document.write( "Question 1178895: construct a 99% confidence interval estimate for a population of 50, a standard deviation of .02 and a sample mean of .995\r
\n" ); document.write( "\n" ); document.write( "Then a 95% interval and how it changes from the 99% interval
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Algebra.Com's Answer #808339 by Boreal(15235)\"\" \"About 
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99% half-interval is z(0.995)*sigma/sqrt(n)=2.576*0.02/sqrt(50)=0.0073
\n" ); document.write( "mean +/- half-interval is (0.9877, 1.0223)
\n" ); document.write( "95% interval is going to be narrower but with less confidence.
\n" ); document.write( "half-interval is 1.96*0.02/sqrt(50)=0.0055
\n" ); document.write( "so interval is (0.9895, 1.0005)
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