Question

Suppose we want to estimate the mean salary μ of all college graduates. We take a sample of 25 graduates and the sample average is $39,000 with a sample standard deviation of $10,000. We construct a 95% confidence interval for the true average salary. What is the upper bound of the confidence interval i.e. what is the upper confidence limit?

$39,000 $43,128 $41,000 $42.422

Answer 1

Solution :

Given that,

= $39000

s =$10000

n = Degrees of freedom = df = n - 1 = 25- 1 = 24

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  =0.05

t ,df = t0.05,24 = 1.711 ( using student t table)

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

= 1.711* (10000 / 25)

= 3422

The 95% confidence interval estimate of the population mean is,

+ E

39000 + 3422

42422

upper bound=42422