Question

A random sample of 28 students at a particular university had a mean age of 22.4 years. If the standard deviation of ages for all university students is known to be 3.1 years ,Find a 90% confidence interval for the mean of all students at that university.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 22.4

Population standard deviation =    = 3.1
Sample size = n =28

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2    * ( /n)

= 1.645* (3.1 /  28 )

= 0.96
At 90% confidence interval
is,

- E < < + E
22.4 - 0.96 <   < 22.4+ 0.96

(21.44 , 23.36)