Sample Size - Wildlife Service

If a wildlife service wishes to estimate the mean number of days of hunting per hunter for all hunters licensed in the state during a given season, with a bound on the error of estimation equal to 2 hunting days at 95% confidence, how many hunters must be included in the survey? Assume that data collected in earlier surveys have shown sigma to be approximately equal to 10.

Here's my answer. I am wondering if it is correct??

Margin of Error = 2

Standard Deviation = 10

z at 95% confidence would be z(0.025) = 1.96

Standard Error = (z * standard deviation) / sqrt(n)

Solving for n:

n = [ (z * standard deviation) / margin of error ]^2 = [ (1.96*10) / 2 ]^2 = 96.04

So, the sample size would have to be n = 96.04 at the 95% confidence level.

Re: Sample Size - Wildlife Service

Quote:

Originally Posted by

**AwesomeHedgehog** If a wildlife service wishes to estimate the mean number of days of hunting per hunter for all hunters licensed in the state during a given season, with a bound on the error of estimation equal to 2 hunting days at 95% confidence, how many hunters must be included in the survey? Assume that data collected in earlier surveys have shown sigma to be approximately equal to 10.

Here's my answer. I am wondering if it is correct??

Margin of Error = 2

Standard Deviation = 10

z at 95% confidence would be z(0.025) = 1.96

Standard Error = (z * standard deviation) / sqrt(n)

Solving for n:

n = [ (z * standard deviation) / margin of error ]^2 = [ (1.96*10) / 2 ]^2 = 96.04

So, the sample size would have to be n = 96.04 at the 95% confidence level.

This is correct.