document.write( "Question 308673: Hello, I am having a very difficult time figuring this statistics problem. I appreciate any help.\r
\n" ); document.write( "\n" ); document.write( "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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Algebra.Com's Answer #220723 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
You know the mean, the standard deviation, the confidence interval, and the number of golfers.
\n" ); document.write( "Plug the data into the formula.
\n" ); document.write( "(\"Xave+%2B-+Z%2Asigma%2Fsqrt%28N%29\")
\n" ); document.write( "At 99%, \"Z=2.58\"
\n" ); document.write( "\"Xave=97\"
\n" ); document.write( "\"N=47\"
\n" ); document.write( "\"sigma=7\"
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