420
Spring STAT 2012
Homework #5
(due Friday, February 24, by 4:00 p.m.)
1.
For the prostate data, fit a model with lpsa as the response and the other
variables as predictors.
a)
Plot the residuals vs. the fitted values. Check the constant variance assumption
for the errors. ( 4.3 (a) )
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library(faraway)
data(prostate)
fit = lm(lpsa~lcavol+lweight+age+lbph+svi+lcp+gleason+pgg45)
plot(fit$fitted.values,fit$residuals)
abline(h=0,lty=2)
The residuals look quite random. There's no clear evidence for a non-constant variance.
b)
Make a histogram and a Normal Q-Q plot for the residuals. Check the normality
assumption for the errors. ( 4.3 (b) )
> hist(fit$residuals)
> qqnorm(fit$residuals)
There's a little evidence for non-normality, but it's not outstanding. Normality assumption
seems to be fine.
> shapiro.test(fit$residuals)
Shapiro-Wilk normality test
data: fit$residuals
W = 0.9911, p-value = 0.7721
2.
A society of bird watchers has collected data from several
towns on stork sighting ( x ) and human births ( y ) to test
the widely expressed belief that storks bring babies.
Assume that ( X , Y ) have a bivariate normal distribution.
The data are given in the table below:
Storks, x
18
16
10
20
14
26
22
Babies, y
27
15
13
21
19
39
27
x = 126,
y = 161,
x 2 = 2,436,
y 2 = 4,175,
x y = 3,150,
( x x ) 2 = 168,
( y y ) 2 = 472,
( x x ) ( y y ) = ( x x ) y = 252.
a)
Test H 0 : = 0 vs. H a : > 0 at the = 0.01 level of significance.
What can you say about the p-value of this test?
r=
( x x )( y y )
(x x )
2
(y y )
t=
Test Statistic:
2
168
r n2
1 r
Rejection Region:
252
=
=
2
0.8949.
472
0.8949 7 2
1 0.8949
4.484.
2
Rejects H 0 if t > t 0.01 ( n 2 = 5 df ) = 3.365.
Reject H 0 .
t > t 0.005 ( 5 ) = 4.032
p-value < 0.005.
P-value = Area to the right of t.
( p-value 0.00325. )
OR
1
2
W = ln
1 + r 1 1 + 0.8949
= ln
= 1.446.
1 r 2 1 0.8949
Under H 0 ,
1
2
W = ln
1+ 0
1 0
=
1
1+ 0
ln
= 0,
2 1 0
2
W =
1
1
=.
n3 4
z=
Test Statistic:
Rejection Region:
W W
W
=
1.446 0
1
= 2.892.
4
Rejects H 0 if z > z 0.01 .
Reject H 0 .
z 0.01 = 2.326.
P-value = right tail = P ( Z > 2.892 ) = 0.0019.
b)
Does it follow from part (a) that storks do bring babies? Explain.
No. Correlation does not imply causation; there may be some other reasons
(other variables, other factors) for the link (association).
c)
Test H 0 : = 0.50 vs. H 1 : > 0.50 at the = 0.05 level of significance.
What can you say about the p-value of this test?
1
2
W = ln
1 + r 1 1 + 0.8949
= ln
= 1.446.
1 r 2 1 0.8949
Under H 0 ,
1
2
W = ln
2
W =
Test Statistic:
Rejection Region:
z 0.05 = 1.645.
1+ 0
1 0
=
1
1 + 0.50
ln
= 0.5493,
2 1 0.50
1
1
=.
n3 4
z=
W W
W
=
1.446 0.5493
1
= 1.79.
4
Rejects H 0 if z > z 0.05 .
Reject H 0 .
P-value = right tail = P ( Z > 1.79 ) = 0.0367.
Test H 0 : = 0.30 vs. H 1 : 0.30 at the = 0.05 level of significance.
What can you say about the p-value of this test?
d)
1
2
W = ln
1 + r 1 1 + 0.8949
= ln
= 1.446.
1 r 2 1 0.8949
Under H 0 ,
1
2
W = ln
2
W =
=
1 0
1
1 + 0.30
ln
= 0.3095,
2 1 0.30
1
1
=.
n3 4
z=
Test Statistic:
1+ 0
W W
=
W
1.446 0.3095
1
= 2.273.
4
Rejects H 0 if z < z 0.025 or z > z 0.025 .
Rejection Region:
Reject H 0 .
z 0.05 = 1.96.
P-value = 2 tail = 2 P ( Z > ) 2.273 = 2 0.0116 = 0.0232.
e)
Construct a 95% confidence interval for
.
a = ln
2 z 2
2 1.96
1+ r
= 2.892
= 0.932.
1 r
n3
4
b = ln
2 z 2
2 1.96
1+ r
+
= 2.892 +
= 4.852.
