Question

A newspaper conducted a statewide survey concerning a proposal to raise taxes to prevent budget cuts to education. The newspaper took a random sample (assume it is an SRS) of 1200 registered voters and found that 580 would vote to raise taxes. Let p represent the proportion of registered voters in the state that would vote to raise taxes.Reference: Ref 20-4How large a sample n would you need to estimate p with margin of error 0.01 with 90% confidence

Answer 1

A sample of 6758 would be needed.

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)`.

The margin of error is:

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

The newspaper took a random sample (assume it is an SRS) of 1200 registered voters and found that 580 would vote to raise taxes.

This means that
`\pi = (580)/(1200) = 0.4833`

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 large a sample n would you need to estimate p with margin of error 0.01 with 90% confidence?

This is n for which M = 0.01. So

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

`0.01 = 1.645\sqrt{(0.4833*0.5167)/(n)}`

`0.01√(n) = 1.645√(0.4833*0.5167)`

`√(n) = (1.645√(0.4833*0.5167))/(0.01)`

`(√(n))^2 = ((1.645√(0.4833*0.5167))/(0.01))^2`

`n = 6757.5`

Rounding up:

A sample of 6758 would be needed.