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Question:
The attached Excel file 2013 NCAA BB Tournament shows the salaries paid to the coaches of 62 of the 68 teams in the 2013 NCAA basketball tournament (not all private schools report their coach's salaries). Consider these 62 salaries to be a sample from the population of salaries of all 346 NCAA Division I basketball coaches.
Question 1. Use the 62 salaries from the TOTAL PAY column to construct a 95% confidence interval for the mean salary of all basketball coaches in NCAA Division I.
xbar = $1,465,752
SD = $1,346,046.2
lower bound of confidence interval ________
upper bound of confidence interval _______
Question 2. Coach Mike Krzyzewski's high salary is an outlier and could be significantly affecting the confidence interval results. Remove Coach Krzyzewski's salary from the data and recalculate the 95% confidence interval using the remaining 61 salaries.
xbar = $1,371,191
SD = $1,130,666.5
lower bound of confidence interval _________
upper bound of confidence interval. ________
Answer:
Question 1:
lower bound of confidence interval = $1,124,027
upper bound of confidence interval = $1,807,477
Question 2:
lower bound of confidence interval = $1,081,512
upper bound of confidence interval = $1,660,870
Step-by-step explanation:
Question 1:
The sample mean salary of 62 couches is
The standard deviation of mean salary is
The confidence interval for the mean salary of all basketball coaches is given by
Where is the sample mean, n is the sample size, s is the sample standard deviation and
is the t-score corresponding to a 95% confidence level.
The t-score corresponding to a 95% confidence level is
Significance level = α = 1 - 0.95 = 0.05/2 = 0.025
Degree of freedom = n - 1 = 62 - 1 = 61
From the t-table at α = 0.025 and DoF = 61
t-score = 1.999
So the required 95% confidence interval is
Question 2:
After removing the Coach Krzyzewski's salary from the data
The sample mean salary of 61 couches is
The standard deviation of the mean salary is
The t-score corresponding to a 95% confidence level is
Significance level = α = 1 - 0.95 = 0.05/2 = 0.025
Degree of freedom = n - 1 = 61 - 1 = 60
From the t-table at α = 0.025 and DoF = 60
t-score = 2.001
So the required 95% confidence interval is