statistics

Question

A wildlife study is designed to find the mean weight of salmon caught by an Alaskan fishing company. As a preliminary study, a random sample of 50 freshly caught salmon is weighed. The sample standard deviation is 2.15 pounds. How large a sample should be taken to be 99% confident that the sample mean is within 0.20 pound of the true mean weight (mu)?

Answer 1

Required sample size is found by rearranging the expression for confidence interval in terms of the margin of error.

ME = std error (z*)

standard error = s/√n

rearranging: √n =s(z*)/ME

n= (σ (z*)/ME)2

z* = 2.575 for 99% CI. ME=.2

n= (2.15*2.575/.2)2

n= 27.68

But you always round up to the next whole number. so n=28