Katrina wants to estimate the proportion of adults who read at least 10 books last year. To do so, she obtains a simple random sample of 100 adults and constructs a $95 \%$ confidence interval. Matthew also wants to estimate the proportion of adults who read at least 10 books last year. He obtains a simple random sample of 400 adults and constructs a $99 \%$ confidence interval. Assuming both Katrina and Matthew obtained the same point estimate, whose estimate will have the smaller margin of error? Justify your answer. Whose estimate will have the smaller margin of error and why? A. Matthew's estimate will have the smaller margin of error because the sample size is larger and the level of confidence is higher. B. Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidence. C. Katrina's estimate will have the smaller margin of error because the sample size is smaller and the level of confidence is lower. D. Katrina's estimate will have the smaller margin of error because the lower level of confidence more than compensates for the smaller sample size.

Step 1 ：The margin of error in a confidence interval is determined by two factors: the sample size and the level of confidence. The larger the sample size, the smaller the margin of error, because a larger sample provides more information about the population and thus reduces uncertainty. On the other hand, a higher level of confidence increases the margin of error, because to be more confident that the true population parameter is within the interval, we need to make the interval wider.

Step 2 ：In this case, Matthew has a larger sample size than Katrina, which would tend to decrease his margin of error. However, he also has a higher level of confidence, which would tend to increase his margin of error. The question is which of these two effects is stronger.

Step 3 ：The formula for the margin of error in a confidence interval for a proportion is: \[E = Z \sqrt{\frac{{p(1-p)}}{n}}\] where: - E is the margin of error - Z is the Z-score, which depends on the level of confidence - p is the point estimate of the proportion - n is the sample size

Step 4 ：Since Katrina and Matthew obtained the same point estimate, p is the same for both of them. Therefore, the difference in their margins of error comes down to the difference in their Z-scores and sample sizes.

Step 5 ：The Z-score for a 95% confidence interval is approximately 1.96, and the Z-score for a 99% confidence interval is approximately 2.58. Therefore, Matthew's Z-score is larger, which would tend to increase his margin of error.

Step 6 ：However, Matthew's sample size is 4 times larger than Katrina's. Since the sample size appears in the denominator of the formula under the square root, quadrupling the sample size will halve the margin of error.

Step 7 ：Therefore, even though Matthew's higher level of confidence increases his margin of error, his larger sample size more than compensates for this, resulting in a smaller overall margin of error.

Step 8 ：So, the answer is B: Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidence.

Step 9 ：Final Answer: \(\boxed{\text{B. Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidence.}}\)