Question

A large university is interested in learning about the average time it takes students to drive to campus. The university sampled 238 students and asked each to provide the amount of time they spent traveling to campus, the average and standard deviation were 21.5 and 4.32 respectively.This variable, travel time, was then used conduct a test of hypothesis. The goal was to determine if the average travel time of all the university's students differed from 20 minutes. Use a level of significance of 0.05. (p value method)

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 20
Alternative Hypothesis, Ha: μ ≠ 20

Rejection Region
This is two tailed test, for α = 0.05 and df = 237
Critical value of t are -1.97 and 1.97.
Hence reject H0 if t < -1.97 or t > 1.97

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (21.5 - 20)/(4.32/sqrt(238))
t = 5.357

P-value Approach
P-value = 0
As P-value < 0.05, reject the null hypothesis.