Sections 6.7, 6.8, 7.7

1
Sections 6.7, 6.8, 7.7 (Note The approach used
here to present the material in these sections is
substantially different from the approach used in
the textbook.)
Recall If X and Y are random variables with E(X)
?X , E(Y) ?Y , Var(X) ?X2 , Var(Y) ?Y2 ,
and Cov(X,Y) ??X?Y , then the least squares
line for predicting Y from X is
?Y (x ?X) ?X
?Y ?X ?X
?Y ? x ?X
y ?Y ?
y ?Y ?
or
a
b
The least squares line is derived in Section 4.2
by minimizing
EY (a bX)2 .
y
Consider a set of observed data (x1 , y1) , (x2
, y2) , , (xn , yn ) . Imagine that we treat
this data as describing a joint p.m.f. for two
random variables X and Y where each points is
assigned a probability of 1/n. Then, we see that
x
2
n
?
plays the role of E(X) ?X ,
plays the role of E(Y) ?Y ,
plays the role of Var(X) ?X2 ,
plays the role of Var(Y) ?Y2 , and
plays the role of Cov(X,Y) ??X?Y .
1 n
x
xi
i 1
n
?
1 n
y
yi
i 1
n
?
sx2
1 n
n 1 n
(xi x)2
i 1
n
?
sy2
1 n
n 1 n
(yi y)2
i 1
we shall complete this equation shortly.
n
?
1 n
(xi x)(yi y)
i 1
3
n
?
(xi x)(yi y)
We define the sample covariance to be c
, and we define the sample correlation to be r
Consequently, the least squares line for
predicting Y from X is This least squares line
minimizes
i 1
n 1
c . sxsy
The sample correlation r is a measure of the
strength and direction of a linear relationship
for the sample in the same way that the
correlation ? is a measure of the strength and
direction of a linear relationship for the two
random variables X and Y.
4
n
?
plays the role of E(X) ?X ,
plays the role of E(Y) ?Y ,
plays the role of Var(X) ?X2 ,
plays the role of Var(Y) ?Y2 , and
plays the role of Cov(X,Y) ??X?Y .
1 n
x
xi
i 1
n
?
1 n
y
yi
i 1
n
?
sx2
1 n
n 1 n
(xi x)2
i 1
n
?
sy2
1 n
n 1 n
(yi y)2
i 1
n
?
1 n
n 1 c n
(xi x)(yi y)
i 1
5
n
?
(xi x)(yi y)
We define the sample covariance to be c
, and we define the sample correlation to be r
Consequently, the least squares line for
predicting Y from X is This least squares line
minimizes
i 1
n 1
c . sxsy
sy (x x) sx
sy x sx
sy r x sx
y y r
y y r
or
a
b
n
?
yi (a bxi)2 .
i 1
The sample correlation r is a measure of the
strength and direction of a linear relationship
for the sample in the same way that the
correlation ? is a measure of the strength and
direction of a linear relationship for the two
random variables X and Y.
6
r 1
r close to 1
r is positive
r is negative
r 1
r close to 1
r close to 0
r close to 0
r is negative
r close to 0
7
Suppose Y1 , Y2 , , Yn are independent with
respective N(?1 , ?2), N(?2 , ?2), , N(?n , ?2)
distributions. Let x1 , x2 , , xn be fixed
values not all equal, and suppose that for i 1,
2, , n, ?i ?0 ?1xi . Then the joint p.d.f.
of Y1 , Y2 , , Yn is
n
?
yi (?0 ?1xi)2 2?2
yi (?0 ?1xi)2 2?2
n ? i 1
exp ??2?
exp ?n (2?)n/2
i 1
for ? lt y1 lt ? , ? lt y2 lt ? , , ? lt yn
lt ?
If we treat this joint p.d.f. as a function L(?0
, ?1), that is, a function of the unknown
parameters ?0 and ?1 , then we can find the
maximum likelihood estimates for ?0 and ?1 by
maximizing the function L(?0 , ?1).
It is clear the function L(?0 , ?1) will be
maximized when is minimized.
n
?
yi (?0 ?1xi)2
i 1
8
The previous result concerning the least squares
line for predicting Y from X with a sample of
data points tells us that the mle of ?1 is ?1
and the mle of ?0 is ?0
n
n
?
?
(xi x)(Yi Y)
(xi x)Yi

Sy R sx


i 1
i 1
n
n
?
?
(xi x)2
(xi x)2
i 1
i 1
n
(xi x)
?
Yi
n
?
i 1
(xj x)2
j 1
x

n
Yi n
n
(xi x)
Sy x sx
?
?
Y R


Yi
n
?
i 1
i 1
(xj x)2
j 1
x
(xi x)
n
1 n
?

