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Solved Questions for Applied Regression Analysis - Assignment | STAT 645, Assignments of Statistics

Ohio State University (OSU) - LimaStatistics

Material Type: Assignment; Class: Applied Regression Analysis; Subject: Statistics; University: Ohio State University - Main Campus; Term: Unknown 2000;

Typology: Assignments

Pre 2010

Uploaded on 07/23/2009

koofers-user-no6
koofers-user-no6 🇺🇸

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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, s{b0} = 2.80394
b1 = 15.0352, s{b1} =0.483087
−0.580157 ± 2.32262(2.80394) −7.093 ≤ β0 ≤ 5.932
15.0352 ± 2.32262(0.483087) 13.913 ≤ β1 ≤ 16.157
c. Yes.
H0: β0=0, β1=14
HA: β0≠0 or β1≠14
[]
[
]
780414XY)Xb(bYSSE(R) 2
jij
2
j10ij =−=+−=
∑
∑
∑∑ , df(R) = 45
SSE(F) = 3416, df(F) = 43.
n = 45, c = 10
The general linear test gives
63.8
43
3416
2
34164780
43
SSE(F)
43-45
SSE(F)-SSE(R)
*
F=÷
−
=÷=
F(0.95;2,43) = 3.21
Therefore, we reject H0.
4.8.
a. F(.95; 2,8) = 4.46, W = 2.987
Xh = 0: 10.2000 ± 2.987(.6633) 8.219 ≤ E{Yh} ≤ 12.181
Xh = 1: 14.2000 ± 2.987(.4690) 12.799 ≤ E{Yh} ≤ 15.601
Xh = 2: 18.2000 ± 2.987(.6633) 16.219 ≤ E{Yh} ≤ 20.181
b. B = t(.99167; 8) = 3.016, yes. Since 4.46>3.016.
c. F(.95; 3,8) = 4.07, S = 3.494
Xh = 0: 10.2000 ± 3.494(1.6248) 4.523 ≤ Yh(new) ≤ 15.877
Xh = 1: 14.2000 ± 3.494(1.5556) 8.765 ≤ Yh(new) ≤ 19.635
Xh = 2: 18.2000 ± 3.494(1.6248) 12.523 ≤ Yh(new) ≤ 23.877
d. B = 3.016, yes 3.016<3.494
4.12.
a. The regression equation is
y = 18.0 x
b. The regression line seems to be a good fit.
pf3
pf4
pf5

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Download Solved Questions for Applied Regression Analysis - Assignment | STAT 645 and more Assignments Statistics in PDF only on Docsity!

HOMEWORK SOLUTION ON CHAPTER 4 AND 5

a. b 0 and b 1 tend to err in opposite directions because of the negative correlation

b. B = t (. 9875; 43) = 2_._ 32262,

b 0 = − 0_._ 580157, s{b 0 } = 2_._ 80394 b 1 = 15_._ 0352 , s{b 1 } =0_._ 483087 − 0_._ 580157 ± 2_._ 32262(2_._ 80394) − 7_._ 093 ≤ β 0 ≤ 5_._ 932 15_._ 0352 ± 2_._ 32262(0_._ 483087) 13_._ 913 ≤ β 1 ≤ 16_._ 157 c. Yes. H 0 : β 0 =0, β 1 = HA: β 0 ≠0 or β 1 ≠ 14

SSE(R) [Y (b bX)] [Y 14X ] 4780

2 ij j

2

= ∑ ∑ ij − 0 + 1 j =∑ ∑ − = , df(R) = 45

SSE(F) = 3416, df(F) = 43. n = 45, c = 10 The general linear test gives

  1. 63 43

SSE(F)

* SSE(R)-SSE(F)

F ÷ =

= ÷ =

F(0.95;2,43) = 3.

Therefore, we reject H 0.

a. F (. 95; 2 , 8) = 4_._ 46, W = 2_._ 987

Xh = 0: 10_._ 2000 ± 2_._ 987(. 6633) 8_._ 219 ≤ E{Yh} ≤ 12_._ 181 Xh = 1: 14_._ 2000 ± 2_._ 987(. 4690) 12_._ 799 ≤ E{Yh} ≤ 15_._ 601

Xh = 2: 18_._ 2000 ± 2_._ 987(. 6633) 16_._ 219 ≤ E{Yh} ≤ 20_._ 181

b. B = t (. 99167; 8) = 3_._ 016, yes. Since 4.46>3.016.

c. F (. 95; 3 , 8) = 4_._ 07, S = 3_._ 494

Xh = 0: 10_._ 2000 ± 3_._ 494(1_._ 6248) 4_._ 523 ≤ Yh (new) ≤ 15_._ 877 Xh = 1: 14_._ 2000 ± 3_._ 494(1_._ 5556) 8_._ 765 ≤ Yh (new) ≤ 19_._ 635

Xh = 2: 18_._ 2000 ± 3_._ 494(1_._ 6248) 12_._ 523 ≤ Yh (new) ≤ 23_._ 877

d. B = 3_._ 016, yes 3.016<3.

a. The regression equation is

y = 18.0 x

b. The regression line seems to be a good fit.

x4.

y4.

5 10 15 20 25 30

600

500

400

300

200

100

Scatterplot of y4.12 vs x4.

c. H 0: β 1 = 17.5 , Ha : β 1 not equal 17..

B = t (. 99; 11) = 2.

If B*>2.72, conclude Ha, otherwise H0.

B*= (18.0283-17.5)/0.0795=6.65>2.72, conclude H a.

d.

Fit SE Fit 98% CI 98% PI 180.28 0.79 (178.12, 182.44) (167.84, 192.72)

a.

Sum of RESI1 = 3.

Problem 5.

a. Y’Y = 2194

b. (^) ⎟⎟ ⎠

X' X

c. (^) ⎟⎟ ⎠

X' Y

Problem 5.

A

Problem 5.

a. (^) ⎟⎟ ⎠

− −

  1. 1 0. 1

1 1 X' X

− 4

1 b X'X X'Y

−

1 H XXX X

e ( I H ) Y

SSE = e’e = 17.

−

  1. 22 0. 22

MSE( ) 2. 2

2 1 s b X'X

Yˆ^ h = X (^) h ' b = 18. 2 for X^ h =^2. s 2 {Yˆh}= MSE( Xh '(X'X) −^1 X h )= 0. 44

b. From part (a), s 2 {b 0 }=0.44,s{b 0 ,b 1 }=-0.22,s{b 1 }= 0. 22

c. The matrix of quadratic form for SSR is

n

H J