Question

A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week. Assuming a standard deviation of 6 hours, what is the required sample size if the error is to be less than ½ hour with a 95% level of confidence?

Answer 1

n=(ZxStdDev/Error)^2 where Z=1.96 (approx =2), then n=(2*6/0.5)^2=576

Answer 2

The Z score can be obtained from a Normal Distribution Table for alpha=0.05 (two tails) or alpha=0.025 (one tail)