Question

A random sample of 432 voters revealed that 100 are in favor of a certain bond issue. A 95 percent confidence interval for the proportion of the population of voters who are in favor of the bond issue is A 100 + 1.96 0.5(0.5) 432 100 + 1.645 0.5(0.5) 432 100 + 1.96 0.231(0.769) 432 0.231 +1.96 0.231(0.769) 432 0.231(0.709) 0.231 +1.6451 432​

Answer 1

The 95% confidence interval is
`0.231 \pm 1.96\sqrt{(0.231*0.769)/(432)}`

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)`.

A random sample of 432 voters revealed that 100 are in favor of a certain bond issue.

This means that
`n = 432, \pi = (100)/(432) = 0.231`

95% confidence level

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

Confidence interval:

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

`0.231 \pm 1.96\sqrt{(0.231*0.769)/(432)}`

The 95% confidence interval is
`0.231 \pm 1.96\sqrt{(0.231*0.769)/(432)}`