In: Statistics and Probability
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?
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 = t0.05,84 = 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)