For a sample proportion, margin of error depends on the confidence level, the sample size, and the sample proportion itself. Which of these can be ruled out using the information in the problem?

Once you know that both samples use the same method, same size, and the same 95% level, only the estimated percentages (18% vs. 49%) are different. How does being near the middle (around half) versus near an extreme (near 0% or 100%) affect variability of a proportion?

In Desmos, type the general formula for each sample’s standard error part (you can ignore $z^*$ since it is the same for both):

• For sample C: sqrt(0.18*(1-0.18)/900)
• For sample D: sqrt(0.49*(1-0.49)/900) Desmos will display numerical values for each expression.

Look at the two outputs. The expression with the larger value corresponds to the sample with greater sampling variability and therefore a larger margin of error at the same confidence level and sample size. Use this comparison to decide which explanation best matches the situation.

For a sample proportion, the (approximate) margin of error at a given confidence level is

where:

• $z^*$ is determined by the confidence level (for 95%, $z^* \approx 1.96$),
• $\hat p$ is the sample proportion,
• $n$ is the sample size.

So margin of error changes when the confidence level, sample size, or the value of $\hat p$ changes.

From the table:

• Both samples C and D have the same sample size $n = 900$.
• The problem says these are 95% margins of error, so the confidence level (and therefore $z^*$) is the same for both.

That means the difference in margins of error must come from the difference in the estimated percent (the value of $\hat p$):

• Sample C: $\hat p = 0.18$ (18%)
• Sample D: $\hat p = 0.49$ (49%)

Look at the part inside the square root: $\hat p(1-\hat p)$. This product is a measure of the variability of the sample proportion.

If you test values between 0 and 1, $\hat p(1-\hat p)$ is smallest when $\hat p$ is very close to 0 or 1, and largest when $\hat p$ is around $0.5$ (50%).

So, for the same sample size and confidence level, the margin of error is largest when the estimated proportion is near 50%.

Sample C's estimate (18%) is far from 50%, while sample D's estimate (49%) is very close to 50%. Because variability $\hat p(1-\hat p)$ and thus the margin of error are greatest near 50%, it makes sense that the margin of error for sample D is larger.

Therefore, the most appropriate reason is: The estimated percentage in sample D is closer to 50%, where sampling variability for proportions is greatest.