Question

6.Suppose the Gallup Organization wants to estimate the population proportion of those who think there should be a law that would ban the possession of handguns. In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law. How large a sample size is needed to be 95% confident with a margin of error of E

Answer 1

A sample of
`n = ((1.96√(0.3696*0.6304))/(E))^2` is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

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 of:

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

In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law.

This means that
`n = 1012, \pi = (374)/(1012) = 0.3696`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

How large a sample size is needed to be 95% confident with a margin of error of E?

A sample size of n is needed, and n is found when M = E.

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

`E = 1.96\sqrt{(0.3696*0.6304)/(n)}`

`E√(n) = 1.96√(0.3696*0.6304)`

`√(n) = (1.96√(0.3696*0.6304))/(E)`

`(√(n))^2 = ((1.96√(0.3696*0.6304))/(E))^2`

`n = ((1.96√(0.3696*0.6304))/(E))^2`

A sample of
`n = ((1.96√(0.3696*0.6304))/(E))^2` is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.