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

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