# 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

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 Click here to see ALL problems on Miscellaneous Word Problems 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(10682)   (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 ```