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 98% confident that the sample proportion will not differ from the true proportion by more than 2%?

a. 30
b. 2,401
c. 6,787
d. 3,394

Explanation:

Step1: Determine the z - value

For a 98% confidence level, the significance level \(\alpha=1 - 0.98=0.02\). Then \(\alpha/2=0.01\). Using the standard normal distribution \(z_{\alpha/2}=z_{0.01}\approx 2.33\) (from the standard normal table).

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 given, we use \(p = 0.5\) (this maximizes the value of \(p(1 - p)\)). The margin of error \(E = 0.02\).
Substitute the values into the formula:

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Answer:

D. 3,394