Question

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The 90% confidence interval for the true average age of all students in the university is

		19.50 to 20.50
		19.36 to 20.38
		19.18 to 20.49
		19.02 to 20.59

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 20

sample standard deviation = s = 4

sample size = n = 64

Degrees of freedom = df = n - 1 = 64 -1 =63

At 90% confidence level

= 1-0.90% =1-0.9 =0.10

/2 =0.10/ 2= 0.05

t/2,df = t0.05,63 = 1.669

t /2,df = 1.669

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

=1.669 * ( 4/ 64)

Margin of error = E =0.835

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

- E < <  + E

20 - 0.835< < 20 +0.835

19.18 < < 20.49

(19.18,20.49)

Answer = 19.18 to 20.49