1 r
n3
4
e 0.932 1
e 0.932 + 1
,
e 4.852 1
e 4.852 + 1
100 ( 1 ) % confidence interval for
ea 1
ea + 1
,
eb 1
eb + 1
,
( 0.4350 , 0.9845 )
:
where a = ln
2 z 2
2 z 2
1+ r
1+ r
, b = ln
+
.
1 r
1 r
n3
n3
3.
Do NOT use a computer for this problem.
Consider the following data set:
x1
a)
16
1
25
11
8
9
9
4
16
13
4
12
18
5
15
24
5
150
X T Y = 470 ,
1350
7
3
( XT X )
11
3
1
2
2
90
10 30
30 110 350 ,
Then X T X =
90 350 1170
12
2
where is are i.i.d. N ( 0, 2 ).
0
1
Y i = 0 + 1 x i 1 + 2 x i 2 + i .,
y
1
Consider the model
x2
17
14
0.775 0.45 0.075
= 0.45 0.45 0.1 ,
0.075 0.1 0.025
( y y ) 2 = 126,
and
( y y ) 2 = 306.
Obtain the least-squares estimates 0 , 1 , and 2 .
= ( XT X )
1
6
0.775 0.45 0.075 150
0.45 0.45 0.1 470 = 9 .
XT Y =
2
0.075 0.1 0.025 1350
b)
Perform the significance of the regression test at a 5% level of significance.
Specify the null and the alternative hypotheses. Report the value of the test
statistic, the critical value(s), and the decision.
n = 10,
p = 3.
Source
SS
df
MS
F
Regression
180
p1=2
90
5
Residuals
126
np=7
18
Total
306
n1=9
H 0 : 1 = 2 = 0.
H a : at least one of 1 and 2 is
significantly different from 0.
Critical Value: F 0.05 ( 2 , 7 ) = 4.74.
Decision:
c)
Reject H 0.
( p-value 0.0448. )
Find the p-value (approximately) of the test H 0 : 1 = 5 vs. H 1 : 1 > 5.
Var ( 1 ) = C 11 s 2 = 0.45 18 = 8.1.
t=
Test Statistic:
95
1.405457.
8.1
n p = 7 d.f.
0.711
<
1.405457
<
1.415
t 0.25 ( 7 )
<
t
<
t 0.10 ( 7 )
p-value = right tail.
0.10 < p-value < 0.25.
( p-value 0.10134. )
p-value 0.10.
d)
Find the p-value (approximately) of the test H 0 : 1 = 11 vs. H 1 : 1 11.
Var ( 1 ) = C 11 s 2 = 0.45 18 = 8.1.
t=
Test Statistic:
9 11
0.70273.
8.1
n p = 7 d.f.
0.711
<
0.70273
<
0.263
t 0.25 ( 7 )
<
t
<
t 0.40 ( 7 )
0.25 < left tail < 0.40.
left tail 0.25.
0.50 < p-value < 0.80.
p-value = 2 tails.
p-value 0.50.
( p-value 0.504916. )
e)
Find the p-value (approximately) of the test H 0 : 2 = 0 vs. H 1 : 2 0.
Var ( 2 ) = C 22 s 2 = 0.025 18 = 0.45.
t=
Test Statistic:
( 2) 0
0.45
2.9814.
n p = 7 d.f.
2.998
<
2.9814
<
2.365
t 0.01 ( 7 )
<
t
<
t 0.025 ( 7 )
p-value = 2 tails.
0.01 < left tail < 0.025
left tail 0.01.
0.02 < p-value < 0.05.
p-value 0.02.
( p-value 0.020474. )
f)
Construct a 90% prediction interval for the value of Y at x 1 = 2 and x 2 = 5.
X 0T = [ 1 2 5 ]
Y 0 = 1 6 + 2 9 + 5 ( 2 ) = 14.
0.775 0.45 0.075 1
X 0T C X 0 = [ 1 2 5 ] 0.45 0.45
0.1 2 = 0.15.
0.075
0.1 0.025 5
[ 1 + X 0T C X 0 ] s 2 = ( 1 + 0.15 ) 18 = 20.7.
t 0.05 ( 7 ) = 1.895.
g)
14 1.895
20.7
14 8.62
Construct a 95% prediction interval for the value of Y at x 1 = 4 and x 2 = 13.
X 0T = [ 1 4 13 ]
X 0T
Y 0 = 1 6 + 4 9 + 13 ( 2 ) = 1 6 .
0.775 0.45 0.075 1
C X 0 = [ 1 4 13 ] 0.45 0.45 0.1 4 = 0.15.
0.075
0.1 0.025 13
[ 1 + X 0T C X 0 ] s 2 = ( 1 + 0.15 ) 18 = 20.7.
t 0.025 ( 7 ) = 2.365.
16 2.365
20.7
16 10.76