Question

a private opinion poll is conducted for a politician to determine what proportion of the population favors decriminalizing marijuana possession. 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 3%?

a. 456
b. 752
c. 14
d. 1,504

Explanation:

Step1: Determine the critical value

For a 90% confidence level, the significance level \(\alpha = 1 - 0.90=0.10\), and \(\alpha/2=0.05\). The critical value \(z_{\alpha/2}\) is such that \(P(Z > z_{\alpha/2})=\alpha/2 = 0.05\). From the standard normal table, \(z_{\alpha/2}=z_{0.05}\approx1.645\).

Step2: Use the formula for sample size in proportion

The formula for sample size \(n\) when estimating a proportion is \(n=\frac{z_{\alpha/2}^{2}\cdot p(1 - p)}{E^{2}}\). When no prior estimate of \(p\) (the proportion) is available, we use \(p = 0.5\) (this gives the maximum value of \(p(1 - p)\)). The margin of error \(E = 0.03\).
Substitute \(z_{\alpha/2}=1.645\), \(p = 0.5\), and \(E=0.03\) into the formula:

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Answer:

B. 752