Determine the sample size required to estimate a population proportion to within 0.032 with 90.8% confidence,?

Determine the sample size required to estimate a population proportion to within 0.032 with 90.8% confidence, assuming that you have no knowledge of the approximate value of the sample proportion. (You will need to use software to get your critical value and should expect to use 4 decimal places of accuracy.)

I've been reading through the book and I don't even know where to start with this problem. I looked through all the lecture notes and can't find an example problem for this one. What formula should I be using? Thank you for all your help.

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  • 6 years ago
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    Rita

    Here is the formula ...

    margin of error = (z*)sqrt[(p)(1 - p)/n]

    z* = 1.68494 [from Standard Normal with (1 - 0.908)/2 = 0.046 area in each tail

    p = 0.5 [note: always use 1/2 if you are not given the proportion]

    Now, simply insert the values from the problem and solve for n ...

    margin of error = (z*)sqrt[(p)(1 - p)/n]

    0.032 = (1.68494)sqrt[(0.5)(1 - 0.5)/n]

    n = 693.1 , but ALWAYS round up ...

    ANSWER = 694

    hope that helps

    • Rita6 years agoReport

      I found my error, again thank you so much. You've been a tremendous help with my statistics class and I appreciate all the help you've given me :)

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  • 4 years ago

    The formula is actually

    n= (Z*/m)^2(p*)(1-p*)

    This was really confusing to me when I looked at this because I couldn't figure out how you came up with your answer of N. Once I looked up this formula in the book it made a lot more sense.

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