Question

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. Katrina's estimate will have the smaller margin of error because the sample size is smaller and the level of confidence is lower.

C. Katrina's estimate will have the smaller margin of error because the lower level of confidence more than compensates for the smaller sample size.

D. Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidence

Answer 1

Answer:

D. 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-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence interval $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is the division by 2 of the upper level of the interval $\pi + z\sqrt{\frac{\pi(1-\pi)}{n}}$ by the lower level $\pi - z\sqrt{\frac{\pi(1-\pi)}{n}}$, so this is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

Both Katrina and Matthew are working with the same population, so $\pi$ is equal for both.

Katrina

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. This means that she has $z = 1.96$ and $n = 100$. So her margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = (1.96)\sqrt{\frac{\pi(1-\pi)}{100}}$

$M = 0.196\sqrt{\pi(1-\pi)}$

Matthew

He obtains a simple random sample of 400 adults and constructs a​ 99% confidence interval.

This means that, for Matthew, $z = 2.575, n = 400$. So his margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = (2.575)\sqrt{\frac{\pi(1-\pi)}{20}}$

$M = 0.12875\sqrt{\pi(1-\pi)}$

Higher confidence levels usually led to higher margins of error. However, Matthew's larger sample size compensates his higher confidence level.

So the correct answer is:

D. Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidence