use the given data to find the minimum sample size required to estimate a population proportion or percentage margin of error: two percentage points; confidence level: 99%; from a prior study, p is estimated by the decimal equivalent of 14%

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The error is 1/2 of the confidence interval. Since we want 99% confidence, we have `z_(alpha/2)=2.58` .

Then the error E is given by`E=2.58sqrt((pq)/n)` where p is the percentage with a given characteristic, and q the percentage without. Here p=.14,q=.86, E=.02 and we want to solve for n:

`.02=2.58sqrt((.14*.86)/n)`

`.007751938=sqrt((.14*.86)/n)`

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The error is 1/2 of the confidence interval. Since we want 99% confidence, we have `z_(alpha/2)=2.58` .

Then the error E is given by`E=2.58sqrt((pq)/n)` where p is the percentage with a given characteristic, and q the percentage without. Here p=.14,q=.86, E=.02 and we want to solve for n:

`.02=2.58sqrt((.14*.86)/n)`

`.007751938=sqrt((.14*.86)/n)`

`16641=n/.1204`   Square both sides; then reciprocate both sides

`n=2004` rounding to the next integer.

Thus the minimum sample size is 2004 to yield an error of 2 percent with a 99% confidence.

 

 

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