Question

A school states that the standard deviation of tuition of their training program for various certifications taught there is $100. You want to prepare a 95% confidence interval of the average tuition for courses. You took a sample of 25 courses and found that the average tuition is $891. Prepare a 95% confidence interval for average tuition for certificate programs at the school.

Answer 1

Solution :

Given that,

= $891

s = $100

n = 25

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

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,25 =2.064

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

= 2.064 * (100 / 25) = 41.2800

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

- E < < + E

891 -41.2800 < <891 + 41.2800

849.7200 < <932.2808

(849.7200 , 932.2800)