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 4105 patients treated with the drug, 111 developed the adverse reaction of
nausea. construct a 90% confidence interval for the proportion of adverse reactions.
a) find the best point estimate of the population proportion p.
0.027
(round to three decimal places as needed.)
b) identify the value of the margin of error e.
e=
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the best - point estimate

The best - point estimate of the population proportion \(p\) is the sample proportion \(\hat{p}\).

$$ \hat{p}=\frac{x}{n} $$

where \(x = 111\) (number of patients with adverse reaction) and \(n=4105\) (total number of patients).

$$ \hat{p}=\frac{111}{4105}\approx0.027 $$

Step2: Find the critical value \(z_{\alpha/2}\) for a \(90\%\) confidence interval

For a \(90\%\) confidence interval, \(\alpha=1 - 0.90 = 0.10\), and \(\alpha/2=0.05\).
From the standard normal distribution table, \(z_{\alpha/2}=z_{0.05}\approx1.645\)

Step3: Calculate the margin of error \(E\)

The formula for the margin of error for a proportion is \(E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)
Substitute \(\hat{p}=0.027\), \(n = 4105\), and \(z_{\alpha/2}=1.645\)

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Answer:

a) The best - point estimate of the population proportion \(p\) is \(0.027\)
b) The margin of error \(E\approx0.004\)