Question

In: Statistics and Probability

John Calipari, head basketball coach for the 2012 national champion University of Kentucky Wildcats, is the...

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. The following sample shows the head basketball coach's salary for a sample of 10 schools playing NCAA Division I basketball. Salary data are in millions of dollars.

Coach's salary for a sample of 10 schools
University Coach's Salary
Indiana 2.2
Xavier 0.5
Texas 2.4
Connecticut 2.7
West Virginia 2.0
Syracuse 1.5
Murray State 0.2
Florida State 1.5
South Dakota State 0.1
Vermont 0.2

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 I basketball.

b. Use the data to estimate the population standard deviation for the annual salary for head basketball coaches.

c. What is the 95% confidence interval for the population variance?

d. What is the 95% confidence interval for the population standard deviation?

e. What is the 95% confidence interval for the population mean?

Solutions

Expert Solution

(a)

From the given data, the following statistics are calculated as follows:

n = Sample Size = 10

= SampleMean = 13.3/10 = 1.33

s = Sample SD = 1.0023

Estimate of Population mean = Sample Mean = 1.33

(b) Estimate of Population standard deviation = Sample standard deviation = 1.0023

(c)

ndf = n - 1 = 10- 1 = 9

Confidence interval:

Low End:

From Table:

Low End:

High End:

From Table:

High End:

So,

Confidence interval:

0.4753 < < 3.3482

(c)

ndf = n - 1 = 10- 1 = 9

Confidence interval:

Low End:

From Table:

Low End:

High End:

From Table:

High End:

So,

Confidence interval:

0.6894 < < 1.8294

(d)

SE = s/

= 1.0023/ = 0.3170

=0.05

ndf = 10 - 1 = 9

From Table, critical values of t = 2.2622

Confidence interval:

t SE

= 1.33 (2.2622 X 0.3170) = 1.33 0.7171

= ( 0.6130, 2.0471)


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