2009 Math 2b Homework 8 Solutions
Problem 1.
The stopping distance (
y
) of an automobile on a certain road was studied s
a function of velocity (Brownlee 1960). The data are listed in the following table. Fit
y
and
√
y
as linear functions of velocity and examine the residuals in each case. Which ±t is
better? Can you suggest a physical reason that explains why?
Solution.
We use the models
y
i
=
β
0
+
β
1
x
i
+
±
i
and
√
y
i
=
β
±
0
+
β
±
1
x
i
+
±
±
i
. To calculate our
esimators for the
β
i
we use the formulae given on page 565 in Rice
ˆ
β
=(
X
T
X
)

1
X
T
Y.
Plugging in our data and doing the calculation gives
ˆ
β
=
±

62
.
0454
3
.
4930
²
,
ˆ
β
±
=
±

0
.
8776
0
.
2277
²
.
To compare the quality of the ±ts, we will compare the residuals, but ±rst we must rescale
the data because the
√
y
±t will have smaller residuals simply because it takes smaller values
than the other ±t. To do this, we look at the
√
y
±t, take the square of the values and then
look at the residuals, i.e.
(
√
y
)
2

(
ˆ
β
±
0

ˆ
β
±
1
v
)
2
where we use the values for
β
±
i
found earlier. Calculating the errors in this way gives:
1
.
03
,

1
.
07
,

2
.
92
,
3
.
46
,
8
.
24
,

8
.
32
whose squares add up to 159
.
8238. On the other hand, the
y
±t errors are:
5
.
84
,
3
.
74
,

10
.
59
,

6
.
32
,
4
.
59
,
2
.
75
whose squares add up to 228
.
81. This suggests that the
√
y
±t is better than the
y
±t.
Physically, we know that under constant acceleration
a
, it is true that
v
2
f
=
v
2
0
+2
ay
. In
the case where acceleration is negative and
v
f
= 0, this gives
y
=
v
2
0
2

a

, so
√
y
∝
v
.
Problem 2.
The data in the problem were obtaind in a Stanford study of the survival
time of heart transplant patients. The “mismatch score” is an indication of how well the
transplanted heart should ±t the patient. Consider the regression model
log(Survival Time) =
β
1
+
β
2
(Age) +
β
3
(Mismatch Score) +
error,
where we assume the errors are independent normal with mean 0 and unknown variance
σ
2
.
(
a
) Fit the model i.e. calculate the least squares estimates of
β
1
,
β
2
and
β
3
and calculate
the sample standard deviation
s
.
(
b
) Calculate
R
2
, the coe²cient of determination.
(
c
) Test the hypothesis
β
2
= 0 at the
α
=0
.
1 level.
(
d
) Give a 90% con±dence interval for
β
3
.
(
e
) Give a 90% con±dence interval for the expectation of the logarithm of the survival
time of patients with a mismatch score of 1.00 and age 55.
1
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View Full DocumentSolution. (a)
We have
Y
=
Xβ
+
±
where
Y
=
log(44)
log(550)
.
.
.
log(48)
,X
=
1 36 0
.
00
1 48 0
.
12
1 45 0
.
16
.
.
.
.
.
.
.
.
.
1 53 3
.
05
,
β
=
β
1
β
2
β
3
.
One of the data points has a survival time of 0, so to prevent taking log(0), we replace this
0 with 0
.
5. Therefore
ˆ
β
=(
X
T
X
)

1
X
T
Y
=
6
.
39417

0
.
0746735

0
.
0154375
and
s
2
=

Y

X
ˆ
β

2
157

3
=3
.
24106
.
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 Winter '08
 Makarov,N
 Math, Differential Equations, Statistics, Equations, Normal Distribution, Probability, Regression Analysis, Variance, Null hypothesis, Chisquare distribution

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