SOLUTION: A sample of 106 golfers showed that their average score on a particular golf course was 87.98 with a standard deviation of 5.39. Answer each of the following (SHOW ALL WORK and st

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Question 662733: A sample of 106 golfers showed that their average score on a particular golf course was 87.98 with a standard deviation of 5.39.
Answer each of the following (SHOW ALL WORK and state the FINAL answer to at least TWO Decimal places):
(A) Find the 90% confidence interval of the mean score for all 106 golfers.
(B) Find the 90% confidence interval of the mean score for all golfers if this is a sample of 130 golfers instead of a sample of 106.
(C) Which confidence interval is larger and why?
Thank you for your help!
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A sample of 106 golfers showed that their average score on a particular golf course was 87.98 with a standard deviation of 5.39.
Answer each of the following (SHOW ALL WORK and state the FINAL answer to at least TWO Decimal places):
(A) Find the 90% confidence interval of the mean score for all 106 golfers.
x-bar = 87.98
ME = 1.645*5.39/sqrt(106) = 0.8612
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90% CI: 87.98-0.8612 < u < 87.98+0.8612
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(B) Find the 90% confidence interval of the mean score for all golfers if this is a sample of 130 golfers instead of a sample of 106.
new ME = 1.645*5.39/sqrt(130) = 0.7776
90% CI: 87.98-0.7776 < u < 87.98+0.7776
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(C) Which confidence interval is larger and why?
The 1st is larger because the sample size is smaller.
Note 2 things:
1st: The CI is ALWAYS 2*ME in width
2nd: Since ME and sqrt(n) are indirectly related, ME gets
smaller when n gets larger.
That makes sense because having a larger sample means you
have more information; so the confidence interval can be tighter.
Cheers,
Stan H.