Yi
n
?
i 1
(xj x)2
j 1
9
1. (a)
Suppose we are interested in predicting a
person's height from the person's length of
stride (distance between footprints). The
following data is recorded for a random sample of
5 people Length of Stride (inches) 14 13 21
25 17 Height (inches) 61 54 63 72 59
Find the equation of the least squares line for
predicting a person's height from the person's
length of stride.
120 1.2 . 100
The slope of the least squares line is
The intercept of the least squares line is 61.8
(1.2)(18) 40.2 .
The least squares line can be written y 40.2
1.2x .
10
(b)
Suppose we assume that the height of humans has a
normal distribution with mean ?0 ?1x and
variance ?2, where x is the length of stride.
Find the maximum likelihood estimators for ?0 and
?1 .
120 1.2 . 100
The mle of ?1 is
The mle of ?0 is 61.8 (1.2)(18) 40.2 .
11
2. (a)
Suppose Y1 , Y2 , , Yn are independent with
respective N(?1 , ?2), N(?2 , ?2), , N(?n , ?2)
distributions. Let x1 , x2 , , xn be fixed
values not all equal, and suppose that for i 1,
2, , n, ?i ?0 ?1xi .
Use Theorem 5.5-1 (Class Exercise 5.5-1) to find
the distribution of the maximum likelihood
estimator of ?1.
n
(xi x)

?
?1
has a normal distribution with
Yi
n
?
i 1
(xj x)2
j 1
n
(xi x)
n
(xi x)
?
?
(?0 ?1xi)
?0 ?1x ?1(xi x)
mean
n
n
?
?
i 1
i 1
(xj x)2
(xj x)2
j 1
j 1
n
(xi x)
n
(xi x)
n
(xi x)2
?
?
?
?0 ?1x
?1(xi x)
?1
n
n
n
?
?
?
i 1
i 1
i 1
(xj x)2
(xj x)2
(xj x)2
j 1
j 1
j 1
?1
12
2
n
n
(xi x)
(xi x)2
?
?
?2
?2
and variance

2
n
n
?
?
i 1
i 1
(xj x)2
(xj x)2
j 1
j 1
?2
n
?
(xi x)2
i 1
13
2. - continued (b)
Use Theorem 5.5-1 (Class Exercise 5.5-1) to find
the distribution of the maximum likelihood
estimator of ?0.

x
(xi x)
n
1 n
?
?0
has a normal distribution with

Yi
n
?
i 1
(xj x)2
j 1
x
(xi x)
n
1 n
?
mean
(?0 ?1xi)

n
?
i 1
(xj x)2
j 1
x
(xi x)
n
1 n
n
?
?
(?0 ?1xi)
(?0 ?1xi)
n
?
i 1
i 1
(xj x)2
We already found this in part (a).
j 1
n
(xi x)
?
(?0 ?1xi)
?0
?1x
x
n
?
i 1
(xj x)2
j 1
14
?0
?1x ?1x
?0
2
x
(xi x)
n
1 n
?
?2
and variance


n
?
i 1
(xj x)2
j 1
x
x2
2 (xi x)
n
(xi x)2
1 n2
?
?2



2
n
n
?
?
n
i 1
(xj x)2
(xj x)2
j 1
j 1
x2
1 n
?2

n
?
(xi x)2
i 1
15
Suppose we treat the joint p.d.f. as a function
L(?0 , ?1 , ?2), that is, a function of the three
unknown parameters (instead of just two). Then,
analogous to Text Example 6.1-3, we find that the
maximum likelihood estimates for ?0 and ?1 are
the same as previously derived, and that the
maximum likelihood estimator for ?2 is
n


?
Yi (?0 ?1xi)2
i 1
n
Recall If Y1 , Y2 , , Yn are independent with
each having a N(?, ?2) distribution (i.e., a
random sample from a N(?, ?2) distribution),
then Y has a distribution, has
a distribution, and
n
?
Yi
?2 n
N(? , )
i 1
n
(n 1)S2 ?2
?2(n 1)
16
(n 1)S2 ?2
the random variables Y and are
independent.
Analogous results for the more general situation
previously considered can be proven using matrix
algebra. Suppose Y1 , Y2 , , Yn are independent
with respective N(?1 , ?2), N(?2 , ?2), , N(?n
, ?2) distributions. Let x1 , x2 , , xn be
fixed values not all equal, and suppose that for
i 1, 2, , n, ?i ?0 ?1xi . Then ?1 has
a distribution, ?0 has a
distribution,
?2

