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 21 students how many hours they had slept the night before. The average of the sample was 4.72 with a standard deviation of 1.58. When estimating the average amount of sleep with a 90% confidence interval, what is the margin of error? Question 3 options: 1) 0.5947 2) 0.5933 3) 0.3448 4) 0.3602 5) 0.457

Answer 1

Solution :

Given that,

= 4.72

s = 1.58

n = 21

Degrees of freedom = df = n - 1 = 21 - 1 = 20

At 90% confidence level the t iis ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,20 = 1.725

Margin of error = E = t_/2,df * (s /n)

= 1.725 * (1.58 / 21)

= 0.5947

Margin of error = 0.5947

Option 1 ) 0.5947 is correct.