Question

what percentage of eligibible american vote? In 2008, a random sample of 500 american adults was take and we found that 68% of them voted. How many of the 500 adults in the sample voted? Now construct a 95% confidence interval estimate of the population percent of Americans that vote and write a sentence to explain the confidence interval.

Answer 1

340 of the adults in the sample voted.

The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.

Explanation:

In 2008, a random sample of 500 american adults was take and we found that 68% of them voted. How many of the 500 adults in the sample voted?

This is 68% of 500.

So 0.68*500 = 340.

340 of the adults in the sample voted.

Now construct a 95% confidence interval estimate of the population percent of Americans that vote and write a sentence to explain the confidence interval.

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`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 = 500, p = 0.68`

95% confidence interval

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)} = 0.68 - 1.96\sqrt{(0.68*0.32)/(500)} = 0.6391`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.68 + 1.96\sqrt{(0.68*0.32)/(500)} = 0.7209`

The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.