Question 1199227: A question on poll accuracy:
A poll with accuracy of plus or minus 3% means if 50% are in favor of a certain type of reform, the number could be between 47 and 53%.
To obtain an accuracy of plus or minus 1%, determine number needed to interview.
Not sure how to solve.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you are testing for the mean of a sample compared to the test mean.
the z-score formula is z = (x - m) / s
x is the score you are testing against the mean.
m is the mean
s is the standard error when you are looking for the mean of a sample.
when you are testing against a proportion, p is the proportion and q is equal to 1 - p).
the mean is equal to p.
the standard error is equal to sqrt(p * q / n)
p is the proportion.
q is equal to 1 - p
n is the sample size.
the margin of error is equal to (x - m)
when looking for the size of the sample needed, you test for the margin of error.
for your problem, you are looking for a margin of error of 1% = .01
this means that (x - m) = .01
the z-score formula becomes z = (x - m) / s which becomes z = .01 / s
s is the standard error = standard deviation / square root of sample size = sqrt(p * q / n)
n is the sample size
if the mean proportion is equal to .5, then p = .5 and q = (1 - p) = (1 - .5) = .5
you have p = .5 and q = .5 which makes standard error = sqrt(p * q / n) which makes standard error = sqrt(.5 * .5 / n) which makes standard error = sqrt(.25 / n).
your z-score formula becomes z = (1 - m) / sqrt(.25 / n)
xince (x - m0 = .01, your z-score formula becomes z = .01 / sqrt(.25 / n)
you also need to know what the confidence interval needs to be.
this was not specified.
the normal confidence interval is 95%, so we'll use that for now.
the critical z-score for a 95% confidence interval is plus or minus z = 1.96
you only need to work with 1.96 because the normal distribution curve is symmetric about the mean and the margin of error on one side of the mean will be equal to the margin or error on the other side of the mean as well.
your z-score formula becomes 1.96 = .01 / sqrt(.25/n)
solve for sqrt(.25/n) to get sqrt(.25/n) = .01 / 1.96
square both sides to get .25/n = (.01/1.96)^2
solve for n to get n = .25 / (.01/1.96)^2 = 9604.
if this was done correctly, your sample size needs to be at least 9604.
test this using the z-score formula.
you get z = (x-m) / s becomes z = .01 / sqrt(.5 * .5 / 9604).
this becomes z = .01 / sqrt(.25/9604) which becomes z = 1.96
this confirms the standard error calculation is correct foor 95% confidence interval.
as long as the standard error is the same and the confidence interval is 95%, you should get a margin of error of plus or minus .01.
the standard error is equal to sqrt(.5 * .5 / 9604) = .0051020408.
to test this, use the z-score formula again.
when the mean is .5, the z-score formula becomes:
1.96 = (x - .5) / .0051020408
1.96 is the critical z-score at 95% confidence interval.
.5 is the mean
.005..... is the standard error.
solve for x to get:
x = 1.96 * .0051020408 + .5 = .51
on the low side of the confidence interval, you get:
x = -1.96 * .0051020408 + .5 = .49
your margin of error is .01.
this is what it looks like on a graph.
as stated before, as long as the standard error is equal to .0051020408 and the confidence interval is 95%, the margin of error will be the same.
for example, if m is equal to .8, you get:
1.96 = (x - .8) / .0051020408
solve for x to get x = 1.96 * .0051020408 + .8 = .81
on the low side, you get x = -1.96 * .0051020408 + .8 = .79
note that if the sample size is not an integer, you need to round to the next higher integer for your final answer.
let me know if you have any questions.
theo
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