Question

Student Debt – Vermont: The average student loan debt of a U.S. college student at the end of 4 years of college is estimated to be about $22,600. You take a random sample of 136 college students in the state of Vermont and find the mean debt is $23,500 with a standard deviation of $2,200. We want to construct a 90% confidence interval for the mean debt for all Vermont college students.

(a) What is the point estimate for the mean debt of all Vermont college students?


(b) What is the critical value of t (denoted t_α/2) for a 90% confidence interval? Use the value from the table or, if using software, round to 3 decimal places.


(c) What is the margin of error (E) for a 90% confidence interval? Round your answer to the nearest whole dollar.


(d) Construct the 90% confidence interval for the mean debt of all Vermont college students. Round your answers to the nearest whole dollar.


(e) Based on your answer to (d), are you 90% confident that the mean debt of all Vermont college students is greater than the quoted national average of $22,600 and why?

Answer 1

a)

point estimate =sample mean, xbar = 23500

b)

sample standard deviation, s = 2200
sample size, n = 136
degrees of freedom, df = n - 1 = 135

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.656

Critical value = 1.656

c)

ME = tc * s/sqrt(n)
ME = 1.656 * 2200/sqrt(136)
ME = 312

d)

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (23500 - 1.656 * 2200/sqrt(136) , 23500 + 1.656 * 2200/sqrt(136))
CI = (23188 , 23812)

e)
yes, because confidenc einterval doe snot contain 22600