A sample of 1000 college students at NC State University were randomly selected for a survey. Among the survey participants, 102 students suggested that classes begin at 8 AM instead of 8:30 AM. The sample proportion is 0.102.What is the margin of error for a 99% confidence interval for this sample? Give your answer to three decimal places.

The margin of error is 0.025 = 2.5 percentage points

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

99% confidence level

So , z is the value of Z that has a pvalue of , so .

The sample proportion is 0.102.

1000 college students at NC State University were randomly selected for a survey.

So

What is the margin of error for a 99% confidence interval for this sample?

step-by-step explanation: