Question

8. In a recent year the fuel economy of all passenger vehicles was 19.8 mpg. A trade organization sampled 50 passenger vehicles for fuel economy and obtained a sample mean of 20.1 mpg with standard deviation 2.45 mpg. The sample mean 20.1 exceeds 19.8, but perhaps the increase is only a result of sampling error. a. Perform the relevant test of hypotheses at the 20% level of significance using the critical value approach. b. C. Compute the observed significance of the test. Perform the test at the 20% level of significance using the p-value approach. You need not repeat the first three steps, already done in part (a).

Answer 1

a)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 19.8
Alternative Hypothesis: μ > 19.8

Rejection Region
This is right tailed test, for α = 0.2 and df = 49
Critical value of t is 0.849.
Hence reject H0 if t > 0.849

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (20.1 - 19.8)/(2.45/sqrt(50))
t = 0.866

Rejection Region Approach
As the value of test statistic, t is outside critical value range, reject the null hypothesis

b)
Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (20.1 - 19.8)/(2.45/sqrt(50))
t = 0.866

c)

P-value Approach
P-value = 0.1954
As P-value < 0.2, reject the null hypothesis.