SOLUTION: a sample of 85 golfers showed that their average score on a particular golfers course was 91.31 with a standard deviation of 5.37. show all work. A) find the 99% confidence inte

Question 612153: a sample of 85 golfers showed that their average score on a particular golfers course was 91.31 with a standard deviation of 5.37. show all work.
A) find the 99% confidence interval of the mean score for all 85 golfers.
B) find the 99% confidence interval of the mean score for all golfers if this is a sample of 120 golfers instead of a sample of 85.
C) which onfidence interval is smaller and why?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a sample of 85 golfers showed that their average score on a particular golfers course was 91.31 with a standard deviation of 5.37. show all work.
A) find the 99% confidence interval of the mean score for all 85 golfers.
(89.775,92.845)
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B) find the 99% confidence interval of the mean score for all golfers if this is a sample of 120 golfers instead of a sample of 85.
(90.027,92.593)
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C) which onfidence interval is smaller and why?
Interval B is slightly smaller.
ME = t*s/sqrt(n)
Notice that ME and sqrt(n) are indirectly related.
As n gets larger, ME gets smaller.
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Cheers,
Stan H.
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