Classwork 2 sol - Yaxing Yang PDF

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Yaxing Yang...

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Solution to the Classwork2 (Assignment 7) a)i)

Analysis of Maximum Likelihood Estimates Paramete r

D F

Estimat e

Standar d Error

Wald ChiSquare

Pr > ChiS q

Intercept

1

0.1276

0.7313

0.0305

0.8615

vision

1

1

1.6692

0.8815

3.5858

0.0583

vision

2

1

0.1645

0.7784

0.0447

0.8326

drive

1

1

-0.8852

0.6681

1.7552

0.1852

Log(p/1-p) = 0.1276 + 1.6692* I{vision=1}+0.1645*I{vision=2}– 0.8852 *I{drive=1} Where p = Pr ( accident = 1).

Deviance and Pearson Goodness-of-Fit Statistics Criterion Value DF Value/DF Pr > ChiSq Deviance 2.6498 2 1.3249 0.2658 Pearson 2.6166 2 1.3083 0.2703 H0: The model is fitted well H1: The model is not fitted well Deviance Statistic= 1.3249 and its p-value is 0.2658. Pearson Statistic= 1.3083 and its p-value is 0.2703. Sine both p-value >0.05, the model is fitted well.

ii)

Odds Ratio Estimates Effect

Point Estimate

90% Wald Confidence Limits

vision 1 vs 3

5.308

1.245

22.627

vision 2 vs 3

1.179

0.328

4.241

drive 1 vs 0

0.413

0.137

1.238

Estimation of Odd Ratio for a smoker verse a non-smoke is 0.413, Confidence Interval= (0.137,1.238) iii)

Contrast Estimation and Testing Results by Row Contrast

Type

Ro w

Estimat e

Standar d Error

bad vision vs EXP norm vision

1

4.5029

3.7138

Alph a

Confidence Limits

0.05 0.8942 22.6744

Wald ChiSquare

Pr > ChiS q

3.3285

0.0681

Estimation of Odd Ratio= 4.5029 H0: Odd ratio is equal to 1 H1: Odd ratio is not equal to 1 Test Statistic= 3.3285, P-value= 0.0681>0.05. We conclude that the odd ratio is equal to 1. iv) Contrast Estimation and Testing Results by Row Contrast

Type

Ro w

Estimat e

Standar d Error

Alph a

Confidence Limits

Wald ChiSquare

Pr > ChiS q

drive=1 & vision=1

PRO B

1

0.7133

0.1488

0.05

0.3741

0.9120

1.5690

0.2104

drive=1 & vision=2

PRO B

1

0.3559

0.1414

0.05

0.1416

0.6493

0.9241

0.3364

drive=1 & vision=3

PRO B

1

0.3192

0.1356

0.05

0.1212

0.6144

1.4730

0.2249

Prob. (having an accident for vision=1 & drive=1)= 0.7133 Prob. (having an accident for vision=2 & drive=1)= 0.3559 Prob. (having an accident for vision=3 & drive=1)= 0.3192 The worse the vision , the higher the probability of having accident. b)i) Analysis of Maximum Likelihood Estimates Parameter

D F

Estimat e

Standar d Error

Wald ChiSquare

Pr > ChiS q

Intercept

1

1.0986

1.1547

0.9052

0.3414

vision

1

1

0.9808

1.5679

0.3913

0.5316

vision

2

1

-1.3218

1.3354

0.9796

0.3223

drive

1

1

-2.3514

1.4058

2.7978

0.0944

drive*visio n

1 1

1

0.9651

1.9624

0.2418

0.6229

Analysis of Maximum Likelihood Estimates Parameter

drive*visio n

1 2

D F

Estimat e

Standar d Error

Wald ChiSquare

Pr > ChiS q

1

2.5745

1.7106

2.2651

0.1323

Log(p/1-p) = 1.0986 + 0.9808*I{ vision=1}– 1.3218 *I{ vision=1} – 2.3514I{ drive=1} + 0.9651 *I{drive=1,vision=1}+ 2.5745 *I{drive=1,vision=1} Where p = Pr(accident=1) ii) Contrast Estimation and Testing Results by Row Contrast

Type

Ro w

Estimat e

Standar d Error

Alph a

bad vs good

EXP

1

2.6667

4.1811

0.01

Confidence Limits 0.0470

151.3

Wald ChiSquare

Pr > ChiS q

0.3913

0.5316

Estimation of odd ratio= 2.6667 H0: Odd ratio is equal to 1 H1: Odd ratio is not equal to 1 Test Statistic= 0.3913, p-value= 0.5316 So, we cannot rejeect H0 and we conclude that odd ratio is equal to 1. (Here we assume drive=1, the answer for assuming drive=0 is also right). c)

Analysis of Maximum Likelihood Estimates Paramete r

D F

Estimat e

Standar d Error

Wald ChiSquare

Pr > ChiS q

Intercept

1

0.8747

1.0644

0.6753

0.4112

1

-2.6399

1.7296

2.3296

0.1269

1 -0.00261

0.0222

0.0138

0.9063

1

0.0347

0.9588

0.3275

drive

1

age age*drive

1

0.0340

Contrast Estimation and Testing Results by Row Wald ChiSquare

Pr > ChiS q

0.6102

1.0914

0.2962

0.0546

0.6400

1.6948

0.1930

0.3700

0.8982

1.4164

0.2340

Contrast

Type

Ro w

Estimat e

Standar d Error

Alph a

Confidence Limits

40 & drive=1

PRO B

1

0.3749

0.1147

0.05

0.1869

20 & drive=1

PRO B

1

0.2426

0.1607

0.05

20 & drive=0

PRO B

1

0.6948

0.1466

0.05

(3)>(1)>(2).

Since the coefficient of drive is negtive and the coefficient of age*drive is positive, we conclude from the model that the subject receiving drive education has a lower possibility for having an accident than those without receiving drive education. Older driver with drive education has higher possibility for having an accident than young ones with drive education....