Question

A rental car company wants to investigate whether the type of car rented affects the length of the rental period. An experiment is run for one week at a particular location, and 10 rental contracts are selected at random for each car type. The results are show in the following table.

Type of Car	Observations
Subcompact	3,5,3,7,6,5,3,2,1,6
Compact	1,3,4,7,5,6,3,2,1,7
Midsize	4,1,3,5,7,1,2,4,2,7
Full Size	3,5,7,5,10,3,4,7,2,7

Use Tukey’s multiple comparison method to compare the pairs of treatment means. Present the results of these comparisons in a clear and concise way.

Answer 1

One-way ANOVA: C1 versus C2

Source DF SS MS F P
C2 3 16.67 5.56 1.11 0.358
Error 36 180.30 5.01
Total 39 196.98

Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev --+---------+---------+---------+-------
compact 10 3.900 2.283 (----------*-----------)
Fullsize 10 5.300 2.452 (-----------*-----------)
Midsize 10 3.600 2.221 (-----------*-----------)
Subcompact 10 4.100 1.969 (-----------*-----------)
--+---------+---------+---------+-------
2.4 3.6 4.8 6.0

Pooled StDev = 2.238

Grouping Information Using Tukey Method

C2 N Mean Grouping
Fullsize 10 5.300 A
Subcompact 10 4.100 A
compact 10 3.900 A
Midsize 10 3.600 A

Since same letter is shared in 4 type of car so Means that do not share a letter are significantly different.

Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of C2

Individual confidence level = 98.93%

C2 = compact subtracted from:

C2 Lower Center Upper --------+---------+---------+---------+-
Fullsize -1.296 1.400 4.096 (----------*---------)
Midsize -2.996 -0.300 2.396 (----------*----------)
Subcompact -2.496 0.200 2.896 (----------*----------)
--------+---------+---------+---------+-
-2.5 0.0 2.5 5.0

C2 = Fullsize subtracted from:

C2 Lower Center Upper --------+---------+---------+---------+-
Midsize -4.396 -1.700 0.996 (----------*----------)
Subcompact -3.896 -1.200 1.496 (----------*----------)
--------+---------+---------+---------+-
-2.5 0.0 2.5 5.0

C2 = Midsize subtracted from:

C2 Lower Center Upper --------+---------+---------+---------+-
Subcompact -2.196 0.500 3.196 (----------*----------)
--------+---------+---------+---------+-
-2.5 0.0 2.5 5.0

Moreover all the confidence intervals contain zero hence there are insignificant difference between the pairs of treatment means.

Note: It is noted that since from ANOVA table it is observed that p-value=0.358>0.05, hence there is are insignificant difference between the pairs of treatment means. Therefore it is not required to perform post hoc test. Post hoc test is making sence we ANOVA test reject the null hypothesis i.e. treatment means are all equal.