Question

2)Adverse drug reactions to legally prescribed medications are among the leading causes of drug-related death in the United States. Suppose you investigate drug- related deaths in your city and find that 223 out of 250 deaths were caused by legally prescribed drugs. Construct a 99% confidence interval for the proportion of drug-related deaths that were caused by legally prescribed drugs. Round to the thousandths place then convert your interval into a percent.​

Answer 1

Answer:

The 99% confidence interval for the proportion of drug-related deaths that were caused by legally prescribed drugs is (0.841, 0.943).

As a percentage, the confidence interval is (84.1%, 94.3%).

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 $1 - \frac{\alpha}{2}$.

223 out of 250 deaths were caused by legally prescribed drugs.

This means that:

$n = 250, \pi = \frac{223}{250} = 0.892$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.892 - 2.575\sqrt{\frac{0.892*0.108}{250}} = 0.841$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.892 + 2.575\sqrt{\frac{0.892*0.108}{250}} = 0.943$

The 99% confidence interval for the proportion of drug-related deaths that were caused by legally prescribed drugs is (0.841, 0.943).

As a percent:

Multiply the proportions by 100%.

0.841*100% = 84.1%

0.943*100%=  94.3%

As a percentage, the confidence interval is (84.1%, 94.3%).