An investigator wishes to develop a 90 percent confidence interval for number of

Question

An investigator wishes to develop a 90 percent confidence interval for number of hours devoted to studying for classes each week. He would like his interval to have a margin of error no greater than 15 minutes (.25 of an hour). He takes a preliminary sample of 35 students and obtains a mean = 12.2 hours with a standard deviation of 2.6 hours. Historically – from earlier studies – the standard deviation ranged from 2.4 to 2.6 hours based on ‘good sized’ samples. What size sample will be required to yield a mean that has a margin of error at 15 minutes.

Explanation / Answer

Compute Sample Size

n = (Z a/2 * S.D / ME ) ^2

Z/2 at 0.1% LOS is = 1.6449 ( From Standard Normal Table )

Standard Deviation ( S.D) = 156

ME =15

n = ( 1.6449*156/15) ^2

= (256.604/15 ) ^2

= 292.648 ~ 293

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