Question

a researcher at a major hospital wishes to estimate the proportion of the adult population of the united states that has high blood pressure. how large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 5%?

a. 542
b. 164
c. 9
d. 271

Explanation:

Step1: Determine the confidence level and margin of error

The confidence level is \(90\%\), so \(\alpha = 1 - 0.90=0.10\), and \(\alpha/2 = 0.05\). From the standard normal distribution table, \(z_{\alpha/2}=z_{0.05} = 1.645\). The margin of error \(E = 0.05\). When no prior estimate of the proportion \(p\) is given, we use \(p = 0.5\) (this gives the maximum value of \(p(1 - p)\)).

Step2: Use the formula for sample size in proportion

The formula for sample size \(n\) when estimating a proportion is \(n=\frac{z_{\alpha/2}^{2}\times p(1 - p)}{E^{2}}\). Substitute \(z_{\alpha/2}=1.645\), \(p = 0.5\), \(1-p=0.5\), and \(E = 0.05\) into the formula:

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Answer:

D. 271