How do you find the minimal sample size having a margin of error of .03, confidence level of 95%, and p is estimated by the decimal equivalent of 59%? For stats

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The margin of error E is given by:

`E=z_(alpha/2)sqrt((hat(p)(1-hat(p)))/n)` where `hat(p)` is the sample proportion, `alpha/2` is determined by the confidence level, and `n` is the sample size.

With confidence level of 95% we have `z_(alpha/2)=1.96` , `E=.03` , `hat(p)=.59` and `(1-hat(p))=.41` . Solve for n:

`.03=1.96sqrt((.59(.41))/n)`

`.015306=sqrt((.2419)/n)`

`2.3428=.2419/n`

`n=1032.54`

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The margin of error E is given by:

`E=z_(alpha/2)sqrt((hat(p)(1-hat(p)))/n)` where `hat(p)` is the sample proportion, `alpha/2` is determined by the confidence level, and `n` is the sample size.

With confidence level of 95% we have `z_(alpha/2)=1.96` , `E=.03` , `hat(p)=.59` and `(1-hat(p))=.41` . Solve for n:

`.03=1.96sqrt((.59(.41))/n)`

`.015306=sqrt((.2419)/n)`

`2.3428=.2419/n`

`n=1032.54`

So you need a sample size of at least 1033.

 

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