SOLUTION: A sample of 102 golfers showed that their average score on a particular golf course was 88.55 with a standard deviation of 4.27. Answer each of the following (show all work and

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Question 483101: A sample of 102 golfers showed that their average score on a particular golf course was 88.55 with a standard deviation of 4.27.
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 102 golfers.
(B) Find the 95% confidence interval of the mean score for all golfers if this is a sample of 135 golfers instead of a sample of 102.
(C) Which confidence interval is smaller and why
Thank you very much!
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A sample of 102 golfers showed that their average score on a particular golf course was 88.55 with a standard deviation of 4.27.
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Answer each of the following (show all work
and state the final answer to at least two decimal places.):
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(A) Find the 95% confidence interval of the mean score for all 102 golfers.
x-bar = 88.55
ME = 1.96*4.25/sqrt(102) = 0.8248
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95% CI: 88.55-0.8248 < u < 88.55+0.8248
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(B) Find the 95% confidence interval of the mean score for all golfers if this is a sample of 135 golfers instead of a sample of 102.
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Same procedure but put 135 in place of 102 in the calculation.
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(C) Which confidence interval is smaller and why
The Confidence Interval is always twice as wide as
the margin of error.
And ME = z*s/sqrt(n) shows that ME and n are indirectly related.
So, ME gets wider as sample size gets smaller.
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Your sample size got larger (from 102 up to 135, so ME
became smaller. So the CI became narrower for n = 135
then it was for n = 102.
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Cheers,
Stan H.