Question

The average student loan debt of a U.S. college student at the end of 4 years of college is estimated to be about $24,100. You take a random sample of 136 college students in the state of Vermont and find the mean debt is $25,000 with a standard deviation of $2,400. 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 tα/2) for a 90% confidence interval? Use the value from the table or, if using software, round to 3 decimal places. tα/2 =

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

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

(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 $24,100 and why?

a. No, because $24,100 is above the lower limit of the confidence interval for Vermont students.

b. Yes, because $24,100 is above the lower limit of the confidence interval for Vermont students.

c. Yes, because $24,100 is below the lower limit of the confidence interval for Vermont students.

d. No, because $24,100 is below the lower limit of the confidence interval for Vermont students.

(f) We are never told whether or not the parent population is normally distributed. Why could we use the above method to find the confidence interval?

a. Because the sample size is less than 100.

b. Because the margin of error is positive.

c. Because the sample size is greater than 30.

d. Because the margin of error is less than 30.

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = = $ 25,000

sample standard deviation = s = $ 2,400

sample size = n = 136

Degrees of freedom = df = n - 1 = 136 - 1 = 135

b) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t_0.05,135 = 1.656

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

=1.656 * ( 2400 / 136)

Margin of error = E = $ 341

d) The 90% confidence interval estimate of the population mean is,

- E < < + E

25000 - 341 < < 25000 + 341

( $ 24,659 < < $ 25,341 )

e) b. Yes, because $24,100 is above the lower limit of the confidence interval for Vermont students.

f) c. Because the sample size is greater than 30