Stat 333 Spring 2004 2/17/2004
Homework 2 Solution
Problem B, page 97, text
2c. (1) When X= 5.0, ˆ
Y= 3. From the ANOVA table in previous question (Homework 1), we know
s2= MSE = 2.4. Then
est. V ar(ˆ
Y) = d
V ar(ˆ
Y) = s2
n=2.4
20 = 0.12
(Note: est. V ar(ˆ
Y) and d
V ar(ˆ
Y) have the same meaning, which is the estimate of the variance of ˆ
Y.)
Then the 95% confidence interval for ˆ
Yis
ˆ
Y±t18,.025 ·qd
V ar(ˆ
Y) = 3 ±2.101 ·√0.12 = (2.27,3.73)
(2) When X= 9.0, ˆ
Y= 5, and
d
V ar(ˆ
Y) = s21
n+(X0−¯
X)2
P(Xi−¯
X)2= 2.41
20 +16
160.0= 0.36
Then the 95% confidence interval for ˆ
Yis
ˆ
Y±t18,.025 ·qd
V ar(ˆ
Y) = 5 ±2.101√0.36 = (3.74,6.26)
3. a. First, perform the lack-of-fit test.
Source df SS MS F
Regression 1 40.0 40.0
Residual 18 43.2 2.4
Lack of fit 3 1.2 0.4 0.143
Pure Error 15 42.0 2.8
Total, corrected 19 83.2
Since F= 0.143 <3.29 = F3,15,.05, we do not reject H0:no lack of fit.
b. Since the variance of Yiseems to be constant according to the scatter plot, the confidence limits
calculated in 2c are applicable.
c. The model seem appropriate.
4. a. The ANOVA table is the same as that in 3a. So there is no lack of fit for this model.
b. Since it looks like that the variance of Yiis depending on level of Y, the confidence limits calculated
in 2c may not be applicable.
c. We may still consider the first-order model, i.e.,
Z=β0+β1X+
But here Zis some transformation of Y. (We’ll discuss this later.)
Homework 2 Solution 1Ting-Li Lin