SOLUTION: A researcher wishes to estimate with 99% confidence, the population proportion of adults who say chocolate is their favorite ice cream flavor. Her estimate must be accurate within

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Question 1118093: A researcher wishes to estimate with 99% confidence, the population proportion of adults who say chocolate is their favorite ice cream flavor. Her estimate must be accurate within 2% of the population proportion.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i presume you're looking for the sample size that would make your range of estimates at 99% confidence interval be within plus or minus 2% of the desired proportion.

i came up with a formula.

that formula is n = (p * q * z^2) / (plus or minus .02 * p^2).

if you want your confidence level to be 99%, then your z-score will have to be plus or minus 2.575829303.

that confidence level means your alpha is .5% on either end of the normal distribution.

that's 1% outside of the interval with 99% inside the interval.

that makes the formula n = (p * q * 2.575829303^2) / (plus or minus .02 * p ^ 2).

to apply this formula, you need to know what p is.

once you know p, you can solve for q, because q = 1 - p.

this formula should work with any valid p.

i tested with p = .3 and found the formula to be good.

i'll test with p = .9 to see if the formula is still good.

when p = .9, q = .1.

the formula for z is:

z = (x-m) / s

in a proportion, s = sqrt(p * q / n).

if you want your estimates to be accurate within 2% of the population proportion, then your range needs to be plus or minus .02 * p.

this means that (x-m) will be either .02 * p or -.02 * p.

assuming (x-m) = .02 * p, then your formula for z becomes z = .02 * p / s.

since p = .9, the formula becomes z = .02 * .9 / s.

this can be simplified to z = .018 / s.

solve for s to get s = .018 / z.

since s = sqrt(p * q / n), then the formula becomes:

sqrt(.9 * .1 / n) = .018 / z.

this simplifies to sqrt(.09 / n) = .018 / z

square both sides of this equation to get .09 / n = .018^2 / z^2.

solve for n to get n = .09 * z^2 / .018^2.

since z = 2.575829303, this formula becomes n = .09 * 2.575829303^2 / .018^2.

this results in n = 1843.026833 which can be rounded to 1844.

if we did this right, we now know that the minimum sample size has to be 1844 when the desired population proportion is equal to .9.

when n = 1844, you get:

p = .9
q = .1
n = 1844.
z = plus or minus 2.575829303.

the formula for z is z = (x-m) / s

s = sqrt(p*q/n) becomes s = sqrt(.9*.1/1844) which becomes s = .0069861965.

you want to solve for x when z is plus or minus 2.575829303.

on the plus side, z = (x-m)/s becomes 2.575829303 = (x-.9) / .0069861965.

solve for x to get x = .9179952496.

on the minus side, -z = (x-m)/s becomes -2.575829303 = (x-.9) / .0069861965.

solve for x to get x = .8820047504.

your range of answers at 99% confidence level is .8820047504 to .9179952496.

your population proportion is .9

(.9 - .8820047504) / .9 = .0199947218 which rounds to .02.

(.9179952496 - .9) / .9 = .0199947218 which rounds to .02.

your estimate is within plus or minus 2% of the population proportion of .9.

in order to achieve this accuracy, your sample size had to be a minimum of 1844.

depending on the proportion, the required minimum sample size will be different.

the solution to determining the sample size required is:

n = (p * q * 2.575829303^2) / (plus or minus .02 * p ^ 2).

p is the population proportion.
q is 1 - p

this formula is for the estimate to be within 2% of the desired proportion at confidence level of 99%.