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Math Help - Sample Size - Wildlife Service

  1. #1
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    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.
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  2. #2
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    Re: Sample Size - Wildlife Service

    Quote Originally Posted by AwesomeHedgehog View Post
    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.
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