You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable estimate for the population proportion.

Question

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable estimate for the population proportion. You would like to be 98% confident that your estimate is within 0.1% of the true population proportion. How large of a sample size is required?

n =

Do not round mid-calculation. However, use a critical value accurate to three decimal places.

Answer 1

Hi Britt,

You may want to double check your error percentage of 0.1%, because sample sizes usually are not as high as this one is.

The formula to compute sample size is:

n=p(1-p)(z_a/2/e)²

Let's break this down:

n=sample size, which we are trying to find

p is the estimate for the population proportion, which we were not given here. When that is the case, it is appropriate to estimate p=0.5. Thus:

p=0.5

1-p=0.5

z_a/2 should actually read zsub alpha/2, but my computer does not have Greek letters. This is our critical value for a 98% confidence interval. It's available at z-score tables online. It's 2.330. This is somewhat atypical. Most confidence intervals I've seen are 95%; 1.96 critical value. Anyway:

z_a/2=2.330

e=acceptable error, this was given in the problem as 0.1%; converting to decimal:

e=0.001

Substituting:

n=(0.5)(0.5)(2.330/0.001)²

n=1357225

Again, this seems very large for a sample size, and it's impractical in the real world to sample that many people. The low error value is what drives it up, since that's in a quotient. Again, you may want to double check the error value you were given. I hope this helps.