Statistics 500 – Fall 2009
Solutions to Homework 6
1
1. Economic growth and Democratic presidential vote
(a)
1
4.149
ˆ
0.0108
385.256
β
==
,
0
8.288
32.18
ˆ
0.0108
0.4671
17
17
=−
=
(b)
0.0108 0
4.29
0.03653 15
385.26
t
−
( )
17
2
4.29
0.001
p
value
P t
−=
><
There is very strong evidence that the linear slope relating growth rate to the Democratic vote is not
0.
(c)
()
ˆ
0.4671 0.0108 5
0.521
Y
=+
=
(
)
2
15,0.995
5 32.18 17
0.03653
1
ˆ
0.521 2.946 0.0143
0.479,0.563
15
17
385.26
Yt
⎛⎞
−
±+
=
±
=
⎜⎟
⎝⎠
(d)
(
)
2
15,0.995
5 32.18 17
0.03653
1
ˆ
1
0.521 2.946 0.0514
0.37,0.67
15
17
385.26
−
+
=
±
=
The prediction interval is much wider because it concerns a prediction for one year.
(e)
Source
d.f.
SS
MS
F
Model
1
0.04468
0.04468
18.34
Error
15
0.03653
0.00244
Total
16
0.0812
Yes, the F statistic is, at least within round off error, 4.29
2
(f) R
2
= 0.04468 / 0.0812 = 0.55 or 55%
Yes, this is a model with R
2
larger than 50%.
(g)
(
)
15,0.975
0.03653
1
(0.4671 0.0108*1.9)
1
0.488 2.13 0.0508
0.38,0.60
15
17
t
+±
+
=
±
=
I don’t think this is a very precise prediction.
Comparing to my suggested criteria:
The 95% prediction is almost as wide as the range of the data
It is much wider than what is needed to predict close races. The prediction interval is over twice as
wide as the desired 0.10 width.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentStatistics 500 – Fall 2009
Solutions to Homework 6
2
2. Snow gauge calibration
a) X is density, Y is gain.
The values of X are fixed by the experimenter’s choice of block
densities.
The values of Y are measured with error.
The error variation is associated with gain, so
gain is the Y variable.
b) The proposed model was:
ε
β
+
+
=
density
Gain
1
0
.
My SAS code is:
data snow;
infile 'snow1.txt';
input density gain;
/* I added a line with density = 0.2 and gain = . to the data file */
proc glm;
model gain = density;
estimate 'gain at 0.2' intercept 1 density 0.2;
output out = resids r = resid p = yhat stdp = stderr stdi = stdobs
lclm = lline uclm = uline lcl = lpred ucl = upred;
proc print;
where gain = .;
run;
I got:
Standard
Parameter
Estimate
Error
t Value
Pr > t
Intercept
348.4059986
13.40911904
25.98
<.0001
density
579.9308681
33.49515224
17.31
<.0001
So, the estimated intercept is 348 and the estimated slope is 579.
•
Notice the wording of the question.
A regression of gain on density is talking about a
regression using Y=gain and X=density.
Sometimes folks have misinterpreted the wording
and reversed X and Y.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Staff
 Statistics, Regression Analysis, 50%, 55%, Ynew, 0.2

Click to edit the document details