N( ?1 , )
n
?
(xi x)2
i 1
x2

1 n
N( ?0 , )
?2

n
?
(xi x)2
i 1
17
n


?
Yi (?0 ?1xi)2
?2(n 2)
i 1
has a distribution, random
variables and are independent,
and random variables and are
independent.
?2
n


?
Yi (?0 ?1xi)2

i 1
?1
?2
n


?
Yi (?0 ?1xi)2

i 1
?0
?2
18



For each i 1, 2, , n, we define the random
variable Yi ?0 ?1xi , that is, Yi is the
predicted value corresponding to xi . With
appropriate algebra, it can be shown that


n
n
n
?
?
?
(Yi Y)2
(Yi Y)2
(Yi Yi)2
i 1
i 1
i 1
This is called the total sum of squares and is
denoted SST.
This is called the regression sum of squares and
is denoted SSR.
This is called the error (residual) sum of
squares and is denoted SSE.
Since, as we have noted, SSE / ?2 has a ?2(n 2)
distribution, we say that the df (degrees of
freedom) associated with SSE is n 2 .
If Y1 , Y2 , , Yn all have the same mean, that
is, if ?1 0, then SST / ?2 has a ?2(n 1)
distribution consequently, the df associated
with SST is n 1 .
If Y1 , Y2 , , Yn all have the same mean, that
is, if ?1 0, then it can be shown that SSR and
SSE are independent, and that SSR / ?2 has a
?2(1) distribution consequently, the df
associated with SSR is 1 .
19
3.
Suppose Y1 , Y2 , , Yn are independent with
respective N(?1 , ?2), N(?2 , ?2), , N(?n , ?2)
distributions. Let x1 , x2 , , xn be fixed
values not all equal, and suppose that for i 1,
2, , n, ?i ?0 ?1xi . Prove
that SST SSR SSE .




First, we observe that for i 1, 2, , n, Yi
?0 ?1xi
Y ?1(xi x)
2


n
n
?
?
(Yi Y)2
?1(xi x) (Yi Y) ?1(xi x)
SST
i 1
i 1


n
?
?1(xi x)2
(Yi Y) ?1(xi x)2
i 1


2?1(xi x)(Yi Y) ?1(xi x)


n
n
?
?
?1(xi x)2
Yi Y ?1(xi x)2
i 1
i 1


n
n
?
?
2 ?1(xi x)(Yi Y) ?12(xi x)2

i 1
i 1
20


n
n
?
?
(Yi Y)2
(Yi Yi)2
i 1
i 1


n
n
?
?
2 ?1 (xi x)(Yi Y) ?12 (xi x)2

i 1
i 1
SSR SSE
2
n
n
?
?
(xi x)(Yi Y)
(xi x)(Yi Y)
n
n
?
?
2 (xi x)(Yi Y)
(xi x)2
i 1
i 1
2
n
n
?
?
(xi x)2
(xi x)2
i 1
i 1
i 1
i 1
SSR SSE
21
A mean square is a sum of squares divided by its
degrees of freedom.
SSE n 2
The error (residual) mean square is MSE .
SSR 1
The regression mean square is MSR .
Variation in Y explained by the linear
relationship with X is based on SSR.
Total variation in Y is based on SST.
y
Variation in Y explained by random error is based
on SSE.
22
If Y1 , Y2 , , Yn all have the same mean, that
is, if ?1 0, then E(SSR / ?2) and E If ?1
? 0, then E(SSR / ?2)
1
SSE / ?2 n 2
1 .


n
n
?
?
(Yi Y)2
?1(xi x)2
i 1
i 1
E ?2
E ?2
n
n
?
?
(xi x)2
(xi x)2



E(?12)
Var(?1) E(?1)2
i 1
i 1
?2
?2
n
?
(xi x)2
1 ?12 gt 1 .
i 1
?2
23
SSE / ?2 n 2
This suggests that if ?1 0, then the ratio of
SSR / ?2 to is expected to be close to one, but
if ?1 ? 0, then this ratio is expected to be
larger than one.
If ?1 0, then since SSR and SSE are
independent, then the ratio of SSR / ?2 to
must have an distribution.
SSE / ?2 n 2
f( , )
1
n 2
Consequently, we can perform the hypothesis
test H0 ?1 0 vs. H1 ?1 ? 0 . by using
the test statistic F We reject H0 (in favor
of H1) when
The linear relationship (correlation) between X
and Y is not significant.
The linear relationship (correlation) between X
and Y is significant.
SSR / ?2
MSR . MSE
SSE / ?2 n 2
f gt f?(1, n 2) .
24
The calculations in a regression leading up to
the f statistic are often organized into an
analysis of variance (ANOVA) table such as the
following
Source df SS MS f p-value
Regression
Error
Total
SSR
1
MSR
MSR / MSE
SSE
n 2
MSE
n 1
SST
It can be shown that the squared sample
correlation is R2 which is often called the
proportion of variation in Y explained by X. The
standard error of estimate is defined to be SXY
SSR . SST
(This is done in Class Exercise 7.)
? MSE .
25
1. - continued (c)
Find the sums of squares SSR, SSE, and SST then
construct the ANOVA table and perform the
corresponding f test with ? 0.05, find and
interpret the squared sample correlation, and
find the standard error of estimate.


