Question

question 4
1 pts
two researchers, researcher a and researcher b, independently survey random samples of 1,000 and 500 college students, respectively, to estimate the proportion of all college students who prefer online courses to in - person courses. both researchers use the data they obtain to construct a 99% confidence interval. these intervals are given in the table below. assuming the researchers did not make any calculation errors when constructing their confidence intervals, which one of the following statements is correct?

researcher
99% confidence interval
a
0.480 to 0.562
b
0.447 to 0.563

researcher as interval is more accurate than researcher bs interval since researcher a surveyed a larger sample.
the smaller sample size is the one that resulted in a narrower confidence interval width.
we have enough information here to verify that the intervals were computed correctly.
the widths of the two intervals are different because of differences in population size.
changing the level of confidence to 95% will result in wider intervals.

Explanation:

Step1: Calculate the width of each confidence interval

The width of a confidence interval is calculated as \( \text{Upper limit}-\text{Lower limit} \).
For Researcher A: \( 0.562 - 0.480=0.082 \)
For Researcher B: \( 0.563 - 0.447 = 0.116 \)

Step2: Analyze each option

• Option 1:

The formula for the margin of error \( E = z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\) (where \(z\) is the critical value, \(\hat{p}\) is the sample proportion and \(n\) is the sample size). A larger sample size (\(n\)) leads to a smaller margin of error (more accurate). Researcher A has \(n = 1000\) and Researcher B has \(n=500\). Since \(n_A>n_B\), Researcher A's interval is more accurate.

• Option 2:

As calculated above, Researcher A (larger \(n\)) has a narrower width (\(0.082\)) compared to Researcher B (\(0.116\)). So, this option is wrong.

• Option 3:

To verify the intervals, we need \(\hat{p}\) (sample proportion) which is \(\frac{\text{Lower limit}+\text{Upper limit}}{2}\). But we don't have information about the sampling method (other than random) and \(\hat{p}\) values to check if \(E = z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\) holds exactly. So, we can't verify.

• Option 4:

The formula for margin of error does not involve population size (\(N\)) (when \(n/N\leq0.1\)). Here \(n = 1000\) and \(n = 500\) (assuming \(N\) is large, as we are dealing with college students in general). So, population size is not the reason for different widths.

• Option 5:

A lower confidence level (e.g., \(95\%\) instead of \(99\%\)) leads to a smaller \(z\) - value. Since \(E = z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\), the margin of error decreases and the interval gets narrower.

Answer:

Researcher A's interval is more accurate than Researcher B's interval since Researcher A surveyed a larger sample.