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

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%).

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)`.

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

This means that:

`n = 250, \pi = (223)/(250) = 0.892`

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)} = 0.892 - 2.575\sqrt{(0.892*0.108)/(250)} = 0.841`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.892 + 2.575\sqrt{(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%).