n
n
?
?
SSR
(Yi Y)2
(xi x)2
?12
(1.2)2(100) 144
i 1
i 1
n
?
SST
(Yi Y)2
174.8
i 1
SSE
SST SSR
174.8 144 30.8
26
Source df SS MS f p-value
Regression
Error
Total
144
1
144
14.03
0.025lt p lt 0.05
30.8
3
10.267
4
174.8
Since f 14.03 gt f0.05(1,3) 10.13, we reject
H0. We conclude that the slope in the linear
regression of height on stride length is
different from zero (0.025lt p-value lt 0.05), and
the results suggest that this slope is
positive. (Note We could alternatively conclude
that the linear relationship or correlation is
significant, and that the results suggest a
positive linear relationship or correlation.)
R2 144 / 174.8 82.4 of the variation in
height is explained by stride length. The
standard error of estimate is SXY ? 10.267
3.20 inches.
27
8. (a)
The prediction of grip strength from age for
right-handed males is of interest. It is assumed
that for any age x in years, where 10 ? x ? 25, Y
grip strength in pounds has a N(?x , ?2)
distribution where ?x ?0 ?1x . For a
random sample of right-handed males, the
following data are recorded Age (years) 15 17
19 11 16 22 17 25 12 14 25 23 Grip
Strength (lbs.) 50 54 66 46 58 54 64 80
46 70 76 80
Obtain the calculations below from a calculator
and from SPSS.
x
n
y
12
18
62
n
n
?
?
(yi y)2
(xi x)2
256
1728
i 1
i 1
n
?
(xi x)(yi y)
r
512
0.770
i 1
28
(No Transcript)
29
8. - continued (b) (c)
Find the equation of the least squares line from
a calculator and from SPSS.

?1 ?0
2

The least squares line can be written
y 26 2x .

26
Write a one-sentence interpretation of the slope
in the least squares line and a one-sentence
interpretation of the intercept in the least
squares line.
Grip strength appears to increase on average by
about 2 pounds with each increase of one year in
age.
The intercept is the mean grip strength at age
zero, which makes no sense in this situation.
30
Find r2, and write a one-sentence interpretation.
(d) (e)
r2 0.593
About 59.3 of the variation in grip strength is
explained by age.
Find the standard error of estimate.
s 8.390
31
8. - continued (f)
Construct the ANOVA table and perform the
corresponding f test with ? 0.05.


n
n
?
?
SSR
(Yi Y)2
(xi x)2
?12
(2)2(256) 1024
i 1
i 1
n
?
SST
(Yi Y)2
1728
i 1
SSE
SST SSR
1728 1024 704
Source df SS MS f p-value
Regression
Error
Total
1024
1
1024
14.55
p lt 0.01
704
10
70.4
11
1728
The test statistic is f
14.55
32
The critical region with ? 0.05 is
f ? 4.96 .
The p-value is
smaller than 0.01 (from the table) or p 0.003
(from the SPSS output).
4.96 f0.05(1,10)
0
Since f 14.55 gt f0.05(1,10) 4.96, we reject
H0. We conclude that the slope in the linear
regression of grip strength on age is different
from zero (p lt 0.01), and the results suggest
that this slope is positive. (Note We could
alternatively conclude that the linear
relationship or correlation is significant, and
that the results suggest a positive linear
relationship or correlation.)
33
4.
Suppose Y1 , Y2 , , Yn are independent with
respective N(?1 , ?2), N(?2 , ?2), , N(?n , ?2)
distributions. Let x1 , x2 , , xn be fixed
values not all equal, and suppose that for i 1,
2, , n, ?i ?0 ?1xi . Show that
the maximum likelihood estimator of ?2 is not
unbiased then, find a constant multiple of this
maximum likelihood estimator which is an unbiased
estimator of ?2.
n


n


?
?
Yi (?0 ?1xi)2
Yi (?0 ?1xi)2
?2 n
?2 (n 2) n
E

E

i 1
i 1
n
?2
n


?
Yi (?0 ?1xi)2
n n 2
An unbiased estimator of ?2 is

i 1
n

n
?
(Yi Yi)2
SSE n 2

MSE
i 1
n 2
34

?1

?1

?
?1

?1

n
MSE
?
(xi x)2
n
?
(xi x)2
i 1
SSE / ?2 n 2
i 1
has a distribution.
t(n 2)


