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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document.write( " (Note: this is a t-CI) \n" );
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Algebra.Com's Answer #532008 by ewatrrr(24785) You can put this solution on YOUR website! Re TY \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 \n" ); document.write( " |