An environmental research firm wants to estimate the proportion of cars in Jackson County, Missouri that would be considered SUVs. Using the county's vehicle registration records they randomly select 220 vehicles. From that list they determined that 90 of them were SUVs. The sample proportion is 0.409, and the 95% confidence interval for the proportion of cars in Jackson County that are SUVs is (0.344, 0.474).
Select all of the following which would produce a confidence interval with a larger margin of error.
If you created a 90% confidence interval instead of the 95% confidence interval the margin of error would? Answer 1
If you had a random sample of 98 vehicles instead of 220 the margin of error would? Answer 2
If you created a 99% confidence interval instead of the 95% confidence interval the margin of error would?
Answer
If you had a random sample of 320 vehicles instead of 220 the margin of error would?

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

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is given by:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

This means that a higher confidence level(higher value of z) leads to a larger margin of error, and a larger sample(higher value of n) leads to a lower margin of error.

Select all of the following which would produce a confidence interval with a larger margin of error.

Higher confidence level or smaller sample. So the correct options are:

If you had a random sample of 98 vehicles instead of 220.

If you created a 99% confidence interval instead of the 95% confidence interval.