Question

Use the following information:

n = 158, x = 68, a = 0.05
To find how large of a sample is needed in order to have a margin of error of 6%.
NOTE: Round your answer to the nearest WHOLE number?

Answer 1

We can use the formula for margin of error:

ME = z*(sigma/sqrt(n))

where z is the z-score corresponding to the confidence level (95% confidence level corresponds to z = 1.96), sigma is the population standard deviation (unknown in this case), and n is the sample size.

To find the sample size needed for a margin of error of 6%, we can rearrange the formula as follows:

n = (z*sigma/ME)^2

Since sigma is unknown, we can use the sample standard deviation as an estimate. We have x = 68 and n = 158, so we can calculate the sample standard deviation as follows:

s = sqrt((1/n)*sum(xi - x)^2) where xi represents each observation in the sample

s = sqrt((1/158)*sum((xi - 68)^2))

s = 16.38

Now we can substitute the values into the formula for n:

n = (z*sigma/ME)^2

n = (1.96*16.38/0.06)^2

n ≈ 2,285.22

Rounding to the nearest whole number, we need a sample size of 2,285.