Question

The postal service sorts mail as Priority Mail Express, Priority Mail, First-Class Mail, or Standard Mail. Over a period of 3 weeks, 18 of each type were mailed from the Network Distribution Center in Atlanta, Georgia, to Des Moines, Iowa. The total delivery time in days was recorded. Minitab was used to perform the ANOVA. The results follow:
Source DF SS MS F P
Factor 3 1.83 0.61 2.77 0.048
Error 68 15.28 0.22
Total 71 17.11
Level N Mean StDev
Priority Mail Express 18 2.873 0.477
Priority Mail 18 3.040 0.574
First-Class Mail 18 3.273 0.462
Standard Mail 18 3.228 0.360
Using the ANOVA results, compare the average delivery times of the four different types of mail.
Identify the null hypothesis and the alternate hypothesis.
Null hypothesis:
A) H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4
B) H0: μ1 = μ2 = μ3 = μ4
A) or B)
Alternate hypothesis:
A) H1: At least one mean is different.
B) H1: All means are equal.
A) or B)
What is the decision rule? Use the 0.05 significance level. (Round your answer to 2 decimal places.)
Reject H0 if F>
Use the 0.05 significance level to test if this evidence suggests a difference in the means for the different types of mail.
H0 is (rejected/not rejected)

Answer 1

Answer:

Null hypothesis:

B) H0: μ1 = μ2 = μ3 = μ4

Alternate hypothesis:

A) H1: At least one mean is different.

Reject H0 if F>2.74

H0 is rejected

Step-by-step explanation:

Null hypothesis always contain equality so, null hypothesis would be

B) H0: μ1 = μ2 = μ3 = μ4.

Alternate hypothesis doesn't contain equality so, alternate hypothesis would be

A) H1: At least one mean is different.

Decision rule will be  "reject null hypothesis if calculated F is greater than critical value of F".

Critical value of F is F0.05(3,68)=2.74.

So, decision rule will be Reject H0 if F>2.74.

As, calculated F 2.77 is greater than 2.74 so, we reject our null hypothesis and conclude that there is sufficient evidence to suggests a difference in the means for the different types of mail at 5% level of significance.