How to find contrast and conduct Bonferonni test? I am trying to understand contrast and Bonferroni by working on a problem. Can someone tell me if I'm on the right track with the work I've already done?
A sample of highway gas mileages has been taken for each of the four new car models.

Model A: 30, 25, 31, 26, 25
Model B: 33, 28, 27
Model C: 28, 36, 32, 35, 33, 33
Model D: 33, 37

(a) Test at the 5% level if the mileages are the same for all four models. Use the test statistic.
I am not quite sure what the "test statistic" is. Does anyone understand the wording? My guess was using ANOVA but I could be wrong.
Conducting ANOVA, I get F = 4.93744 and critical value = 3.59 so reject H0.
(b) Use a contrast to test at the 5% level if Model A is the same as the other 3 models. Note: You should do
only one test.
so I did the following:
$[1(27.4) - 1(29.333)]^2/[1/5 * 1/3] = 7.005$
Since I am comparing two categories, my DF = 1. α = 0.05
But I am stuck on how to conclude the result of this test.
(c) Suppose you do a Bonferroni to test at the 5% level. Show the results of Model D against each of the other
models. Which models, if any, are different from Model D?
Here, my understanding is that we divide α by the number of comparisons. Since we have 3 comparisons, we need to divide α by 3. But how do I go about the actual test? Can someone mention briefly the procedure?
 A: First, I verified your (correct) F-test in Minitab statistical software, with
the result shown below. I believe 'test statistic' refers to the F-statistic
(also called variance ratio statistic), 4.94. You can reject at the 5% level
because the P-value is 0.018. (You could also reject at the 2% level.) I got the
critical value at the 5% level to be 3.49 and $4.94 > 3.49.$ (Is 3.59 a typo?)
Analysis of Variance

Source  DF  Adj SS  Adj MS  F-Value  P-Value
Factor   3   124.3  41.433     4.94    0.018
Error   12   100.7   8.392
Total   15   225.0

Next, a general response to your question about the Bonferroni method.
If you check D vs A, D vs B, and D vs C, that will be three comparisons.
If you want a 'family' error rate no larger than $.05$ for the three,
then do each of the three tests with an error rate of $.05/3.$ 
I got that D and A differ significantly, by this method. Note that the
means for these two groups are farthest apart.
Notes: 
(1) In many kinds of software, Tukey HSD comparisons are not available for groups
with such relatively different numbers of replications.
(2) For part (b), the standard for whether the contrast is significant would be
different depending on whether this particular comparison (of A vs the avg
of the other three) was 'pre-planned' (specified in the analytic
protocol before data were available) or 'ad hoc' (inspired by looking
at the data). You do not say which standard is relevant.
