HOMEWORK SOLUTION FOR CHAPTER 9 AND 10
STAT 645
9.17
a. X1, X3
b. .10
c. X1, X3
d. X1, X3
e. The results of the three selection procedures are very consistent.
10.7
a.
Scatterplot of e(y|x2,x3) vs e(x1|x2,x3)
30
e(y|x2,x3)
20
10
0
-10
-20
-15
-10
-5
0
e(x
Stat 645 Homework 3 Solution
Chapter 2
Problem 2.1
(a) If the SLR is appropriate, the conclusion is warranted. The confidence interval
doesnt include zero which provides evidence that the slope is greater than zero.
Since the confidence level is 95%, the
Stat 645 Homework 2
Chapter 1
Problem 1.7
The model is Yi = 0 + 1 X i + i , where 0 = 100 , 1 = 20 , 2 = 25
(a) We can not decide the probability since we do not know the distribution of i
(b) In this case, we know that
i Yi ( 0 + 1 X i )
=
~ N (0,1) . T
STAT 645 HOMEWORK SOLUTION FOR CHANPTER 7 AND 8
Chapter 7
Problem 7.4
a. The AVNOVA table is:
Source
X1
X3|X1
X2|X1, X3
Error
Total
DF
1
1
1
48
51
Seq SS
MS
136366
136366
2033565
2033565
6675
6675
985530 20531.88
3162136 62002.67
b. H0: 2 = 0
HA: 2 0
SSR(
HOMEWORK SOLUTION ON CHAPTER 4 AND 5
4.3.
a. b0 and b1 tend to err in opposite directions because of the negative correlation
b. B = t(.9875; 43) = 2.32262,
b0 = 0.580157, scfw_b0 = 2.80394
b1 = 15.0352, scfw_b1 =0.483087
0.580157 2.32262(2.80394) 7.093 0
HOMEWORK 4 SOLUTION
3.7
a. Stem-and-Leaf Display: x
Stem-and-leaf of x
Leaf Unit = 1.0
8
15
21
29
(8)
23
16
4
4
5
5
6
6
7
N
= 60
11122334
5677788
123344
56777999
00013334
5556889
001223
I agree that the plot is consistent with the random selection of wome
Question ‘I
:1} Output from Mlnﬂab
One-Sample T: (:1
Teat of mu = 12 vs c 12
95%
Upper
variable N Mean Stnev SE Mean Bound T P
C1 29 11.9350 ﬂ.GBT2 6.0195 12.913? -ﬁ.? 0.225
W SEMEHH = «7% : Mm ,
b} Hurﬂﬂl W- ”051“”
C] “Emu: Prm:n.21530.ﬁi:m, we mm “re—11