Stat 5122
Solutions to Homework #4
Dr. Simonsen
Due Wednesday, September 21, 2005.
1.
Consider the following SAS output giving 5 confidence intervals for the mean of Y. If you wanted
to guarantee that the
joint
coverage of the five confidence intervals was at least 95%, what level
would you use when forming each interval, using the Bonferroni correction? Compute this
adjusted confidence interval for X = 5.
(Note that some observations have been omitted from the
output.)
Analysis of Variance
Sum of
Mean
Source
DF
Squares
Square
F Value
Pr > F
Model
1
16183
805.62
<.0001
Error
16
321.39597
20.08725
Corrected Total 17
16504
Root MSE
4.48188
RSquare 0.9805
Dependent Mean 64.00000
Adj RSq 0.9793
Coeff Var
7.00294
Parameter Estimates
Parameter
Standard
Variable
Estimate
t Value
Pr > t
Intercept
2.32215
2.56435
0.91
0.3786
x
14.73826
0.51926
28.38
Output Statistics
Dep Var
Predicted
Std Error
Obs
y
Value
Mean Predict
95% CL Mean
Residual
3
5
78.0000
71.3691
1.0878
69.0630 73.6752
6.6309
4
10.0000
12.4161
2.1021
7.9598 16.8724
2.4161
6
62.0000
56.6309
54.3248 58.9370
5.3691
8
39.0000
41.8926
1.3125
39.1103 44.6750
2.8926
10
2
33.0000
27.1544
1.6737
23.6064 30.7024
5.8456
To obtain a joint coverage probability of at least 95% for g = 5 intervals, we use the Bonferroni correction
and construct individual confidence intervals with coverage probability 1
α
/g = 10.05/5 = 1 – 0.01 =
0.99.
Thus we would construct
99%
confidence intervals for each value of X.
[Note to grader:
They can either say 99% confidence level or
α
= 0.01 in answer to “what level”.]
From the SAS output we see that df
E
= n – 2 = 16, and
{ }
ˆˆ
1.0878,
71.3691
hh
sY
Y
==
when X
h
= 5.
To
obtain the 99% CI we use t
c
= t(1
α
/2g, n2) = t(0.995, 16) = 2.921.
Thus the 99% CI for the mean when
X = 5 is
71.3691
±
2.921×1.0878 = 71.3691
±
3.1775 = [
68.1916, 74.5466
]
2.
Based on the following small data set, construct the design matrix,
X
, its transpose
X’
,
and the matrices
X’X
,
(
X’X
)
1
,
X’Y
, and
b
= (
X’X
)
1
X’Y
.
(If you have trouble with
matrix multiplication, see pages 45 of Topic 3.)
X
Y
2
1
4
2
6
5
8
7
10
9
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14
16
18
11
0
⎡⎤
⎢⎥
=
⎣⎦
X
,
,
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24681
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2
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6
8 10
30
220
0
⎡
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==
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1
2
11
3
220
30
1.1
0.15
1
10
20
30
5
3
1
0.15
0.025
5 220 30
20
40
−
−
−−
⎡
′
=
⎢
−
×−
⎣
⎤
⎥
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=
⎢
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1
2
12579
2
4
5
2
4
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8 10
2 8 30 56 90
186
7
9
++++
⎡
⎤
′
⎢
⎥
+
+++
⎣
⎦
XY
1
1.1
0.15
24
26.4 27.9
1.5
0.15
0.025
186
3.6
4.65
1.05
−
⎡
⎤
⎡
⎤
⎡
′′
=
=
⎢
⎥
⎢
⎥
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+
⎣
⎦
⎣
⎦
⎣
bX
Y
For the following 5 problems use the Commercial properties data described in KNNL with problem 6.18
on page 251 — 252. (Note:
this dataset is not in NKNW.
It replaces the Mathematicians salary dataset.)
3.
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