Please Help Me!!!

Question

You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you​ survey? Assume that you want to be

95​% confident that the sample percentage is within 4.5 percentage points of the true population percentage. Complete parts​ (a) and​ (b) below.

Answer 1

Hello Nicole!

It looks like we're missing some information... What are the parts a) and b) of the problem?

Here's what we do know:

We're asked how many passengers we must survey, i.e. we want to know what sample size, n, we need.

We're dealing with a confidence interval and we want it to be within 4.5% of the real value, i.e. our maximal margin of error, E, is 4.5%.

Our critical value for this confidence level of 95%, z_c, is 1.96 (you'll want to memorize those common z_c values.)

Finally, because our survey will look at an "either/or" situation and our resultant rate is in a percentage value, we know we can treat this as a binomial experiment problem. And I expect that your parts a) & b) are for you to find n when you are a) given a point estimate for p, and b) NOT given point estimate for p.

To find our sample size n for a confidence interval, we have the following formulas:

n=pq(z_c/E)² when we know our point estimate for p, and

n=(1/4)(z_c/E)² when we DON'T know our point estimate for p,

where n is our sample size, p is our point estimate, q is 1-p, and E is our maximal margin of error.

Also remember, anytime you're estimating for n, you'll want to round up your answer to the nearest whole number.

Hope this helps!