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

Answer:

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).

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

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

This means that $n = 4547, \pi = \frac{114}{4547} = 0.0251$

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 $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.0251 - 2.575\sqrt{\frac{0.0251*0.9749}{4547}} = 0.0191$

The upper limit of this interval is:

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

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