?0

?0
?0

?0

x2
x2
1 n
1 n
?
MSE


n
n
?
?
(xi x)2
(xi x)2
SSE / ?2 n 2
i 1
i 1
has a distribution.
t(n 2)
35
5.
Suppose Y1 , Y2 , , Yn are independent with
respective N(?1 , ?2), N(?2 , ?2), , N(?n , ?2)
distributions. Let x1 , x2 , , xn be fixed
values not all equal, and suppose that for i 1,
2, , n, ?i ?0 ?1xi .
For a given value x0 , use Theorem 5.5-1 (Class
Exercise 5.5-1) to find the distribution of Y
x0 ?0 ?1x0 (the predicted value of Y
corresponding to the value x0).




(xi x)
(x0 x)



n
Yi n
n
?
?
Y ?1(x0 x)
Y x0 ?0 ?1x0


Yi
n
?
(xj x)2
i 1
i 1
j 1
(xi x)
(x0 x)
n
1 n
?
has a normal distribution with

Yi
n
?
(xj x)2
i 1
j 1
36


mean
E(?0 ?1x0)
?0 ?1x0
We can find this with algebra analogous to that
in Class Exercise 2(b).
2
(x0 x)
n
1 n
(xi x)
?
?2
and variance


n
?
(x0 x)2
(xj x)2
1 n
i 1
?2

n
?
j 1
(xi x)2
i 1
In Class Exercise 2(b), this was just x .
In Class Exercise 2(b), this was a minus sign.
37



For a given value x0 , we define Y x0 ?0
?1x0 to be the predicted value of Y
corresponding to the value x0 this predicted
value has a
(x0 x)2
1 n
N( , )
?0 ?1x0
?2
distribution, and

n
?
(xi x)2
n


i 1
?
Yi (?0 ?1xi)2



random variables and are independent.
Y x0 ?0 ?1x0
i 1
?2





?0 ?1x0
(?0 ?1x0)

?0 ?1x0
(?0 ?1x0)

(x0 x)2
(x0 x)2
1 n
1 n
?
MSE


n
n
?
?
(xi x)2
(xi x)2
SSE / ?2 n 2
i 1
i 1
has a distribution.
t(n 2)
We assume minx1 , x2 , , xn ? x0 ? maxx1 ,
x2 , , xn, since prediction outside the range
of the data may not be valid.
38
6. (a)
Suppose Y1 , Y2 , , Yn are independent with
respective N(?1 , ?2), N(?2 , ?2), , N(?n , ?2)
distributions. Let x1 , x2 , , xn be fixed
values not all equal, and suppose that for i 1,
2, , n, ?i ?0 ?1xi .
Derive a 100(1 ?) confidence interval for the
slope ?1 .

?1

?1
? t?/2(n 2) 1 ?
P t?/2(n 2) ?
MSE
n
?
(xi x)2
i 1
MSE
MSE


P ?1 t?/2(n 2) ? ?1 ? ?1 t?/2(n
2)
n
n
?
?
(xi x)2
(xi x)2
i 1
i 1
1 ?
39
If ?1(0) is a hypothesized value for ?1 , then
derive the test statistic and rejection regions
corresponding to the one sided and two sided
hypothesis tests for testing H0 ?1 ?1(0) with
significance level ?.
(b)

?1

?1(0)
The test statistic is T
MSE
n
?
(xi x)2
i 1
For H1 ?1 lt ?1(0) , the rejection region is
t ? t?(n 2) .
For H1 ?1 gt ?1(0) , the rejection region is
t ? t?(n 2) .
For H1 ?1 ? ?1(0) , the rejection region is
t ? t?/2(n 2) .
40
8. - continued (g)
Perform the t test for H0 ?1 0.8 vs. H1 ?1 ?
0.8 with ? 0.05.

?1

0.8
2 0.8
The test statistic is t


2.288 .
MSE
70.4
n
256
?
(xi x)2
i 1
The two-sided critical region with ? 0.05 is
t ? 2.228 .
2.228 t0.025(10)
2.228
The p-value is
between 0.02 and 0.05 (from the table).
41
Since t 2.288 gt t0.025(10) 2.228, we reject
H0. We conclude that the slope in the linear
regression of grip strength on age is different
from 0.8 lbs. (0.02 lt p lt 0.05), and the results
suggest that this slope is greater than 0.8
lbs. (Note We could alternatively conclude that
the average change in grip strength is different
from 0.8 lbs. per year, and that this change is
greater than 0.8 lbs. per year.)
42
8. - continued (h)
Considering the results of the hypothesis tests
in parts (f) and (g), explain why a 95
confidence interval for the slope in the
regression would be of interest. Then find and
interpret the confidence interval.
Since rejecting H0 in part (f) suggests that the
hypothesized zero slope is not correct, and
rejecting H0 in part (g) suggests that the
hypothesized slope of 0.8 is not correct, a 95
confidence interval will provide us with some
information about the value of the slope, which
estimates the average change in grip strength
with an increase of one year in age.
MSE
MSE


