Question

A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval.

Answer 1

The 95% confidence interval for the true proportion of all voters in the state who favor approval is (0.4384, 0.5050).

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)`.

For this problem, we have that:

`n = 865, \pi = (408)/(865) = 0.4717`

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`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.4717 - 1.96\sqrt{(0.4717*0.5283)/(865)} = 0.4384`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.4717 + 1.96\sqrt{(0.4717*0.5283)/(865)} = 0.5050`

The 95% confidence interval for the true proportion of all voters in the state who favor approval is (0.4384, 0.5050).