Question

A researcher wants to estimate the proportion of city residents who favor spending city funds to promote tourism. Would the standard error of the sample proportion \(\hat{p}\) be smaller for random samples of size \(n=100\) or random samples of size \(n=200 ?\)

Short answer

The standard error of the sample proportion \(\hat{p}\) would be smaller for random samples of size \(n=200\).

Step-by-step solution

Understand the formula for standard error

The formula to calculate the standard error of the sample proportion \(\hat{p}\) is \(\sqrt{\frac{p(1 - p)}{n}}\), where \(p\) is the sample proportion and \(n\) is the sample size.

Apply the formula for \(n=100\)

To find out whether the standard error of the sample proportion would be smaller for \(n=100\) or \(n=200\), apply this formula for both. Although the value of \(p\) is not given, it won't affect the comparison since it would be the same for both calculations. For \(n=100\), the standard error would be \(\sqrt{\frac{p(1 - p)}{100}}\).

Apply the formula for \(n=200\)

Applying the formula for \(n=200\), the standard error would be \(\sqrt{\frac{p(1 - p)}{200}}\). Notice that this value would be smaller than the standard error for \(n=100\) because \(n\) has a larger value in the denominator, making the whole fraction and therefore the standard error smaller.

Compare the standard errors

From Step 2 and Step 3, it can be concluded that the standard error of the sample proportion \(\hat{p}\) would be smaller for random samples of size \(n=200\) when compared to random samples of size \(n=100\). This is because the standard error decreases as the sample size \(n\) increases.