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 4107 patients treated with the drug, 111 developed the adverse reaction of nausea. construct a 90% confidence interval for the proportion of adverse reactions
0.027
a) find the best point estimate of the population proportion p
(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 \(x = 11\) (number of patients who developed side - effects) and \(n=410\) (total number of patients).

$$ \hat{p}=\frac{11}{410}\approx0.027 $$

Step2: Find the critical value \(z_{\alpha/2}\)

For a \(90\%\) confidence level, \(\alpha=1 - 0.90=0.10\), 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 \(\hat{p}=0.027\), \(n = 410\), 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 \(E = 0.013\)