Question

For a random sample of 17 recent business school graduates beginning their first job, the mean starting salary was found to be $43,500, and the sample standard deviation was $8,500. Assuming the population is normally distributed, calculate the lower confidence limit (LCL) of the population mean with α = 0.10.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 43500

sample standard deviation = s = 8500

sample size = n = 17

Degrees of freedom = df = n - 1 = 17 - 1 = 16

At 95% confidence level the t is ,

= 0.10

t _,df = t_0.010,16 = 1.337

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

= 1.337 * (8500 / 17)

= 2756

The lower confidence limit (LCL) of the population mean with α = 0.10. is,

- E = 43500 - 8500 = 40744

lower confidence limit = 40744