Question

A random sample of 20 students at a university showed an average age of 20.9 years and a sample standard deviation of 2.5 years. The 98% confidence interval for the true average age of all students in the university is?

Enter in the upper limit of your confidence interval.

Answer 1

Solution :

Given that,

= 20.9

s = 2.5

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,19 =2.539

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

= 2.539 * ( 2.5/ 20)

= 1.42

Margin of error = 1.42

The 98% confidence interval estimate of the population mean is,

+ E

20.9 + 1.42

= 22.32

The upper limit = 22.32