt{n}}\\ =30\pm 1.65\frac{10.2}{\sqrt{42}}\\ =\left(30-2.5969,30+2.5969\right)\\ =\left(27.4031,32.5969\right)\end{array}$

Therefore, the 90% confidence interval for $\mu$, the average annual profit per employee for the profit value 30 thousand dollars, is $\left(27.4031,32.5969\right)$.

The 90% confidence interval for $\mu$, the average annual profit per employee for the profit value 40 thousand dollars, is calculated as follows:

$\begin{array}{c}\overline{x}\pm z_c\frac{\sigma }{\sqrt{n}}=40\pm 1.65\frac{10.2}{\sqrt{42}}\\ =\left(40-2.5969,40+2.5969\right)\\ =\left(37.4031,42.5969\right)\end{array}$

Therefore, the 90% confidence interval for $\mu$, the average annual profit per employee for the profit value 40 thousand dollars, is $\left(37.4031,42.5969\right)$.