Question 1206969: Students at a certain school were surveyed, and it was estimated that 15% of college students abstain from drinking alcohol. To estimate this proportion in your school, how large a random sample would you need to estimate it to within 0.06 with probability 0.99, if before conducting the study (a) you are unwilling to predict the proportion value at your school and (b) you use the results from the surveyed school as a guideline.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! p = .15 = proportion that satisfy the requirement.
q = .85 = 1 - p = proportion that don't satisfy the requirement.
moe = .06 = margin of error.
ci = .99 = confidence interval = 99%.
s = sqrt(p * q / n) = standard error.
n = sample size.
z = critical z-score at 99% confidence interval.
this leaves a tail outside the confidence interval called alpha.
if it's a two-tail confidence interval, than half of the tail is to the left of the confidence interval and half is to the right.
those tails are equal to half the alpha = .01 / 2 = .005.
s = sqrt(p * q / n) = sqrt(.15 * .85 / n) = standard error.
you will need to solve for n.
z = (x - m) / s = z-score formula.
at 99% two tail confidence interval, z = plus or minus 2.5758 rounded to 4 decimal places.
since the normal distribution is symmetric about the mean, you only have to solve for the high side margin of error.
the low side margin of error will be the same.
margin of error = (x - m) when you use the critical z-score.
z = (x - m) / s becomes 2.5758 = .06 / sqrt(.15 * .85 / n)
solve for sqrt(.15 * .85 / n) to get:
sqrt(.15 * .85 / n) = .06 / 2.5758.
square both sides of this equation to get.
.15 * .85 / n = (.06 / 2.2758)^2.
solve for n to get:
n = (.15 * .85) / (.06 / 2.2758)^2 = 234.9805748.
when n = 234.9805748, s becomes sqrt(.15 * .85 / 234.9805748) = .023293734 = standard of error.
z-score formula of z = (x - m) / s becomes 2.5758 = (x - .15) / .023293734.
.15 is the population mean proportion that was given.
solve for x to get x = .21.
the margin of error = (x - m) = .21 - .15) = .06, as desired.
the formula says that the sample size is 234.9805748.
this needs to be rounded to the next highest integer to get n = 235.
that's the smallest sample size that will get a margin of error less than .06.
that should be your solution.
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