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.

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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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}}}

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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