Question

random sample found that forty percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017.

Answer 1

The 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017 is (0.274, 0.526).

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 = 100, \pi = 0.4`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.4 - 2.575\sqrt{(0.4*0.6)/(100)} = 0.274`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.4 + 2.575\sqrt{(0.4*0.6)/(100)} = 0.526`

The 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017 is (0.274, 0.526).