Question

In March of 2018, a survey asked 801 US adults whether they had at least one subscription to a video-streaming service. Of the 801 participants in the survey, 457 indicated they subscribed to at least one video-streaming service. What is the correct 95 percent confidence interval for the proportion of all US adults who would say they subscribe to a video-streaming service?

Answer 1

The 95% confidence interval for the proportion of all US adults who would say they subscribe to a video-streaming service is (0.5362, 0.6048).

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)`.

Of the 801 participants in the survey, 457 indicated they subscribed to at least one video-streaming service.

This means that
`n = 801, \pi = (457)/(801) = 0.5705`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.5705 - 1.96\sqrt{(0.5705*0.4295)/(801)} = 0.5362`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.5705 + 1.96\sqrt{(0.5705*0.4295)/(801)} = 0.6048`

The 95% confidence interval for the proportion of all US adults who would say they subscribe to a video-streaming service is (0.5362, 0.6048).