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Math Help - Margin of Error

  1. #1
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    Margin of Error

    I've been agonizing over this problem involving a sample mean and margin of error. I feel like the sample size I got as my answer is too small?

    Question is:

    How large a sample is needed to estimate a population proportion to within a margin of error of 4 % at the 98 % confidence level?

    First, I calculated the critical value as 2.33 and then plugged the margin of error, .4, and 2.33 into the formula:

    n = (z* divided by E)^2 x p-hat x q-hat or (2.33 divided by .4) ^2 x 1/2 x 1/2

    I got the 1/2 since p-hat and q-hat are both unknowns so you have to use the most conservative estimate, or 1/2.

    Anyway, my answer is 8.49 or sample size of 9

    I feel like this is way too small of a sample size! What'd I do wrong?
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  2. #2
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    update: I think it's supposed to be 849, not 8.49
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Since we have that (1-\alpha)=.98, \alpha must equal 0.02 and \frac{\alpha}{2}=.01. So, the z value we are looking for is z_{\frac{\alpha}{2}}=z_{.01}=2.33. We then require that

    2.33\sqrt{\frac{pq}{n}}=.4.

    Since the we do not know anything about p we use p=.5.

    I have not computed the value, but if you just solve for n you should find your answer. However, I think your answer is correct.
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