Question

A 2015 Gallup survey asked respondents to consider several different foods and beverages and to indicate whether these were things that they actively tried to include in their diet, actively tried to avoid in their diet, or did not think about at all. Of the 1009 adults surveyed, 616 indicated that they actively tried to avoid drinking regular soda or pop. Assume that the sample was an SRS.

Suppose we computed a large sample 99% confidence interval for the proportion of all American adults who actively try to avoid drinking regular soda or pop.
This 99% confidence interval:
would have a larger margin of error than a 90% confidence interval.
would have a smaller margin of error than a 90% confidence interval.
could have either a smaller or a larger margin of error than a 90% confidence interval. This varies from sample to sample.
would have the exact same margin of error as a 90% confidence interval.

Answer 1

would have a larger margin of error than a 90% confidence interval.

Explanation:

Margin of error in statistics can be defined as a small amount that is allowed for in case of miscalculation or change of circumstances.

For a statistical data margin of error can be expressed as;

M.E = zr/√n

Where;

Given that;

r = Standard deviation

z = z score at a particular confidence interval

n = sample size

z at 99% = 2.58

z at 90% = 1.645

Since z at 99% is higher than z at 90% Confidence interval, the Margin of error M.E at 99% confidence interval will be higher than that of 90% confidence interval.