document.write( "Question 1162704: The standard deviation of a normal population is 10.you take a sample of 25 item from this population and compute a 95% confidence interval. In order to compute the confidence interval, you will use? \n" ); document.write( "
Algebra.Com's Answer #786594 by Theo(13342)\"\" \"About 
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sd = 10
\n" ); document.write( "n = 25
\n" ); document.write( "se = sd / sqrt(n) = 10 / sqrt(25) = 10 / 5 = 2
\n" ); document.write( "95% confidence interval requires two ended alpha to be equal to .05.
\n" ); document.write( ".05/2 means alpha os .025 on each end.
\n" ); document.write( "z-score required will be plus or minus the z-score that has .025 area to the left of it.
\n" ); document.write( "that critical z-score will be plus or minus 1.959963986
\n" ); document.write( "to find the associated raw score, you would need the mean.
\n" ); document.write( "unfortunately, the mean is missing from the problem statement.
\n" ); document.write( "i think the answer to your question is that you would use the standard error, otherwise known at the standard deviation of the distribution of sample means.
\n" ); document.write( "that's the formula i showed you above of se = sd / sqrt(n).\r
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