?1 t?/2(n 2) ? ?1 ? ?1 t?/2(n 2)
n
n
?
?
(xi x)2
(xi x)2
i 1
i 1
70.4 256
2 ? (2.228)
0.832 and 3.168
43
We are 95 confident that the slope in the
regression to predict grip strength from age is
between 0.832 and 3.168 lbs.
44
6. - continued (c)
Derive a 100(1 ?) confidence interval for the
intercept ?0 .

?0

?0
? t?/2(n 2) 1 ?
P t?/2(n 2) ?
x2
1 n
MSE

n
?
(xi x)2
i 1
x2
1 n

MSE

P ?0 t?/2(n 2)
? ?0 ?
n
?
(xi x)2
i 1
x2
1 n

MSE

?0 t?/2(n 2)
n
1 ?
?
(xi x)2
i 1
45
If ?0(0) is a hypothesized value for ?0 , then
derive the test statistic and rejection regions
corresponding to the one sided and two sided
hypothesis tests for testing H0 ?0 ?0(0) with
significance level ?.
(d)

?0

?0(0)
The test statistic is T
x2
1 n
MSE

n
?
(xi x)2
i 1
For H1 ?0 lt ?0(0) , the rejection region is
t ? t?(n 2) .
For H1 ?0 gt ?0(0) , the rejection region is
t ? t?(n 2) .
For H1 ?0 ? ?0(0) , the rejection region is
t ? t?/2(n 2) .
46
8. - continued (i)
Perform the t test for H0 ?0 0 vs. H1 ?0 ? 0
with ? 0.05.

?0

?0
The test statistic is t

x2
1 n
MSE

n
?
(xi x)2
i 1
26

0

2.668
1 12
182 256
70.4

The two-sided critical region with ? 0.05 is
t ? 2.228 .
The p-value is
2.228 t0.025(10)
2.228
between 0.02 and 0.05 (from the table) or p
0.024 (from the SPSS output).
47
Since t 2.668 gt t0.025(10) 2.228, we reject
H0. We conclude that the intercept in the linear
regression of grip strength on age is different
from zero (0.02 lt p lt 0.05), and the results
suggest that this intercept is positive.
48
8. - continued (j)
Considering the results of the hypothesis test in
part (i), explain why a 95 confidence interval
for the intercept in the regression would be of
interest. Then find and interpret the confidence
interval.
Since rejecting H0 in part (i) suggests that the
hypothesized zero intercept is not correct, a 95
confidence interval will provide us with some
information about the value of the intercept.
x2
1 n

MSE

?0 t?/2(n 2)
n
?
(xi x)2
i 1
? ?0 ?
x2
1 n

MSE

?0 t?/2(n 2)
n
?
(xi x)2
i 1
49
26 ? (2.228)
1 12
182 256
70.4

4.288 and 47.712
We are 95 confident that the intercept in the
regression to predict grip strength from age is
between 4.288 and 47.712 lbs.
50
6. - continued (e)
For a given value x0 , we can call E(Y x0) ?0
?1x0 the mean of Y corresponding to the value
x0 , and an unbiased estimator of this mean is Y
x0 ?0 ?1x0 (from Class Exercise 5).
Derive a 100(1 ?) confidence interval for E(Y
x0) ?0 ?1x0 .






?0 ?1x0
(?0 ?1x0)
? t?/2(n 2) 1 ?
P t?/2(n 2) ?
(x0 x)2
1 n
MSE

n
?
(xi x)2
i 1
51
(x0 x)2


1 n
P ?0 ?1x0 t?/2(n 2)
MSE

n
?
(xi x)2
i 1
? ?0 ?1x0 ?
(x0 x)2
1 n


MSE

?0 ?1x0 t?/2(n 2)
n
?
(xi x)2
i 1
1 ?
52
6. - continued (f)
If ?0 is a hypothesized value for E(Y x0) ?0
?1x0 , then derive the test statistic and
rejection regions corresponding to the one sided
and two sided hypothesis tests for testing H0
E(Y x0) ?0 with significance level ?.



