SOLUTION: A sample of 83 golfers showed that their average score on a particular golf course was 84.81 with a standard deviation of 5.99. Answer each of the following (state the final ans

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Question 616136: A sample of 83 golfers showed that their average score on a particular golf course was 84.81 with a standard deviation of 5.99.
Answer each of the following (state the final answer to at least two decimal places.):
(A) Find the 90% confidence interval of the mean score for all 83 golfers.
(B) Find the 90% confidence interval of the mean score for all golfers if this is a sample of 130 golfers instead of a sample of 83.
(C) Which confidence interval is larger and why?
Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!

Hi,
A sample of 83 golfers showed that their average score on a particular golf course was
84.81 with a standard deviation of 5.99.
ME = 1.645[5.99/sqrt(83)] = 1.0816
CI: 84.81-1.0816< u < 84.81+1.0816
CI: 83.7284 < u < 85.8916
ME = 1.645[5.99/sqrt(130)] = .8642
CI: 84.81-.8642 < u < 84.81+.8642
CI: 83.9458 < u < 85.6742
(C) Which confidence interval is larger and why?
Smaller sample has the larger confidence Interval.
the larger the sample the smaller the confidence interval is.