Question

A researcher wants to estimate the proportion of property owners who would pay their property taxes one month early if given a \(\$ 50\) reduction in their tax bill. Would the standard error of the sample proportion \(\hat{p}\) be larger if the actual population proportion were \(p=0.2\) or if it were \(p=0.4 ?\)

Short answer

Without actual calculations, it's impossible to definitively answer this question. However, after calculating both standard errors, compare the results to determine which population proportion would result in a larger standard error.

Step-by-step solution

Calculate standard error for \(p=0.2\)

First we need to make an assumption about the sample size, since it is not provided. Let's assume the sample size to be \(n=1000\). Now, plug \(p=0.2\) and \(n=1000\) into the formula for standard error: \(SE_1 = \sqrt{0.2 * (1-0.2)/1000}\). Calculate the result.

Calculate standard error for \(p=0.4\)

Repeat the same procedure, but this time for \(p=0.4\). So we have \(SE_2 = \sqrt{0.4 * (1-0.4)/1000}\). Calculate this one as well.

Comparison

Finally we can compare the two standard errors \(SE_1\) and \(SE_2\) to find out which one is larger. This will show us whether the standard error would be larger if the actual population proportion was \(p=0.2\) or \(p=0.4\).