Question

A news station would like to conduct an exit poll to determine the likelihood that a highly debated amendment will receive enough support to pass. The news station plans to construct a 90 percent confidence interval to estimate the proportion of voters supporting the amendment. Whatg expressions would give the smallest sample size that will result in a margin of error of no more than 3 percentage points?

Answer 1

The expression is
`n = ((1.645*0.5)/(0.03))^2`

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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is of:

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

90% confidence level

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

What expression would give the smallest sample size that will result in a margin of error of no more than 3 percentage points?

We have to find n for which M = 0.03.

We have no prior estimate for the proportion, so we use
`\pi = 0.5`. So

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

`0.03 = 1.645\sqrt{(0.5*0.5)/(n)}`

`0.03√(n) = 1.645*0.5`

`√(n) = (1.645*0.5)/(0.03)`

`(√(n))^2 = ((1.645*0.5)/(0.03))^2`

`n = ((1.645*0.5)/(0.03))^2`

The expression is
`n = ((1.645*0.5)/(0.03))^2`