SOLUTION: Could someone please assist in helping me with this?... A sample of 142 golfers showed that their average score on a particular golf course was 90.46 with a standard deviation o

Question 623123: Could someone please assist in helping me with this?...
A sample of 142 golfers showed that their average score on a particular golf course was 90.46 with a standard deviation of 3.84.
Answer each of the following (show all work
and state the final answer to at least two decimal places.):
(A) Find the 95% confidence interval of the mean score for all 142 golfers.
(B) Find the 95% confidence interval of the mean score for all golfers if this is a sample of 120 golfers instead of a sample of 142.
(C) Which confidence interval is larger and why?
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

 

Hi,

(A) Find the 95% confidence interval of the mean score for all 142 golfers.

ME = 1.96[3.84/sqrt(142)]   and   90.46 - ME <  < 90.46 + ME

(B) Find the 95% confidence interval of the mean score for all golfers if this is a sample of 120 golfers 

ME = 1.96[3.84/sqrt(120)]  and   90.46 - ME <  < 90.46 + ME

(C) Which confidence interval is larger and why?  for the smaller sample as then ME is larger

	

 

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