Question

Use the sample data and confidence level given below to complete parts​ (a) through​ (d). A drug is used to help prevent blood clots in certain patients. In clinical​ trials, among 4547 patients treated with the​ drug, 114 developed the adverse reaction of nausea. Construct a 99​% confidence interval for the proportion of adverse reactions. ​a) Find the best point estimate of the population proportion p.

Answer 1

The best point estimate of the population proportion p is 0.0251.

The 99​% confidence interval for the proportion of adverse reactions is (0.0191, 0.0311).

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)`.

In clinical​ trials, among 4547 patients treated with the​ drug, 114 developed the adverse reaction of nausea.

This means that
`n = 4547, \pi = (114)/(4547) = 0.0251`

The best point estimate of the population proportion p is 0.0251.

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)} = 0.0251 - 2.575\sqrt{(0.0251*0.9749)/(4547)} = 0.0191`

The upper limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.0251 + 2.575\sqrt{(0.0251*0.9749)/(4547)} = 0.0311`

The 99​% confidence interval for the proportion of adverse reactions is (0.0191, 0.0311).