Question

John Calipari, head basketball coach for the 2012 national champion University of Kentucky Wildcats, is the highest paid coach in college basketball with an annual salary of $5.4 million (USA Today, March 29, 2012). The sample below shows the head basketball coaches salary for a sample of 10 schools playing NCAA Division 1 basketball. Salary data are in millions of dollars. University Coach’s Salary University Coach’s Salary Indiana 2.2 Syracuse 1.5 Xavier .5 Murray State .2 Texas 2.4 Florida State 1.5 Connecticut 2.7 South Dakota State .1 West Virginia 2.0 Vermont .2 a. Use the sample mean for the 10 schools to estimate the population mean annual salary for head basketball coaches at colleges and universities playing NCAA Division 1 basketball (to 2 decimals). b. Use the data to estimate the population standard deviation for the annual salary for head basketball coaches (to 4 decimals). c. What is the 95% confidence interval for the population variance (to 2 decimals)

Answer 1

Answer:

A) sample mean = $1.36 million

B) standard deviation = $0.9189 million

C) confidence interval = ($1.93 million , $0.79 million)

*since the sample size is very small, the confidence interval is not valid.

Step-by-step explanation:

samples:

1. $2.7 million
2. $2.4 million
3. $2.2 million
4. $2 million
5. $1.5 million
6. $1.5 million
7. $0.5 million
8. $0.5 million
9. $0.2 million
10. $0.1 million

sample mean = $1.36 million

the standard deviation:

• $2.7 million - $1.36 million = 1.34² = 1.7956
• $2.4 million - $1.36 million = 1.04² = 1.0816
• $2.2 million - $1.36 million = 0.84² = 0.7056
• $2 million - $1.36 million = 0.64² = 0.4096
• $1.5 million - $1.36 million = 0.14² = 0.0196
• $1.5 million - $1.36 million = 0.14² = 0.0196
• $0.5 million - $1.36 million = -0.86² = 0.7396
• $0.5 million - $1.36 million = -0.86² = 0.7396
• $0.2 million - $1.36 million = -1.16² = 1.3456
• $0.1 million - $1.36 million = -1.26² = 1.5876
• total $8.444 million / 10 = $0.8444 million

standard deviation = √0.8444 = 0.9189

95% confidence interval = mean +/- 1.96 standard deviations/√n:

$1.36 million + [(1.96 x $0.9189 million)/√10] = $1.36 million + $0.57 million = $1.93 million

$1.36 million - $0.57 million = $0.79 million

Answer 2

Following are the calculation to the given points:

The average annual pay for head basketball coaches at NCAA Division 1 colleges and universities:

$\to \bar{x}= \Sigma_{i=1}^{10} \ \frac{x_i}{n}$

       $=\frac{2.2 +0.5+2.4 +2.7 +2.0+1.5 +0.2+1.5 +0.1+0.2}{10}\\\\ = 1.33$

Therefore, the mean= 1.33  

Calculating the standard deviation:  

$\to \sigma= \sqrt{\frac{1}{n} \Sigma (x_i -\bar{x})^2}$

       $= \sqrt{\frac{1}{10} {(2.2-1.33)^2 +(0.5 - 1.33)^2 + ... +(0.2 - 1.33)^2}}\\\\= 0.9508$

As just a result, our projected standard deviation for annual compensation for head basketball coaches is 0.9508.

Calculating the sample variance:  

$\to s^2=\frac{1}{n-1} \Sigma (x_i -\bar{x})^2$

        $=\frac{1}{10-1} {(2.2-1.33)^2 +(0.5-1.33)^2 +..........+(0.2-1.33)^2} \\\\=1.005$

Calculating the degrees of freedom:

 $\to df = n-1 = 10-1 = 9$

The formula to compute interval estimate of a population variance is as follows:  

$\to \frac{(n-1)s^2}{ \chi^2_{\frac{\alpha}{2}} } \leq \sigma^2 \leq \frac{ (n-1) s^2}{ \chi^2_{(1- \frac{\alpha}{2})}}$

From the chi-square distribution tables the critical value of

$\to \chi^2_{\frac{0.05}{2}}\ \ is\ \ 19.0228$

From the chi-square distribution tables the critical value of

$\to \chi^2_{(1-(\frac{0.05}{2}))} = 2.7004$

Thus,

$\to \frac{(10-1)1.005}{19.0228} \leq \sigma^2 \leq \frac{(10 - 1)1.005}{2.7004}\\\\ \to 0.48 \leq \sigma^2 \leq 3.35$

Therefore, the $95\%$ confidence interval for the population variance is lies between $0.48 \ and \ 3.35$

Calculating the sample variance:

$\to s^2 =\frac{1}{n-1} \Sigma (x_i-\bar{x})^2$

        $=\frac{1}{10-1} {(2.2 -1.33)^2 +(0.5-1.33)^2 +...+(0.2 - 1.33)^2}\\\\=1.005$

Calculating the degrees of freedom:

$\to df = n-1 = 10-1 = 9$

The formula to compute interval estimate of a population variance is as follows:

$\to \sqrt{\frac{(n-1)s^2}{ \chi^2_{\frac{\alpha}{2}}}} \leq \sigma \leq \sqrt{\frac{(n-1) s^2}{ \chi^2_{(1- \frac{\alpha}{2})}}$

From the chi-square distribution tables the critical value of $\chi^2_{\frac{0.05}{2}} = 19.0228$

From the chi-square distribution tables the critical value of$\chi^2_{(1-(\frac{0.05}{2}))} = 2.7004$

Thus,

$\to \sqrt{\frac{(10-1)1.005}{19.0228}} \leq \sigma \leq \sqrt{\frac{(10 - 1)1.005}{2.7004}}\\\\ \to \sqrt{0.48} \leq \sigma \leq \sqrt{3.35}\\\\ \to 0.69 \leq \sigma \leq 1.83$

Therefore, the $95\%$ confidence interval for the population variance is lies between $0.69 \ and\ 1.83$

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