代写Exercise 2代做Python编程

日期：2024-07-16 01:05

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]