Question

1. It is known that the average tuition per semester at colleges in Virginia is $3712. Suppose we take a sample of 15 colleges and find that the average tuition per semester is $4512 with a sample standard deviation of $595. We are interested in testing if the average amount of tuition is increasing.
   1. Using the sample of 15 people, what would the 95% confidence interval be for the population mean?
   2. What are the null and alternative hypotheses?
   3. What is the critical value at 95% confidence?
   4. Calculate the test statistic.
   5. Find the p-value
   6. What conclusion would be made here at the 95% confidence level?
   7. Would my conclusion change if I changed alpha to .01? Show reasoning

Answer 1

x̅ = 4512, s = 595, n = 15

a)

95% Confidence interval :

At α = 0.05 and df = n-1 = 14, two tailed critical value, t-crit = T.INV.2T(0.05, 14) = 2.145

Lower Bound = x̅ - t-crit*s/√n = 4512 - 2.145 * 595/√15 = 4182.5

Upper Bound = x̅ + t-crit*s/√n = 4512 + 2.145 * 595/√15 = 4841.5

4182.5 < µ < 4841.5

b)

Null and Alternative hypothesis:

Ho : µ ≤ 3712

H1 : µ > 3712

c)

df = n-1 = 14

Critical value :

Right tailed critical value, t-crit = ABS(T.INV(0.05, 14)) = 1.761

Reject Ho if t > 1.761

d)

Test statistic:

t = (x̅- µ)/(s/√n) = (4512 - 3712)/(595/√15) = 5.2074

e)

p-value :

Right tailed p-value = T.DIST.RT(5.2074, 14) = 0.0001

f)

As p-value < α, Reject the null hypothesis

There is enough evidence to conclude that the average amount of tuition is increasing at 95% confidence level.

g) No, it would not change.

As p-value < 0.01, Reject the null hypothesis.