Question 1188020: A federal report indicated that 27% of children ages 2 to 5 years had a good diet—an increase
over previous years. How large a sample is needed to estimate the true proportion of children with good diets within 2% with 95% confidence?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! p = .27
q = .73
s = sqrt(.27 * .73 / n)
p is the mean proportion.
q is 1 minus the mean proportion.
i think that s is the standard error of the mean proportion.
that's because it is used as the standard error in the calculation, even though they call it the standard deviation.
very confusing, but it works.
at 95% two tailed confidence interval, the critical alpha is .025 on each end.
the critical z-score becomes plus or minus 1.96.
the z-score formula is z = (x - m) / s
z is the z-score.
x - m is the sample mean proportion minus the population mean proportion.
s is the standard error.
since we want the margin of error to be plus or minus 2% of the mean, then (x - m) must be equal to .02 * the mean proportion.
.02 * .27 = .0054.
the z-score formula becomes:
1.96 = .0054 / sqrt(.27 * .73 / n)
multiply both sides of this equation by sqrt(.27 * .73 / n) and divide both sides of this equation by 1.96 to get:
sqrt(.27 * .73 / n) = .0054 / 1.96
square both sides of this equation to get:
.27 * .73 / n = (.0054 / 1.96) ^ 2
multiply both sides of this equation by n and divide both sides of this equation by .(.0054 / 1.96) ^ 2 to get:
.27 * .73 / (.0054 / 1.96) ^ 2 = n
solve for n to get:
n = 25966.37037.
that is the sample size required for the margin of error to be plus or minus 2% of the mean.
with that sample size, the standard error becomes sqrt(.27 * .73 / that) = .002755102.
with a mean of .27 and that standard error and a critical z-score of 1.96, the z-score formula becomes:
1.96 = (x - .27) / .002755102.
solve for x to get:
x = 1.96 * .002755102 + .27 = .2754.
the difference between .2754 and .27 is .0054.
.0054 / .27 = .02, as predicted with that sample size.
that's on the high side of the confidence interval.
on the low side of the confidence interval, the z-score formula becomes:
-1.96 = (x - .27) / .002755102.
solve for x to get:
x = -1.96 * .002755102 + .27 = .2646.
the difference between .27 and .2646 is .0054 again, which, when divided by .0025 = .02.
if you multiply the population mean proportion by the sample size and you multiply the standard error by the sample size, your formula of:
p = .27
q = 1 - .27 = .73
s = sqrt(p * q / n) becomes:
m = .27 * n
s = sqrt(p * q * n)
n = 25966.37037, same as we calculated above.
m = .27 * n = 7010.92
s = sqrt(.27 * .73 * 25966.37037) = 71.54.
on the high side of the confidence interval, the critical z-score formula is:
1.96 = (x - 7010.92) / 71.54
solve for x to get:
x = 1.96 * 71.54 + 7010.92 = 7151.1384
7151.1384 minus 7010.92 = 140.2184 / 7010.92 = .02 which is off the mean by 2% of the mean.
on the low side of the confidence interval, the critical z-score formula is:
-1.96 = (x - 7010.92) / 71.54.
solve for x to get:
x = -1.96 * 71.54 + 7010.92 = 6870.7016
7010.92 minus 6870.7016 = 140.2184 / 7010.92 = .02 which is off the mean by 2% of the mean.
the formulas works.
your solution is that the sample size needs to be at least equal to 25966.37037 so that the margin of error will be at most 2% of the mean.
a sample size larger than that will give you a margin of error less than 2% of the mean.
for example, if the sample size is 30,000, then:
s = sqrt(.27 * .73 * 30000) = 76.896 rounded to 3 decimal places.
m = .27 * 30,000 = 8100.
the high side value of x will be 76.896 * 1.96 + 8100 = 8250.716 rounded to 3 decimal places.
8250.716 minus 8100 = 150.716 rounded to 3 decimal places
150.716 / 8100 = .0186 which rounds to .019 which is less than .02.
the margin of error is less than 2% of the mean in this case, because the sample size was larger than 25,966.37037.
let me know if you have any questions.
theo
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