Question

A professor at a university wants to estimate the average number of hours of sleep students get during exam week. On the first day of exams, she asked 27 students how many hours they had slept the night before. The average of the sample was 3.79 with a standard deviation of 1.224. When estimating the average amount of sleep with a 99% confidence interval, what is the margin of error?

1) 0.3549

2) 0.6546

3) 0.5839

4) 0.2356

5) 0.6527

Answer 1

solution:- option 2) 0.6546

explanation:-

given that mean = 3.79 , standard deviation = 1.224

n = 27

degree of freedom df = n - 1 = 27 - 1 = 26

the value of 99% confidence with df from t table is t = 2.779

margin of error formula

=> t * standard deviation/sqrt(n)

=> 2.779 * 1.224/sqrt(27)

=> 0.6546