Question

From a random sample of 85 teens, it is found that on average they spend 31.8 hours each week online with a standard deviation of 5.91 hours. What is the 90% confidence interval for the amount of time they spend online each week?

Answer 1

Solution :

Given that,

= 31.8

s = 5.91

n = 85

Degrees of freedom = df = n - 1 = 85 - 1 = 84

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,84 = 1.663

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

= 1.663 * (5.91 / 85)

= 1.07

The 90% confidence interval estimate of the population mean is,

- E < < + E

31.8 - 1.07 < < 31.8 + 1.07

30.73 < < 32.87

(30.73 , 32.87)