Solutions Manual to Advanced Regression Models with SAS and R [1 ed.]

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Solutions Manual to Advanced Regression Models with SAS and R [1 ed.]

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SOLUTIONS MANUAL FOR Korosteleva, O. (2018). Advanced Regression Models with SAS and R, CRC Press By OLGA KOROSTELEVA Department of Mathematics and Statistics California State University, Long Beach

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TABLE OF CONTENTS CHAPTER 1 ……………………………………………………………………………………. 3 CHAPTER 2 ……………………………………………………………………………………. 24 CHAPTER 3 ……………………………………………………………………………………. 58 CHAPTER 4 ……………………………………………………………………………………. 92 CHAPTER 5 ……………………………………………………………………………………. 131 CHAPTER 6 ……………………………………………………………………………………. 163 CHAPTER 7 ……………………………………………………………………………………. 187 CHAPTER 8 ……………………………………………………………………………………. 218 CHAPTER 9 ……………………………………………………………………………………. 284 CHAPTER 10 …………………………………………………………………………………. 315

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CHAPTER 1 EXERCISE 1.1. Show that the normal distribution belongs to the exponential family of distributions. 𝑓(𝑦, 𝜇, 𝜎 ) = √

exp −

(

)

= exp − ln(2𝜋𝜎 ) −

(𝑦 − 2𝑦𝜇 + 𝜇 ) . Let 𝜃 = 𝜇

and 𝜙 = 𝜎 . Then, we can write 𝑓(𝑦, 𝜃, 𝜙) = exp − ln(2𝜋𝜙 ) −

= exp

− ln(2𝜋𝜙) −

( )

= exp

(𝑦 − 2𝑦𝜃 + 𝜃 )

+ ℎ(𝑦, 𝜙) where 𝑐(𝜃) =

, and

1 𝑦 ℎ(𝑦, 𝜙) = − ln(2𝜋𝜙) − . 2 2𝜙

EXERCISE 1.2. (a) Verify normality of the response variable, then fit the linear regression model to the data. State the fitted model. Give estimates for all parameters. In SAS: data weightloss; input drug$ age gender$ EWL cards; A 49 F 14.2 A 54 M 25.4 A A 34 F 15.9 A 51 F 17.4 A A 44 M 8.4 A 56 M 11.2 A B 51 M 21.9 B 44 F 23.6 B B 26 M 14.1 B 56 F 24.6 B B 52 F 15.7 B 54 F 23.7 ;

@@; 37 54 44 53 28

F F M F F

14.1 22.8 17.3 23.8 17.8

A A A B B

43 45 47 55 34

F F M M M

20.0 16.7 20.5 7.4 27.8

A A A B B

57 36 44 30 43

M M F F M

11.7 12.7 6.7 23.1 10.6

A A B B B

48 57 52 47 55

M M F M M

16.6 15.0 29.4 16.8 26.8

/*running normality check*/ proc univariate; var EWL; histogram/normal; run;

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Goodness-of-Fit Tests for Normal Distribution Test Statistic p Value Kolmogorov-Smirnov D 0.10216310 Pr > D >0.150 Cramer-von Mises W-Sq 0.05103595 Pr > W-Sq >0.250 Anderson-Darling A-Sq 0.28788730 Pr > A-Sq >0.250

Based on the large p-values of the normality tests and the histogram, we can conclude that the response variable follows a normal distribution. /*fitting general linear model*/ proc genmod; class drug(ref="A") gender; model EWL = drug age gender / dist=normal link=identity; run; Log Likelihood -98.4395

Parameter Intercept drug drug age gender gender Scale

DF

B A F M

1 1 0 1 1 0 1

Analysis Of Maximum Likelihood Parameter Estimates Estimate Standard Wald 95% Confidence Wald Chi- Pr > ChiSq Error Limits Square 9.2146 5.3301 -1.2322 19.6614 2.99 0.0838 4.8103 1.8697 1.1456 8.4749 6.62 0.0101 0.0000 0.0000 0.0000 0.0000 . . 0.1102 0.1067 -0.0988 0.3192 1.07 0.3015 2.7235 1.8664 -0.9346 6.3815 2.13 0.1445 0.0000 0.0000 0.0000 0.0000 . . 5.2451 0.6556 4.1054 6.7012

The fitted model is 𝐸 (𝐸𝑊𝐿) = 9.2146 + 4.8103 ∙ 𝑑𝑟𝑢𝑔𝐵 + 0.1102 ∙ 𝑎𝑔𝑒 + 2.7235 ∙ 𝑓𝑒𝑚𝑎𝑙𝑒 , and 𝜎 = 5.2451. In R: weightloss.data0.250

P-values for the normality tests are all in excess of 0.05, indicating that normality holds. The histogram also displays a distribution close to bell-shaped. 7

/*fitting general linear model*/ proc genmod; class bodystyle(ref="hatchback") country(ref="Japan") leather(ref="no"); model priceK=bodystyle country hwy doors leather/dist=normal link=identity; run; Log Likelihood -67.2613

Parameter Intercept bodystyle bodystyle bodystyle country country country hwy doors leather leather Scale

Analysis Of Maximum Likelihood Parameter Estimates DF Estimate Standard Wald 95% Confidence Wald Chi- Pr > ChiSq Error Limits Square 1 5.1353 4.6900 -4.0570 14.3276 1.20 0.2735 coupe 1 2.2698 1.6836 -1.0301 5.5696 1.82 0.1776 sedan 1 6.4107 1.5477 3.3772 9.4441 17.16