Question

John Calipari, head basketball coach for the national champion University of Kentucky Wildcats, is the highest paid coach in college basketball with an annual salary of million (USA Today, March 29, 2012). The following sample shows the head basketball coach's salary for a sample of schools playing NCAA Division basketball. Salary data are in millions of dollars.

University Coach's Salary University Coach's Salary
Indian 2.2 Syracuse 1.5
Xavier .5 Murry State .2
Texas 2.4 Florida State 1.1
Connecticut 2.7 South Dekota 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 decimal).
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)?
D. what is the 95% confidence interval for the population standard deviation (to 2 decimals)?

Answer 1

A) sample mean = $1.36 million

B) standard deviation = $0.9189 million

C) variance confidence interval = ($0.40 million, $2.81 million)

D) standard deviation confidence interval = ($1.93 million , $0.79 million)

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

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

variance 0.8444

standard deviation = √0.8444 = 0.9189

in order to calculate the confidence interval for the population variance we are going to use a chi-square distribution with 2.5% on each tail ⇒ table values 2.7004 and 19.023 enclose 95% of the distribution.

[(n - 1) x variance] / 2.7004 = (9 x 0.8444) / 2.7004 = 2.81

[(n - 1) x variance] / 19.023 = (9 x 0.8444) / 19.023 = 0.40

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