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Question #1:
(a) The least-square estimate of the regression line when Y regressed on X is:
Y = − 1 .69956 + 0. 83991 X
The change in the mean response when X increases by one unit is just β 1
Therefore, the estimate of the change in the mean response when X increases
by one point is
β 1
(b) The scatter plot is given on the computer output on pages 2. The estimated
regression line seems ok, but other complicated model might fit better.
(c) Note that
1 − α = 0. 99 →
α
= 0. 005 , df = n − 2 = 20 − 2 = 18 → t α
2
A 99% confidence interval for β 1 is:
β 1
± t
- 005
se(
β 1
or (0. 42528 , 1 .25454). This confidence interval is also given in the computer
output on page 1. Therefore, we are 99% sure that the unknown β 1 is between
0.42528 to 1.25454. The confidence interval does not include zero, therefore, we
reject H 0 : β 1 = 0 versus H 1 : β 1
= 0 at α = 1 − 0 .99 = 0.01. The director of
admission might be interested in β = 0, since it implies that there is no linear
relationship between X and Y. By rejecting H 0 , we conclude that there is a
linear relationship between X and Y. Therefore, X is useful to predict Y. (Note
that a more complicated model might predict Y better. We should check this
part in our research.)
(d) If x = 5.0, we have
Y = − 1 .69956 + 0.83991(5) = 2. 49999
(e) S is an unbiased estimate of σ. Then
S = ˆσ =
M S
Res
Note that σ is population standard deviation of random variable Y and it is
expressed in the same unit that Y is expressed. Since S is an estimate of Y , it
has the same unit as σ and Y.
(f ) If x 0 = 4.7, we have
Y = − 1 .69956 + 0.83991(4.7) = 2. 248017
In addition, ¯x = 5 and
1 − α = 0. 05 →
α
= 0. 025 , df = n − 2 = 20 − 2 = 18 → t α
2
Therefore, a 95% confidence interval for μ Y
when x 0
= 4.7 is
[
2
]
or (2. 024386 , 2 .471648)
(g) A 95% confidence interval for Y when x 0
= 4.7 is
[
2
]
or (1. 3071 , 3 .1889)
(h) Yes. The prediction interval in part (f ) is wider than the confidence interval
in part (g). Note that the prediction interval in part (h) is an interval for a
single future observation while the confidence interval in part (f ) is an interval
for average of Y. Since the last one has less variation, it has shorter confidence
interval too.
(i) The test statistic for testing H 0
: β 1
= 0 is
T S =
β 1
se(
β 1
The p-value, for H 1
: β 1
= 0, is
P − value = 2P (T S > 5 .83) < 0. 0001
Since P − value < α = 0.05, we reject H 0
: β 1
(j) The test statistic for testing H 0
: β 0
= 0 is
T S =
β 0
se(
β 0
The p-value, for H 1 : β 0
= 0, is
P − value = 2P (T S > − 2 .34) = 0. 0311
Question #3:
(a) The least-square estimate of the regression line when Heart Wight (Hwt)
regressed on Body Weight (Bwt) for female is:
Hwt = 2.98131 + 2.63641 Bwt
The least-square estimate of the regression line when Heart Wight (Hwt) re-
gressed on Body Weight (Bwt) for male is:
Hwt = − 1 .18409 + 4.31268 Bwt
If you plot the estimated regression function and the data for male and female,
you will see that the regression line for male is ok, but the regression line for
female is not a good model. When the line is not good, we cannot make an
inference about population based on the line. For male, we might conclude that
heart weight is proportional to body weight.
(b) Note that
y¯ Female
= ¯y F
= 9. 20213 and ¯y Male
= ¯y M
In addition
β 0
= 9. 20213 and
β 1
Therefore, we have
β 0
= 9.20213 = ¯y F
and
β 0
β 1
= 9.20213 + 2.12055 = 11.32268 = ¯y M
Recall that our regression model is:
Y = β 0
X + E or μ Y
= β 0
X
If we put Z = 0 for female and Z = 1 for male, we have
μ F = β 0
- β 1 (0) = β 0 and μ M = β 0
- β 1 (1) = β 0
- β 1
Therefore,
μˆ F
β 0 and μˆ M
β 0
β 1
In addition, the estimate of population mean (μ) is sample mean (¯y). Then
μˆ F = ¯y F and μˆ M = ¯y M
So the above arguments imply that
β 0
= ¯y F
and
β 0
β 1
= ¯y M
Therefore, the result of comparisons is what we should expect from our model.
(c) The test statistic for testing H 0 : μ M = μ F or H 0 : μ M − μ F = 0 is
T S =
Y
M
Y
F
S
P
1
n M
1
n F
where
S
P
(n M
− 1)S
2
M
− 1)S
2
F
n M
Therefore, the test statistics is:
T S =
1
97
1
47
In addition, the test statistic for testing H 0
: β 1
= 0 is
T S =
β 1
se(
β 1
The value of two test are exactly the same. Recall that our model implies that
μ F
= β 0
and μ M
= β 0
Therefore, β 1
= 0 implies that μ F
= μ M
. So, H 0
: β 1
= 0 is equivalent
to H 0
: μ M
= μ F
. The two test statistics are exactly the same since both
null hypothesis are the same. This assignment show that the usual t-test for
comparing two means that we learned in an elementary statistical course (such
as STAT 151) is a special case of regression model. In general, experimental
design is special case of regression analysis when all independent variables are
indicator variables. We will talk about indicator variable such as Z in future.
Question #4:
Recall that
β 1
n ∑
i=
c i
Y
i
where
n ∑
i=
c i
Then
Cov
Y ,
β 1
= Cov
Y ,
n ∑
i=
c i
Y
i
n ∑
i=
c i
Cov
Y , Y
i
