Knowing a sample percent `hat(p)` , we find the confidence interval for the population percent `p` by:

`hat(p)-z_(alpha/2)sqrt((hat(p)(1-hat(p)))/n)<p<hat(p)+z_(alpha/2)sqrt((hat(p)(1-hat(p)))/n)`

where `alpha` depends on the level of confidence, and `n` is the sample size.

Holding everything else the same, if you increase `n` , the factor `sqrt((hat(p)(1-hat(p)))/n)` gets smaller (the larger the...

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Knowing a sample percent `hat(p)` , we find the confidence interval for the population percent `p` by:

`hat(p)-z_(alpha/2)sqrt((hat(p)(1-hat(p)))/n)<p<hat(p)+z_(alpha/2)sqrt((hat(p)(1-hat(p)))/n)`

where `alpha` depends on the level of confidence, and `n` is the sample size.

Holding everything else the same, if you increase `n` , the factor `sqrt((hat(p)(1-hat(p)))/n)` gets smaller (the larger the denominator the smaller the fraction; the smaller the radicand the smaller the root).

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**Increasing the sample size shrinks the confidence interval.**

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** If looking at a sample mean and trying to determine a population mean we would have:

`bar(x)-z_(alpha/2)sigma/sqrt(n)<mu<bar(x)+z_(alpha/2)sigma/sqrt(n)`

where `bar(x)` is the sample mean, `alpha` depends on the confidence level desired, `sigma` is the population standard deviation, and `n` the sample size. Again, notice that increasing `n` reduces the term `z_(alpha/2)sigma/sqrt(n)` , thus decreasing the interval.

** This should make intuitive sense. If I poll 1000 people, I can be relatively certain of the projection of the poll results to the population. But if I then poll 5000 people, the second poll should decrease my uncertainty -- it is far more unlikely that I chose 5000 people away from the population parameter than 1000 peiople.

**Further Reading**