Question

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n=83, x=45, 98 percent

Could you please explain the steps and how to get to an answer? Thank you!

Answer 1

The 98% confidence interval for the population proportion p is (0.4149, 0.6695).

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 = 83, \pi = (45)/(83) = 0.5422`

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)} = 0.5422 - 2.327\sqrt{(0.5422*0.4578)/(83)} = 0.4149`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.5422 + 2.327\sqrt{(0.5422*0.4578)/(83)} = 0.6695`

The 98% confidence interval for the population proportion p is (0.4149, 0.6695).