Stat 572 Final Examination Spring 2008
1. false. r=b1∗sx/sy= (−3/4) ∗2/6 = −0.25
2. true.
3. false. The variance equals the mean.
4. false. It depends on the sign of the coefficient
for x1.
5. true.
6. false. The standard deviation 2.5 is about vari-
ability in the data around the mean value, not
about our uncertainty about the mean value.
7. true.
8. true.
9. (a) From model 1, the prediction is the same
at all doses: 21.50 + 0.95 ∗235 = 244.75 g.
From model 2, the predicted weight is 1.12+
1.00 ∗235 = 236.12 g in the control group,
1.12 + 8.90 + 1.00 ∗235 = 245.02 g in the
low dose group, 1.12 + 19.1 + 1.00 ∗235 =
255.22 g in the high dose group. With
model 3, the predicted weight is 4.50 +
0.99 ∗235 = 237.15 g in the control group,
4.50 + 0.99 ∗235 −17.36 + 0.12 ∗235 =
247.99 g in the low dose group, 4.50+0.99∗
235 + 25.07 −0.03 ∗235 = 255.17 g in the
high dose group.
(b) Same slope in all groups: 0.95 (no units:
g/g) in model 1. In model 2, the slope is
still assumed to be the same in all groups:
1.00. This model says that the final weight
is expected to be equal to the initial weight,
plus a value that depends on which treat-
ment was received. Model 3: slopes are
no longer assumed to be equal. Estimated
to be 0.99 −0.03 = 0.96 in the high dose
group. With this model, we estimate that
with the high dose, small rats gain more
weight on average than large rats. For in-
stance, a 180g rat gains (4.50 + 25.07 +
0.96 ∗180) −180 = 22.37 g while a 240g
rat would gain on average (4.50 + 25.07 +
0.96 ∗240) −240 = 19.97 g.
(c) Confidence interval: use ˆ
b1±tSEˆ
b1where
tis the quantile from the t distribution
with df= 30 −4. This is easier in models
1 and 2, where we read the standard er-
ror of the common slope directly from the
output: 0.09 in model 1, 0.05 in model
2. In model 3, we do not have the slope’s
standard error in the high group directly.
We have the SE of the control group slope
(0.08) and the SE of the difference between
the control group slope and the high group
slope (0.11). With more output, we could
get the correlation between these two es-
timates, then we could use a formula to
get the SE of the high group slope (we did
not see this formula in class in 2009). An
alternative would be to set the ’high dose’
group as the baseline group, and re-run
the analysis so that the output provides
the SE of the slope in the high group di-
rectly.
(d) Model 2 is most appropriate: the f test
comparing model 2 and model 3 says that
the slopes in the 3 groups are not signif-
icantly different (p=.4676). Also, the f
test comparing model 2 and model 1 says
that the intercepts in the 3 different groups
are different, with very strong evidence (p=
4.510−9).
(e) We would need to consider the treatment
as a numerical variable (dosage d= 0, 0.10
and 0.25) rather than a categorical vari-
able (control, low, high). The model de-
scribed can be written as:
final weight = (β0+β1d)∗initial weight+e
Therefore, to fit this model, I would use
linear regression with the following pre-
dictors: dosage (numerical), initial weight,
and their interaction dosage:initial weight.
10. (a) Using model 1: estimated probability that
the tumor is malignant:
p= (1 + exp(−(5.1820 −0.9461 ∗3)))−1
= 0.9124 if the score is 3, and p= (1 +
exp(−(5.1820 −0.9461 ∗6)))−1= 0.3788 if
the score is 6. Using model 2:
1
