SOLUTION: 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.
Answer each of the following (show all work
and s
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Question 366127: 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.
Answer each of the following (show all work
and state the final answer to at least two decimal places.):
(A) Find the 98% confidence interval of the mean score for all 89 golfers.
(B) Find the 98% confidence interval of the mean score for all golfers if this is a sample of 135 golfers instead of a sample of 89.
(C) Which confidence interval is larger and why? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 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.
Answer each of the following (show all work
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
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(B) Find the 98% confidence interval of the mean score for all golfers if this is a sample of 135 golfers instead of a sample of 89.
ME = 2.3263*6.47/sqrt(135) = 1.2954
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CI: 90.89-1.2954 < u < 90.89+1.2954
CI: 89.5946 < u < 92.1854
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(C) Which confidence interval is larger and why?
The width of a CI is always 2*ME
Since the ME is larger for n = 89, its CI is larger.
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Cheers,
Stan H.
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