SOLUTION: A sample of 95 golfers showed that their average score on a particular golf course was 84.37 with a standard deviation of 6.31.
Answer each of the following (show all work
and
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Question 672937: A sample of 95 golfers showed that their average score on a particular golf course was 84.37 with a standard deviation of 6.31.
Answer each of the following (show all work
and state the final answer to at least two decimal places.):
(A) Find the 99% confidence interval of the mean score for all 95 golfers.
(B) Find the 99% confidence interval of the mean score for all golfers if this is a sample of 145 golfers instead of a sample of 95.
(C) Which confidence interval is smaller and why?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A sample of 95 golfers showed that their average score on a particular golf course was 84.37 with a standard deviation of 6.31.
Answer each of the following (show all work
and state the final answer to at least two decimal places.):
(A) Find the 99% confidence interval of the mean score for all 95 golfers.
ME = z*s/sqrt(n) = 2.5758*6.31/sqrt(95) = 1.668
x-bar = 84.37
----
99% CI:84.37-1.668 < u < 84.37+1.668
-----------------------------------
(B) Find the 99% confidence interval of the mean score for all golfers if this is a sample of 145 golfers instead of a sample of 95.
ME = 2.5758*6.31/sqrt(145) = 1.350
----
99% CI: 84.37-1.35 < u < 84.37+1.35
-------------
(C) Which confidence interval is smaller and why?
ALL CI's have a width of 2*ME. When sample size is 145 ME is smaller
than when sample size is 95. The "B-CI" is smaller than the "A-CI".
======================
Cheers,
Stan H.
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