# Find the minimum sample size required to estimate the population proportion. if the required margin of error is 0.05 or less at 95% confidence level? The answer is n=196. How do you get it? For a population proportion, the standard error is given by :

`E=z_(alpha/2)sqrt((hat(p)hat(q))/n)` where:

`E` is the standard error

`z_(alpha/2)` gives the area (probability) that the error occurs

`hat(p)` is the sample proportion

`hat(q)=1-hat(p)`

`n` is the sample size.

Solving for n we get: `n=(hat(p)hat(q))/((E/z_(alpha/2))^2)`

Generally, if the problem is presented...

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For a population proportion, the standard error is given by :

`E=z_(alpha/2)sqrt((hat(p)hat(q))/n)` where:

`E` is the standard error

`z_(alpha/2)` gives the area (probability) that the error occurs

`hat(p)` is the sample proportion

`hat(q)=1-hat(p)`

`n` is the sample size.

Solving for n we get: `n=(hat(p)hat(q))/((E/z_(alpha/2))^2)`

Generally, if the problem is presented as above (you are not given `hat(p)` ) then you you make `hat(p)=.5` , since this results in the largest possible sample size. In that case you would get:

`n=(.5*.5)/((.05/1.96)^2)~~384.16` or n=385

Since you gave the answer as n=196, I assume that this is a subproblem, and that you know that `hat(p)=.85` or `hat(p)=.15`

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Given `hat(p)=.85("or"hat(p)=.15),E=.05,alpha=.05` :

`z_(alpha/2)=1.96` from the standard normal table

Then `n=((.85)(.15))/((.05/1.96)^2)~~195.92` so n=196

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