Question

In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weigh more than 180 pounds? Assume that we want 96% confidence that the error is no more than 4 percentage points.

Answer 1

:'( ahhhhhhhhhhhhhhhhhhhhhhhhhhh

Answer 2

For large sample confidence intervals about the mean you have:

xBar ± z * sx / sqrt(n)

where xBar is the sample mean
z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α
sx is the sample standard deviation
n is the sample size

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For sample size calcualtions we need the following. The width of the interval, from one end point to the center is:

z * sx / sqrt(n) = w

and solving for n gives:

n = (z * sx / w) ^ 2

remember that n needs to be an integer. Always take the ceiling, i.e., round up. If you round down then the width of the interval will not be correct, it will be too wide. By rounding up, the interval will be more narrow than asked for, but this is a good thing. It means there is more precision in the estimate.


here we have

z = 2.05
w = 0.04 * 152 = 6.08

n = (z * sx / w) ^ 2
n = (2.05 * 26 / 6.08)^2
n = 76.8506

n = 77 since n must be an integer. Always round up!
(Not my work, it would take a long time to go step by step) Thanks Yahoo!

Answer 3

Thank you so much!

Answer 4

You are very welcome!

Answer 5

Btw.. you would need to survey 16,102 folks. I don't think I mentioned that.

Answer 6

@rokotherodent How is the requirement for the error being no more than 4 percentage points taken into your calculation?

Answer 7

Hey kropot, z=2.05, e = 0.04, n= [2.05]^2*.25/0.04^2 = 656.6, n= 657, is that right? since p-hat is not present, i replace with .5*.5 or .25??

Answer 8

At 96% confidence that the mean from a sample of size n contains the population mean, the confidence interval is:
\[(xbar-2.054\frac{\sigma}{\sqrt{n}})\to (xbar-2.054\frac{\sigma}{\sqrt{n}})\]
Subtracting the lower limit from the upper limit of the confidence interval gives:
\[4.108\frac{\sigma}{\sqrt{n}}\]
4% of the population mean = (152 * 4)/100 = 6.08 pounds
\[6.08=4.108\frac{\sigma}{\sqrt{n}}\]
Solving for n gives a sample size of 309.