?0 ?1x0
?0
The test statistic is T
(x0 x)2
1 n
MSE

n
?
(xi x)2
i 1
For H1 E(Y x0) lt ?0 , the rejection region is
t ? t?(n 2) .
For H1 E(Y x0) gt ?0 , the rejection region is
t ? t?(n 2) .
For H1 E(Y x0) ? ?0 , the rejection region is
t ? t?/2(n 2) .
53
8. - continued (k)
Perform the t test for H0 vs. H1 with ?
0.05, that is, H0 E(Y x 20) 80 vs. H1
E(Y x 20) ? 80
The mean grip strength for 20 year old
right-handed males is 80 lbs.
The mean grip strength for 20 year old
right-handed males is different from 80 lbs.
The test statistic is t



?0 ?1x0
?0
66

80


5.304
(x0 x)2
1 n
1 12
(20 18)2 256
70.4
MSE


n
?
(xi x)2
i 1
54
The two-sided critical region with ? 0.05 is
t ? 2.228 .
2.228 t0.025(10)
2.228
The p-value is
less than 0.01 (from the table).
Since t 5.304 gt t0.025(10) 2.228, we reject
H0. We conclude that the mean grip strength for
20 year old right-handed males is different from
80 lbs. (p lt 0.01), and the results suggest that
this mean is less than 80 lbs.
55
8. - continued (l)
Considering the results of the hypothesis test in
part (k), explain why a 95 confidence interval
for the mean grip strength for 20 year old
right-handed males would be of interest. Then
find and interpret the confidence interval.
Since rejecting H0 in part (k) suggests that the
mean grip strength for 20 year old right-handed
males is not 80 lbs., a 95 confidence interval
will provide us with some information about this
mean.
(x0 x)2


1 n
?0 ?1x0 t?/2(n 2)
MSE

n
?
(xi x)2
i 1
? ?0 ?1x0 ?
(x0 x)2
1 n


MSE

?0 ?1x0 t?/2(n 2)
n
?
(xi x)2
i 1
56
66 ? (2.228)
1 12
(20 18)2 256
70.4

60.12 and 71.88
We are 95 confident that the mean grip strength
for 20 year old right-handed males is between
60.12 and 71.88 lbs.
Return to 6, part (g)
57
For a given value x0 , let Y0 be a random
variable independent of Y1 , Y2 , , Yn and
having a N(?0 , ?2) where ?0 ?0 ?1x0 , that
is, Y0 is a new random observation. Use
Theorem 5.5-1 (Class Exercise 5.5-1) to derive a
100(1 ?) prediction interval for the value of
Y0 .
(g)


Y0 (?0 ?1x0)
has a normal distribution with
mean ?0 ?1x0 (?0 ?1x0) 0
(x0 x)2
1 n
and variance ?2 .
?2

n
?
(xi x)2
i 1


Y0 (?0 ?1x0)


Y0 (?0 ?1x0)

(x0 x)2
1 1 n
?
(x0 x)2

1 1 n
n
MSE
?

(xi x)2
n
?
(xi x)2
i 1
i 1
SSE / ?2 n 2
has a distribution.
t(n 2)
58
6. (g) - continued


Y0 (?0 ?1x0)
t?/2(n 2) ?
? t?/2(n 2)
P
(x0 x)2
1 1 n
MSE

n
?
(xi x)2
i 1
1 ?
59
(x0 x)2
1 1 n


P (?0 ?1x0) t?/2(n 2)
MSE

n
?
(xi x)2
i 1
? Y0 ?
(x0 x)2
1 1 n


(?0 ?1x0) t?/2(n 2)
MSE

n
?
(xi x)2
i 1
1 ?
60
8. - continued (m)
Find and interpret a 95 prediction interval for
the grip strength for a 20 year old right-handed
male.
(x0 x)2
1 1 n


(?0 ?1x0) t?/2(n 2)
MSE

n
?
(xi x)2
i 1
? Y0 ?
(x0 x)2
1 1 n


(?0 ?1x0) t?/2(n 2)
MSE

n
?
(xi x)2
i 1
66 ? (2.228)
1 1 12
(20 18)2 256
70.4

61
46.40 and 85.60
We are 95 confident that the grip strength for a
randomly selected 20-year old right-handed male
will be between 46.40 and 85.60 lbs. OR At least
95 of 20-year old right-handed males have a grip
strength between 46.40 and 85.60 lbs.
(n)
For what age group of right-handed males will the
confidence interval for mean grip strength and
the prediction interval for a particular grip
strength both have the smallest length?
18 year olds
62
6. - continued (h)
Consider the two sided hypothesis test H0 ?1 0
vs. H1 ?1 ? 0 with significance level ?, which
is one of the hypothesis tests defined in part
(b). Prove that the square of the t test
statistic for this hypothesis test is equal to
the f test statistic in the one-way ANOVA.
2


n
?
(xi x)2
?1

0
?12
MSR MSE
i 1


MSE
MSE
n
?
(xi x)2
i 1
We see this from the derivation in Class Exercise
3.
63
7.
Suppose Y1 , Y2 , , Yn are independent with
respective N(?1 , ?2), N(?2 , ?2), , N(?n , ?2)
distributions. Let x1 , x2 , , xn be fixed
values not all equal, and suppose that for i 1,
2, , n, ?i ?0 ?1xi . Prove that R2
SSR . SST
n
?
(xi x)2

