Chapter 5 Solutions
5.1. (a)
The slope is 0.882; this means that, on
the average, reading score increases by 0.882
for each onepoint increase in IQ.
(b)
The
intercept (
−
33.4) would correspond to the
expected reading score for a child with an IQ
of 0; neither that reading score nor that IQ
has any meaningful interpretation.
(c)
The
predicted scores for
x
=
90 and
x
=
130 are
−
33
.
4
+
0
.
882
×
90
=
45
.
98 and
−
33
.
4
+
0
.
882
×
130
=
81
.
26.
(d)
This is most easily done by plotting the
points
(
90
,
45
.
98
)
and
(
130
,
81
.
26
)
, and then drawing the line connecting them.
45
50
55
60
65
70
75
80
85
90
90
100
110
120
130
140
Reading score
IQ
5.2.
If the reading score increases by 1 for each IQ point, the professor’s line has slope 1. In
order for an IQ of 100 to correspond to a reading score of 50, the equation must be
reading score = IQ
−
50.
5.3. (a)
Minitab output is below; regardless of the software or calculator used, the formula
should agree with the one given in the text (except for differences due to rounding).
(b)
The
means and standard deviations are
x
.
=
324
.
75 and
s
x
.
=
257
.
66 calories,
y
.
=
2
.
3875 and
s
y
.
=
1
.
1389 kg. The correlation is
r
.
= −
0
.
7786. This yields
b
=
r
·
s
y
/
s
x
.
= −
0
.
00344 and
a
=
y
−
b
x
.
=
3
.
505.
Minitab output
The regression equation is fat = 3.51  0.00344 nea
Predictor
Coef
Stdev
tratio
p
Constant
3.5051
0.3036
11.54
0.000
nea
0.0034415
0.0007414
4.64
0.000
s = 0.7399
Rsq = 60.6%
Rsq(adj) = 57.8%
5.4.
See also the solution to Exercise 4.4.
(a)
The correlation is
r
.
= −
0
.
7485.
(b)
The
regression equation is
ˆ
y
=
31
.
9
−
0
.
304
x
.
(c)
The slope (
−
0.304) tells us that, on the
average, for every onepercent increase in
returning birds, the number of new birds join
ing the colony decreases by 0.304.
(d)
When
x
=
60, we predict
ˆ
y
.
=
13
.
69 new birds will
join the colony.
•
•
•
•
•
•
•
•
•
•
•
•
•
3
6
9
12
15
18
21
35
40
45
50
55
60
65
70
75
80
Count of new birds
Percent of returning birds
96
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Solutions
97
Minitab output
The regression equation is New = 31.9  0.304 PctRtn
Predictor
Coef
Stdev
tratio
p
Constant
31.934
4.838
6.60
0.000
PctRtn
0.30402
0.08122
3.74
0.003
s = 3.667
Rsq = 56.0%
Rsq(adj) = 52.0%
5.5. (a)
Scatterplot at right. Regression gives
ˆ
y
=
132
.
45
+
0
.
402
x
(Minitab output below).
The plot suggests a curved pattern, so a
linear formula is not appropriate for making
predictions.
(b)
r
2
=
0
.
0182. This confirms
what we see in the graph: This line does a
poor job of summarizing the relationship.
(c)
The mean planting rate is
x
=
18
.
82
thousand plants per acre, and the mean
yield is
y
=
140
.
02 bushels
/
acre, which is
approximately equal to 132
+
0
.
402
×
18
.
82
(except for rounding error).
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
100
110
120
130
140
150
160
170
12
16
20
24
28
Yield (bushels per acre)
Plants per acre (thousands)
Minitab output
The regression equation is Yield = 132 + 0.402 Plants
Predictor
Coef
Stdev
tratio
p
Constant
132.45
14.91
8.89
0.000
Plants
0.4020
0.7625
0.53
0.606
s = 16.57
Rsq = 1.8%
Rsq(adj) = 0.0%
5.6.
Correlations close to 1 or
−
1 mean that a line is a good match for the scatterplot, while
correlations close to 0 mean that the scatterplot is spread widely about any line we attempt
to use for prediction. Therefore, we expect that prediction of gas used will be quite accurate,
while stock return predictions are much more uncertain.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '09
 ?
 Regression Analysis, Errors and residuals in statistics

Click to edit the document details