Question

For a random sample of 18 recent business school graduates beginning their first job, the mean starting salary was found to be $31,500, and the sample standard deviation was $6,500. Assuming the population is normally distributed, calculate the lower confidence limit (LCL) of the population mean with α = 0.025.

Answer 1

solution

n =18

Degrees of freedom = df = n - 1 = 18- 1 = 17

confidence level the t is ,

  = 0.025

t /2,df = t0.025,17 = 2.110 ( using student t table)

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

= 2.110 * ( 6500/ 18)

= 3232.6565

The confidence interval estimate of the population mean is,

- E < < + E

31500- 3232.6565

28267.3435

lower limit=28267.3435