Question

In: Statistics and Probability

1.NCAA Football Coach Salaries: A simple random sample of 40 salaries of NCAA football coaches in...

1.NCAA Football Coach Salaries: A simple random sample of 40 salaries of NCAA football coaches in the NCAA has a mean of $415,953. The population standard deviation of all salaries of the NCAA football coaches is $463,364. Use a 5% significance level to test the claim that the mean salary of a football coach in the NCAA is less than $500,000.

Solutions

Expert Solution

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

H0: µ = 500,000 versus Ha: µ < 500,000

This is a lower tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 500,000

Xbar = 415,953

σ = 463,364

n = 40

α = 0.05

Critical value = -1.6449

(by using z-table or excel)

Z = (415,953 - 500,000)/[ 463,364/sqrt(40)]

Z = -1.1472

P-value = 0.1257

(by using Z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that the mean salary of a football coach in the NCAA is less than $500,000.


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