Exercise 7.3 (22). NCAA Football Coach Salaries A simple random sample of 40 salaries of NCAA football coaches has a mean of \$415,953. Assume that $\sigma=\$463,364$.

(a) Find the best point estimate of the mean salary of all NCAA football coaches.

(b) Construct a 95% confidence interval estimate of the mean salary of an NCAA football coach.

(c) Does the confidence interval contain the actual population mean of \$474,477?

#### Solution

(a) The sample mean of \$415,953 is the best point estimate of the mean salary of all NCAA football coaches.

(b) Given that sample size $n = 40$, sample mean $\overline{X}= 415953$ and population standard deviation $\sigma = 463364$.

#### Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

#### Step 2 Given information

Given that sample size $n =40$, sample mean $\overline{X}=415953$ and population standard deviation is $\sigma = 463364$.

#### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is

` $$ \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned} $$ `

where `$E = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$`

, and `$Z_{\alpha/2}$`

is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

#### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is `$Z_{\alpha/2}$`

.

Thus `$Z_{\alpha/2} = Z_{0.025} = 1.96$`

.

#### Step 5 Compute the margin of error

The margin of error for mean is

` $$ \begin{aligned} E & = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\\ & = 1.96 \frac{463364}{\sqrt{40}} \\ & = 143597.991. \end{aligned} $$ `

#### Step 6 Determine the confidence interval

$95$% confidence interval estimate for population mean is

` $$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 415953 - 143597.991 & \leq \mu \leq 415953 + 143597.991\\ 272355.009 & \leq \mu \leq 559550.991. \end{aligned} $$ `

Thus, $95$% confidence interval estimate for population mean is $(272355.009,559550.991)$.

(c) The confidence interval contain the actual population mean of $474,44.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators