Question

Construct a confidence interval of the population proportion at the given level of confidence. x =860​, n =1100​, 96​% confidence.

a) lower bound of the confidence interval is: ________

b) upper bound of the confidence interval is:________

Answer 1

a) 0.7562

b) 0.8074

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 = 1100, \pi = (860)/(1100) = 0.7818`

96% confidence level

So
`\alpha = 0.04`, z is the value of Z that has a pvalue of
`1 - (0.04)/(2) = 0.98`, so
`Z = 2.054`.

a) The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.7818 - 2.054\sqrt{(0.7818*0.2182)/(1100)} = 0.7562`

B) The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.7818 + 2.054\sqrt{(0.7818*0.2182)/(1100)} = 0.8074`