document.write( "Question 1048818: I really need some assistance on this problem:\r
\n" ); document.write( "\n" ); document.write( "You take a sample of 22 from a population of test scores and the mean of your sample is 60. \r
\n" ); document.write( "\n" ); document.write( "(a) You know the standard deviation of the population is 10. What is the 99% confidence interval on the population mean. \r
\n" ); document.write( "\n" ); document.write( "(b) Now assume that you do not know the population standard deviation, but the standard deviation in your sample is 10. What is the 99% confidence interval on the mean now?\r
\n" ); document.write( "\n" ); document.write( "Any help is greatly appreciated!
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Algebra.Com's Answer #664414 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
a)
\n" ); document.write( "For a population with unknown mean and known standard deviation ,
\n" ); document.write( "a confidence interval for the population mean is: CI = x-bar ± \"z%2Asigma%2Fsqrt%28n%29\"
\n" ); document.write( "CI = = 60 ± \"z%2A10%2Fsqrt%2822%29\" 99% confidence level: z = 2.576
\n" ); document.write( "CI = = 60 ± \"2.576%2A10%2Fsqrt%2822%29\"
\n" ); document.write( "b)
\n" ); document.write( "For a population with unknown mean and unknown standard deviation ,
\n" ); document.write( "a confidence interval for the population mean is: CI = x-bar ± \"t%2As%2Fsqrt%28n%29\"
\n" ); document.write( "CI = 60 ± \"t%2A10%2Fsqrt%2822%29\"
\n" ); document.write( "99% confidence level, Degree of Freedom(n-1) = 21: t = 2.83 (use Caculator 0r table)
\n" ); document.write( "CI = 60 ± \"2.831%2A10%2Fsqrt%2822%29\"
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