Question

An advertiser wishes to estimate the proportion of adults in Utah who already own a gym membership. He wishes to create a 90% confidence interval with a margin of error of 0.10. How many people must he sample?

Answer 1

Explanation:

Given that an advertiser wishes to estimate the proportion of adults in Utah who already own a gym membership.

Confidence level wanted = 90%

Critical value Z for 90% = 1.645

p is not known hence for maximum std error we know p =0.5

Let us take p =0.5

Margin of error =
`1.645*\sqrt{(pq)/(n) } =\\1.645*0.5((1)/(√(n) ) ) 0.10\\n=67.65\\n~68`

Sample size can be 68