document.write( "Question 881161: Assume that the average age of students at a local community college is normally distributed. If a sample of size n=12 has a mean of 27.8 with a standard deviation of 4.1, find a 95 percent confidence interval.
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Algebra.Com's Answer #532008 by ewatrrr(24785)\"\" \"About 
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\n" ); document.write( "E = t*s/sqrt(n)
\n" ); document.write( "degree of freedom is 11, t = 2.201 at 95% confidence interval (Use table)
\n" ); document.write( "ME = 2.20(4.1/sqrt(12))
\n" ); document.write( "CI: 27.8-ME < u < 27.8 + ME
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