Question 1 [22 Marks] In the 1974 Motor Trend US magazine, fuel ...

Question 1 [22 Marks]
In the 1974 Motor Trend US magazine, fuel consumption (mpg in miles/US gallon) was related
to several aspects of automobile design and performance for 32 automobiles (1973-74
models). We consider three of the variables, displacement (disp in cu.in), horse power (hp)
and weight (wt, lb/1000). The data is plotted below together with the log transformed
data.
100 300 2 3 4 5
4.5 5.0 5.5 6.0 0.4 0.8 1.2 1.6
mpg
10 15 20 25 30
logmpg
2.4 2.8 3.2
100 300
disp
4.5 5.0 5.5 6.0
logdisp
hp
50 150 250
loghp
4.0 4.5 5.0 5.5
2 3 4 5
wt
0.4 0.8 1.2 1.6
logwt
10 15 20 25 30
50 150 250
2.4 2.8 3.2
4.0 4.5 5.0 5.5
(i) From the graphs it was decided to use the log transformed variables in the analysis.
Explain why you think this may or may not be a good idea.

(ii) The following model was fitted
logmpg = µ + β 1 logdisp + β 2 loghp + β 3 logwt + error
Complete the summary of the fit that follows by replacing the ∗’s.
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.9462 0.2687 18.41

Question 2 [30 Marks]
(This question has been modified as the original question was not suitable for the unit as
presented in 2002.)
The original data set in question 1 also contains the number of cylinders (cyl: 4, 6 or 8) for
each car tested. Since it is quite possible the relationship between mpg and displacement
will also depend on the number of cylinders it was decided to examine this relationship
separately.
Miles/US gallon
10 15 20 25 30
four
six
eight
100 200 300 400
displacement
The following models were fitted and the output from R is reported:
Model 1: (3 intercepts and 3 slopes)
Analysis of Variance Table
Response: mpg
Df Sum Sq Mean Sq F value Pr(>F)
cyl 2 824.78 412.39 73.3221 2.050e-11
disp 1 57.64 57.64 10.2487 0.003591
cyl:disp 2 97.39 48.69 8.6574 0.001313
Residuals 26 146.23 5.62

Model 2: 3 intercepts, 1 common slope (parallel lines).
Analysis of Variance Table
Response: mpg
Df Sum Sq Mean Sq F value Pr(>F)
cyl 3 13741 4580 526.434 |t|)
(Intercept) 40.87196 3.02012 13.533 2.79e-13
cyl6 -21.78997 5.30660 -4.106 0.000354
cyl8 -18.83916 4.61166 -4.085 0.000374
disp -0.13514 0.02791 -4.842 5.10e-05
cyl6:disp 0.13875 0.03635 3.817 0.000753
cyl8:disp 0.11551 0.02955 3.909 0.000592
Residual standard error: 2.372 on 26 degrees of freedom
Multiple R-Squared: 0.8701, Adjusted R-squared: 0.8452
F-statistic: 34.84 on 5 and 26 DF, p-value: 9.968e-11
(a) Write down the equation of the three lines.
(b) Test whether the slopes of the lines for the 6 and 8 cylinder cars are different
from the slope of the line for the 4 cylinder cars.
(c) Verify the 95% CI for the slope of the line for four cylinder cars is given by
(−0.193, −0.078).
(d) If the 95% CI for the slope of the line for six cylinder cars is (−0.044, 0.051)
and that for eight cylinder cars is (−0.040, 0.0003) what can you say about the
relationship of the slopes of the three lines?

(iii) The mpg was estimated for cars with a displacement of 150 cu.in and having 4, 6
and 8 cylinders using R.
$fit
1 2 3
* 19.6228 19.0877
$se.fit
1 2 3
1.44190 1.18564 2.07059
Comment on the results after first replacing the missing value (for 4 cylinder cars).
(iv) With the aid of the following plots comment on the assumptions for the model and
the fit of the line.
Residual Plot
Normal Q−Q Plot
Residual
−2 0 2 4
Sample Quantiles
−2 0 2 4
15 20 25 30
Fitted
−2 −1 0 1 2
Theoretical Quantiles

Question 3 25 Marks]
An animal experiment is designed to investigate whether levorphanol reduces stress as
reflected in the cortical sterone level. There were four treatment groups containing five
animals each. The four treatments consisted of a control (C), levorphanol only (L),
epinephrine only (E), and levorphanol plus epinephrine (LE).
(i) Complete the ANOVA table below.
Analysis of Variance Table
Response: cs
Df Sum Sq Mean Sq F value Pr(>F)
treat * 37.58 * * 2e-04
Residuals 16 16.30 *
(ii) Write down and test a suitable hypothesis concerning treatments.
(iii) Write down three contrasts to test
(a) The main effect of levorphanol (levor)
(b) The main effect of epinephrine (epin)
(c) the interaction between levorphanol and epinephrine (levor:epin).
(iv) Show the contrasts are orthogonal to each other.

(v) Using the following ANOVA table and table of means, write down and test hypotheses
concerning the main effects of levorphanol, epinephrine and their interaction.
Analysis of Variance Table
Response: cs
Df Sum Sq Mean Sq F value Pr(>F)
levor 1 12.832 12.832 12.598 0.002670
epin 1 18.586 18.586 18.246 0.000584
levor:epin 1 6.161 6.161 6.048 0.025692
Residuals 16 16.298 1.019
Tables of means
Grand mean
2.964
levor
0 1
3.765 2.163
epin
0 1
2.000 3.928
levor:epin
epin
levor 0 1
0 2.246 5.284
1 1.754 2.572
Standard errors for differences of means
levor epin levor:epin
0.4514 0.4514 0.6383
replic. 10 10 5
(vi) Write a conclusion.

Question 4 [20 Marks]
An experiment is conducted to determine the effect of three levels of fertilization on the
yield of sugar cane. The experiment is conducted at four locations, randomly selected
from a number of possible locations available for the experiment. Each fertilizer level is
applied to three plots at each location.
(i) Complete the following table
Df Mean Sq Expected Mean Square F P
fertilizer * 12.15514 * * *
location * 4.60598 * * *
fertilizer:location * 0.15544 * * *
Residuals 24 0.05793
(Note the Sum of Squares column has been omitted and is not needed.)
(ii) Write down and test appropriate hypotheses concerning fertilizer treatments, locations
and the interaction between fertilizer treatment and location.
(iii) Estimate the components of variation.
(iv) Calculate the variation of a single observation.
(v) The fertilizer SS is partitioned using polynomial contrasts. The significance levels
are given for the resulting F values in the ANOVA table.
Analysis of Variance Table
Response: yield
Pr(>F)
linear
3.32e-05
quadratic
1.65e-11
location
1.31e-12
location:fertilizer 0.0389
Interpret these results with the aid of the following table of mean yields for the three
fertilizer levels.
1 2 3
16.7567 17.9208 15.9167
(vi) Write a conclusion.