Question

Wisconsin Public Radio wants to duplicate a survey conducted in 2011 that found that 68% of adults living in Wisconsin felt that the country was going in the wrong direction. How many people would need to be surveyed for a 90% confidence interval to ensure the margin of error would be less than 3%? Be sure to show all your work and round appropriately

Answer 1

655 people would need to be surveyed.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

In this question, we have that:

`\pi = 0.68`

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

How many people would need to be surveyed for a 90% confidence interval to ensure the margin of error would be less than 3%?

We need to survey n adults.

n is found when M = 0.03. So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.03 = 1.645\sqrt{(0.68*0.32)/(n)}`

`0.03√(n) = 1.645√(0.68*0.32)`

`√(n) = (1.645√(0.68*0.32))/(0.03)`

`(√(n))^(2) = ((1.645√(0.68*0.32))/(0.03))^(2)`

`n = 654.3`

Rounding up

655 people would need to be surveyed.