Question

1. 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. SHOW WORK

Answer 1

Given that,

= 22.4

s =3.1

n = 28

Degrees of freedom = df = n - 1 =28 - 1 = 27

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,27 = 1.703    ( using student t table)

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

= 1.703 * (3.1 / 28/) = 0.9977

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

- E < < + E

22.4 - 0.9977 < <22.4 + 0.9977

21.4023 < < 23.3977

( 21.4023,23.3977 )