Sy2 sx2 R2 sx2 Sy2
R2
?12
SSR SST
i 1

n
?
(Yi Y)2
i 1
64
8. - continued (o)
Use SPSS to graph the least squares line on a
scatter plot, and comment on how appropriate the
linear model seems to be.
The linear model appears to be reasonable.
65
(No Transcript)
66
9.
Voters in a state are surveyed. Each respondent
is assigned an identification number (ID), and
the following information about each is recorded
sex, area of residence (RES), political party
affiliation (POL), number of children (CHL), age,
yearly income (INC), job satisfaction score (JSS)
where 0 totally dissatisfied and 10 totally
satisfied, weekly hours spent watching television
(TVH), and weekly hours spent listening to radio
(RAD). The resulting data is as follows
ID SEX RES POL CHL
AGE INC JSS TVH RAD 08
M Suburban Republican 5 35 34,000 5
15 15 27 F Urban Democrat
2 20 28,000 2 20 13 34 F Suburban
Independent 4 35 71,000 7 18 17 18 M
Rural Independent 7 41 35,000 8
12 20 04 M Urban Republican
3 39 55,000 4 14 15 14 M Urban
Democrat 3 59 75,000 1 11 18 23 F
Urban Democrat 1 20 26,000 2
10 11 39 M Rural Other 4 52 30,000
9 12 30 54 F Rural Other
2 44 27,000 7 8 14 44 M Urban
Republican 0 46 53,000 0 21 10 17 F
Urban Republican 2 40 45,000 3 12 14
67
12 F Suburban Other 2 34 34,000 5
8 10 26 F Urban Republican
2 24 30,000 2 11 15 11 M Urban
Democrat 4 62 78,000 1 18 11 38 M
Suburban Other 3 44 68,000 7 17 12 09
F Rural Republican 6 44 29,000
9 8 27 29 F Suburban Democrat
4 38 40,000 9 9 25 13 M Rural
Independent 9 47 39,000 6 8 20 24
M Urban Democrat 3 44 60,000 2
27 4 33 F Urban Democrat
1 45 49,000 3 9 12 15 F
Rural Other 2 56 39,000 8 6 27 52
F Suburban Republican 0 32 33,000 5
11 15 35 M Suburban Other 1 54 65.000
3 14 14 30 F Rural Independent
3 41 25,000 7 5 25 47 M Suburban
Republican 6 50 61,000 5 15 15 32 F
Rural Republican 2 59 41,000 8
8 27 41 F Suburban Other 3 44 44,000
3 10 18 10 M Rural Other
3 62 45,000 10 10 23 48 M Suburban
Republican 2 53 64,000 1 13 15 02 M
Rural Democrat 8 59 39,000 8
9 24
68
9. - continued (a)
Do this exercise for homework.
The data is stored in the SPSS data file survey
with income entered in units of thousands of
dollars. The prediction of yearly income from
age is of interest, and the 30 individuals
selected for the data set are treated as a random
sample for simple linear regression. It is
assumed that for any age x in years, where 20 ? x
, Y yearly income has a N(?x , ?2)
distribution with ?x ?0 ?1x .
Use the Analyze gt Regressiongt Linear options in
SPSS to select the variable income for the
Dependent slot and select the variable age for
the Independent(s) section. To have the mean and
standard deviation displayed for the dependent
and independent variables, click on the
Statistics button, and select the Descriptives
option. To add predicted values and residuals to
the data file, click on the Save button, select
the Unstandardized option in the Predicted Values
section, and select the Unstandardized option in
the Residuals section.
69
10.
Use the fact that the random variable SSE has
a distribution ?2 to derive a 100(1 ?)
prediction interval for ?2 .
?2(n2)
SSE P lt lt 1 ?
?2
?21 ?/2(n2)
?2?/2(n2)
?2 P gt gt 1 ?
SSE
1 ?2?/2(n2)
1 ?21 ?/2(n2)
P gt ?2 gt 1 ?
SSE ?2?/2(n2)
SSE ?21 ?/2(n2)
P lt ?2 lt 1 ?
SSE ?2?/2(n2)
SSE ?21 ?/2(n2)