Question 1046798

An example that may help to clarify Your questions
A sample of 89 golfers showed that their average score on a particular golf course was 90.98 
with a standard deviation of 6.47. 
and state the final answer to at least two decimal places.):
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x=bar = 90.98 
(A) Find the 98% confidence interval of the mean score for all 89 golfers.
ME = 2.3263[6.47/sqrt(89)] = 1.5955
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CI: 90.98-1.5955 < u < 90.98+1.5955
CI: 89.3845 < u < 92.5755