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QHello, I am having a very difficult time figuring this statistics problem

Hello, I am having a very difficult time figuring this statistics problem. I appreciate any help.


A study of 47 golfers show that their average score on a particular course was 97. The standard deviation of the sample is 7. Find the 99% confidence interval of the mean score for all golfers.

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#1Sniper619Answered at 2013-04-07 07:07:12
You know the mean, the standard deviation, the confidence interval, and the number of golfers. Plug the data into the formula. ({{{Xave +- Z*sigma/sqrt(N)}}}) At 99%, {{{Z=2.58}}} {{{Xave=97}}} {{{N=47}}} {{{sigma=7}}}
#2FantasyAnswered at 2013-04-07 07:08:39
The 99% confidence level means that you include 3 standard deviations from the mean. In this case, the standard deviation is 7, and 3*7=21. Now, from the mean of 97, both ADD and SUBTRACT 21: 97-21=76 97+21=118 The interval is from 76 to 118. In other words, 99% of the golfers scored between 76 and 118 on this golf course. Dr. Robert J. Rapalje, Retired
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