Thread: confidence interval

A sample of 350 smokers were randomly selected, we know that 112 of the 350 have a preference for the A brand.

a) determine a 90% confidence interval.
b) determine how many individuals we need to have in the sample so the 90% confidence interval have an error of 0.02?

well a) is not a problem... but I can't get the right answer for b), lets see, isn't the error half the amplitude of the interval? and isn't half the amplitude given by the expression:1.65*(p'(1-p')/n)^(1/2) where n is the number of elements in the sample?

Hello,

the amplitude given by the expression:1.65*(p'(1-p')/n)^(1/2) where n is the number of elements in the sample?

Yes, something like that. So you have to solve for n in the equation : this expression = 0.02/2