Question

A random sample found that 30% of 150 Americans were satisfied with the gun control laws in 2018. Compute a 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 Fill in the blanks appropriately. A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is ( , ) (round to 3 decimal places)

Answer 1

A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is (0.213, 0.387).

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 = 150, \pi = 0.3`

98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a pvalue of
`1 - (0.02)/(2) = 0.99`, so
`Z = 2.327`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.3 - 2.327\sqrt{(0.3*0.7)/(150)} = 0.213`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.3 + 2.327\sqrt{(0.3*0.7)/(150)} = 0.387`

A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is (0.213, 0.387).