adults watching election coverage with 99% confidence and a margin of error of +or-7 minutes?

Question

What sample size would we need to estimate the average amount of time U.S. adults watching election coverage with 99% confidence and a margin of error of +or-7 minutes?

            Use z = 2.575.

                   A)  321 U.S. Adults

                   B) 108 U.S. Adults

                    C) 157 U.S. Adults

                    D)  243 U.S. Adults

Answer 1

To find the required sample size for estimating a population mean, we use the formula of:

n = (zσ/E)², where z equals to 2.575 (for 99% confidence, given), E would be the 7 miutes (margin of error), and σ would be 42 minutes (the population standard deviation given in this problem context).

Step 1: Substitue values

n = ((2.575(42))/7)²

n = (15.45)²

n ≈ 238.7

Step 2: Round Up

Sample size must always be rounded up, so that means that the answer 238.7 would be n = 243. So that means that the answer is D) 243 U.S. Adults. This sample size enures the average time U.S. adults spend watching election coverage can be estimated with 99% confidence and a ±7 minute margin of error.