1.Table 1 is a 2×2×2 contingency table which studied the effects of racial charac- teristics on the death penalty verdict.The victims'race (V),defendants'race (D) and death penalty(P)are three binary variables.
Table 1:Death penalty verdict by defendants race and victims race
|
|
Death Penalty
|
|
Victims'Race
|
Defendants'Race
|
Yes
|
No
|
|
White
|
White
|
53
|
414
|
|
Black
|
11
|
37
|
|
Black
|
White
|
1
|
16
|
|
Black
|
4
|
139
|
(a)Hand calculate the expected frequencies,the Pearson chi-square statistic,the degree of freedom under the total independence of V,D,and P.Report the total independence test result at the 5%significance level.Do you think the test result is appropriate? [5 marks]
(b)Hand calculate the expected frequencies,the Pearson chi-square statistic,the degree of freedom under the group independence between (V,D)and P.Re- port the group independence test result at the 5%significance level.[5 marks]
(c)Hand calculate the expected frequencies,the Pearson chi-square statistic,the degree of freedom under the conditional independence between D and P given V.Report the conditional independence test result at the 5%significance level. [5 marks]
(d)Hand calculate the expected frequencies,the Pearson chi-square statistic,the degree of freedom under the marginal independence between V and P.Report the marginal independence test result at the 5%significance level.[5 marks]
(e)Comment on the test results. [5 marks]
[Total:25 marks]
2.The dataset Neuralgia contains five variables:Treatment,Sex,Age,Duration, and Pain.The last variable,Pain=Yes,No,is the response variable.Treatment is a categorical variable with three levels A,B and P.Sex represents the gender of the patients and has two levels F and M.The Age and Duration are two continuous variables.The SAS outputs from the SAS proc logistic are given as follows.
|
Response Profile
|
|
Ordered Value
|
Pain
|
Total Frequency
|
|
1
|
No
|
35
|
|
2
|
Yes
|
25
|
Probability modeled is Pain='No'.
|
Class Level Information
|
|
Class
|
Value
|
Design Variables
|
|
Treatment
|
A
|
|
0
|
|
|
B
|
0
|
1
|
|
|
P
|
0
|
0
|
|
Sex
|
F
|
1
|
|
|
|
M
|
0
|
|
|
Testing Global Null Hypothesis:BETA=0
|
|
Test
|
Chi-Square
|
DF
|
Pr>ChiSo
|
|
Likelihood Ratio
|
32.9074
|
7
|
<.0001
|
|
Score
|
25.6812
|
7
|
0.0006
|
|
Wald
|
14.2879
|
7
|
0.0463
|
|
Analysis of Maximum Likelihood Estimates
|
|
Parameter
|
|
|
DF
|
Fstimate
|
Standard Error
|
Wald Chi-Square
|
Pr>ChiSq
|
|
Intercept
|
|
|
1
|
15.8108
|
6.6978
|
5.5725
|
0.0182
|
|
Treatment
|
A
|
|
|
3.5448
|
1.5135
|
5.4853
|
0.0192
|
|
Treatment
|
B
|
|
|
3.9417
|
1.6123
|
5.9771
|
0.0145
|
|
Sex
|
F
|
|
1
|
2.1392
|
1.3958
|
2.3490
|
0.1254
|
|
Treatment*Sex
|
A
|
F
|
|
-0.7067
|
1.9377
|
0.1330
|
0.7153
|
|
Treatment*Sex
|
B
|
F
|
|
-0.2072
|
1.9361
|
0.0114
|
0.9148
|
|
Age
|
|
|
|
-0.2688
|
0.0996
|
7.2744
|
0.0070
|
|
Duration
|
|
|
|
0.00523
|
0.0333
|
0.0247
|
0.8752
|
(a)The procedure models the probability of which event?Write down the fitted logistic model [4 marks]
(b)What is the null hypothesis for Global Null Hypothesis?Report the global null hypothesis test result at the 5%significance level.Is the sample size sufficiently large? [4 marks]
(c)Consider the following two patients with
(i)Treatment=P,Sex=M,Age=30,Duration=5; (ii)Treatment=B,Sex=F,Age=50,Duration=10.
Write down the coefficients of contrast for patient (i);patient (ii);and (i)vs (ii).Report the odds ratio of (i)vs (ii)as well as the corresponding 95% confidence interval and provide appropriate interpretations. [9 marks]
(d)What is the null hypothesis for the following TEST statement?
test SexF +TreatmentASexF =0,SexF +TreatmentBSexF =0;
Report the test result at the 5%significance level. [4 marks]
(e)Report the result of the Hosmer-Lemeshow lack-of-fit test at the 5%signif- cance level.The lack-of-fit test yields p=0.4390 for the probit link function. Which link function fits better? [4 marks] [Total:25 marks]