Highlights
Question 1
For this question, DO NOT use the survival library in R.
The following table recorded the numbers of days for 20 patients to have stayed at Hospital A for treatment of accidental injuries:
|
Patient No. |
1-4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
17 |
18-20 |
|
Start day |
0 |
1 |
0 |
1 |
0 |
2 |
0 |
1 |
3 |
0 |
1 |
2 |
0 |
0 |
0 |
|
Days of stay |
1 |
1 |
2 |
2 |
3 |
3 |
4 |
4 |
4 |
5 |
6 |
6 |
8 |
8 |
10 |
|
End of stay |
Y |
Y |
Y |
N |
Y |
Y |
N |
N |
Y |
Y |
Y |
Y |
Y |
N |
N |
where
Let T denote the total time (in days) that a patient stays in all hospitals to treat the injury.
Question 2
Let { t i , δ i , i = 1, 2,…, n } be a random sample of survival data, where t i are survival times and δ i are censoring indicators with δ i = 1 if t i is uncensored, or δ i = 0 if censored. The data are modelled by the distribution with the following survival function:
, where α > 0 and β > 0 are two positive parameters.
|
ti |
0.2 |
4.3 |
2.5 |
3.3 |
1.4 |
0.8 |
3.8 |
0.4 |
|
δi |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
Assume (known) α = 1 . Use maxLik library in R and a modified function of what you wrote in part
Question 3
For this question, DO NOT use the survival library in R.
The following table shows the lifetimes of 14 lives after age 60:
|
Life i |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
|
ti |
6 |
6 |
8 |
8 |
8+ |
10 |
10+ |
15+ |
20 |
20 |
25 |
25+ |
30 |
30+ |
|
z(i) |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
where t i represents the lifetime of life i (in years after age 60), with + indicating survival; and z is a covariate to indicate if a life is male ( z = 0) or female ( z = 1) .
A Cox proportional hazards model is adopted to analyse the above data. Let β denote the coefficient of covariate z and b = e β .
Question 4
The times to the first accident by 15 drivers after getting their drivers’ licences are recorded below:
|
i |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
|
ti |
4* |
4 |
6 |
8* |
12 |
12* |
15 |
15 |
18 |
18* |
22 |
24 |
28 |
36 |
36* |
|
z1(i) |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
1 |
0 |
|
z2 (i) |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
where
A Cox proportional hazards model is applied to analyse the above data by the method of partial likelihood. Let b 1 and b 2 denote the coefficients of covariates z 1 and z 2 respectively, a = e b 1 and b = e b 2
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