Question

In a survey, 376 out of 1,078 US adults said they drink at least 4 cups of coffee a day. Find a point estimate (P) for the population proportion of US adults who drink at least 4 cups of coffee a day, then construct a 99% confidence interval for the proportion of adults who drink at least 4 cups of coffee a day.

Answer 1

The point estimate is 0.3488.

The 99% confidence interval for the proportion of adults who drink at least 4 cups of coffee a day is (0.3114, 0.3862).

Explanation:

In a sample with a number n of people surveyed with a point estimate 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 = 1078, \pi = (376)/(1078) = 0.3488`

The point estimate is 0.3488.

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.3488 - 2.575\sqrt{(0.3488*0.6512)/(1078)} = 0.3114`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.3488 + 2.575\sqrt{(0.3488*0.6512)/(1078)} = 0.3862`

The 99% confidence interval for the proportion of adults who drink at least 4 cups of coffee a day is (0.3114, 0.3862).