Chapter 10 Solutions
10.1.
The given model was
µ
y
=
40
.
5
−
2
.
5
x
, with standard deviation
σ
=
2
.
0.
(a)
The slope is
−
2.5.
(b)
When
x
increases by 1,
µ
y
decreases by 2.5. (Or
equivalently, if
x
increases by 2,
µ
y
decreases by 5, etc.)
(c)
When
x
=
10,
µ
y
=
40
.
5
−
2
.
5
(
10
)
=
15
.
5.
(d)
Approximately 95% of observed responses would fall in
the interval
µ
y
±
2
σ
=
15
.
5
±
2
(
2
.
0
)
=
15
.
5
±
4
.
0
=
11
.
5to 19.5.
10.2.
Example 10.4 gave the estimated regression equation as
d
MPG
=−
7
.
796
+
7
.
874LOGMPH,
with
s
=
0
.
9995. In the text following that example, the parameter estimates were rounded
to two decimal places:
d
MPG
7
.
80
+
7
.
87 LOGMPH, with
s
=
1
.
00.
(a)
If the car travels
at 35 mph, then LOGMPH
=
ln 35
.
=
3
.
5553, so we estimate
d
MPG
.
=
20
.
1988 mpg (or
20.18 mpg, using the rounded estimates).
(b)
The residual is 21
.
0
−
d
MPG
.
=
0
.
8012 mpg (or
0.82, using the rounded estimates).
(c)
Because this regression line was based on speeds
between about 12 and 53 mph, estimates near or outside those boundaries would not be very
reliable; they grow less reliable the more we stray above 53 mph. Even the estimated mpg
for 45 mph would be subject to lots of uncertainty because the points at the high end of the
scatterplot exhibited more spread than those at the low end.
Note:
Some students might mistakenly use the common (base10) logarithm instead of
the natural logarithm; if they do so, they will Fnd LOGMPH
.
=
1
.
5441 and
d
MPG
.
=
4
.
36 mpg.
Hopefully, the large residual (16.6 mpg) in part (b) would help them notice their mistake; in
general, we expect residuals to fall in the range
±
3s, if they follow a Normal distribution.
10.3.
Example 10.6 gives the conFdence interval 7.16 to 8.58 for the slope
β
1
. Recall that
slope is the change in
y
(i.e.,
d
MPG) when
x
(i.e., LOGMPH) changes by
+
1.
(a)
If
LOGMPH increases by 1, we expect
d
MPG to change by
β
1
,so the 95% conFdence interval
for the change is (an increase of) 7.16 to 8.58 mpg.
(b)
If LOGMPH decreases by 1, we
expect
d
MPG to change by
−
β
1
,so the 95% conFdence interval for the change is
−
7.16
to
−
8.58 mpg—that is, a decrease of 7.16 to 8.58 mpg.
(c)
If LOGMPH increases by 0.5,
we expect
d
MPG to change by 0
.
5
β
1
,so the 95% conFdence interval for the change is (an
increase of) 3.58 to 4.29 mpg.
10.4.
Example 10.10 gives the 95% prediction interval
ˆ
y
±
t
∗
SE
ˆ
y
as 17.0 to 21.0, or 19
.
0
±
2
.
0.
±or df
=
58, the critical value
t
∗
is very close to 2, so SE
ˆ
y
.
=
1
.
0.
When
x
=
40 mph, SE
ˆ
y
would be larger because predictions are less accurate for values
of
x
near the extremes (relative to the range of speeds used to determine the regression
formula, which was approximately 12 to 53 mph).
Note:
Working from the original data set, SE
ˆy
equals 1.01 when x
=
30 mph, and 1.02
when x
=
40 mph.
266
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10.5. (a)
The plot suggests a linear increase.
(b)
The regression equation is
ˆ
y
=
−
3271
.
9667
+
1
.
65
x
.
(c)
The ftted values and residuals are given in the table below.
Squaring the residuals and summing gives 0.001
6, so the standard error is:
s
=
r
0
.
001
6
n
−
2
=
q
0
.
001
6
.
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 Spring '07
 Guggenberger
 Regression Analysis, IBI, Minitab output

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