Question

In a random sample of 200 people, 154 said that they watched educational television. Find the 90% confidence interval of the true proportion of people who watched educational television.

Answer 1

The 90% confidence interval of the true proportion of people who watched educational television is (0.7117, 0.8283)

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 = 200, \pi = (154)/(200) = 0.77`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.77 - 1.96\sqrt{(0.77*0.23)/(200)} = 0.7117`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.77 + 1.96\sqrt{(0.77*0.23)/(200)} = 0.8283`

The 90% confidence interval of the true proportion of people who watched educational television is (0.7117, 0.8283)