Question

A random sample of 12 college graduates revealed that they worked an average of 5.5 years on the job before being promoted. The sample standard deviation was 1.1 years. Using the 95% level of confidence, what is the confidence interval for the population mean?

a. 5.5 plus-or-minus 2.40

b. 5.5 plus-or-minus 0.62

c. 5.5 plus-or-minus 0.70

d. 5.5 plus-or-minus 2.16

Answer 1

Solution :

Given that,

= 5.5

s =1.1

n = Degrees of freedom = df = n - 1 = 12- 1 = 11

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,11 =2.201 ( using student t table)

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

= 2.201* ( 1.1/ 12)

= 0.70

confidence interval for the population mean is

-+ E

5.5-+0.70

CORRECT OPTION d. 5.5 plus-or-minus 2.16