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Email : - info@statisticsassignmentexperts.com
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Problem 1
Given a hazard function h(t) = c, derive the corresponding survival function S(t)
and probability density function f(t).
Solutions
That is 7 ~ Exponential(c) here.
Problem 2
Censoring is a missing data problem that is specific to survival analysis. Explain
what is meant by a right censored observation that occurred at week 18.
Solutions
A right-censored observation at 18 weeks means that the event of interest (e.g.
death, recurrence of an illness) had not occurred by 18 weeks and no further
information is known. An observation can be censored due to several reasons,
e.g. subject lost during follow-up, event of interest did not happen during the
follow-up period or the subject died from a different cause unrelated to the event
of interest.
Problem 3
Define the Kaplan-Meier estimator of the population survival function. Consider
a sample of survival data with event times:
6+ 66679 10+ 10 11 + 13 16 17+ 19 + 20+ 22
where a indicates a censored observation. Compute the Kaplan-Meier estimator
for these data.
Solutions
Consider a sample consisting of nobservation times (t1, . . . , tn), which contain
distinct event times.
statisticsassignmentexperts.com
Arrange the mdistinct ordered event/survival times in the sample by t(1) < t(2) < . .
. < t(m)where m≤n. Further let r(i)represent the number of individuals who are
alive just before t(i)and s(i)=r(i)−d(i) represent the number of individuals who
survive at least beyond time t(i). Then the Kaplan-Meier estimator is given by
Then for the data at hand the Kaplan-Meier estimator can be summarised by the
table below:
Table 1: Summary table of survival analysis.
Compute the numbers that are missing in the table above.
statisticsassignmentexperts.com
Solutions
• At time 2 we have r(i)= 20 and s(i)= 20 −3 = 17, so we have S(2) = 0.9524 ×17/20
= 0.8095.
• At time 17 we have r(i)= 8 and s(i)= 8 −1 = 7, so we have S(17) = 0.4906 ×7/8 =
0.4293.
• At time 9 the standard error should be
which here equates to 0.6071 ×0.1790 = 0.1087.
• Finally, at time t= 13 the ∗is
Problem 5
Two samples of five patients on two different treatments (A and B) were followed
until relapse of their disease. The survival times were recorded to the nearest
month and were as follows:
Time 3 6 9 + 11 14 7 11 16 + 17 21+
Treatment A A A A A B B B B B
statisticsassignmentexperts.com
Using the log-rank test, decide if there is sufficient evidence to suggest a
significant difference in the(population) survival functions for the two
treatments.
Solutions
For the log-rank test we have
Table 3: Summary table of log-rank test.
statisticsassignmentexperts.com
Problem 6
In a study of peritonitis infection after kidney transplants, the time (from the end
of the first episode) until occurrence of a second episode of peritonitis was
recorded, as well as whether or not a patient had a low or high immune response.
The data are Low immune response subjects:
1+ 1+ 1+ 1 2+ 2 5 8 9 9 12+ 12 12 14+ 15 17 20 23+ 23 26 30+ 39 40+.
High immune response subjects:
1+ 2 3 3 4+ 6 7+ 14 17 17 23 26 40+ 48+ 48+ 49 89+ 101+ 109+ 117+.
statisticsassignmentexperts.com
Figure 1: Kaplan-Meier survival curves plotted separately for high and low
immune response subjects in a study of peritonitis infection after kidney
transplants.
Comment on the plot before you carry out any formal statistical tests. Do you
anticipate to find a difference in the survival times between the two immune
responses?
Fill in the asterisks, *, in the table below and carry out the corresponding log-
rank test of whether or not the population survival distributions are identical for
the two types of immune response.
Table 4: Summary table of log-rank test with some values missing.
statisticsassignmentexperts.com
Solutions
The two survival curves in the plot cross each other. It is therefore reasonable to
expect no significant difference in the distribution of survival times between the
two immune response groups. That is only an inital impression and needs to be
followed up with a formal statistical analysis. For the log-rank test we have
Table 5: Summary table of log-rank test.
statisticsassignmentexperts.com
The log-rank test statistic is
which is not significant when compared to a χ2(1). One can therefore only
conclude that there is no strong evidence against the two survival distributions
being equal in the light of these data, i.e there is no significant difference in the
distribution of time to second episode of peritonitis between the two immune
response status populations.
Thus the exploratory plot and formal statistical analysis agree.
statisticsassignmentexperts.com
Problem 7
Write down the general form of the proportional hazards model and specify the
assumptions required for such a model to hold. Derive the survival function for
this model.
Solutions
The PH model is
h(t, z) = h0(t) exp(z⊤β).
The key assumptions are that the effects of time, t, and explanatory variables, z,
are multiplicative in the above model, i.e. the ratio of the hazards between two
different values of zdoes not depend upon t. One is also assuming that the
logarithm of the hazard ratio (relative to the baseline hazard, h0(t)) is linear and
additive in z. The survival function is given by
statisticsassignmentexperts.com
where H0(t) = {t
0h0(u)du is the baseline cumulative hazard function.
Problem 8
A Cox proportional hazards model is fit in Rto the data from Problem 6 with
treatment as its single explanatory variable. Treatment is coded as 0 for
treatment A and 1 for treatment B. By default, R chooses treatment A as baseline.
This results in the following output:
statisticsassignmentexperts.com
Using the above output, estimate the hazard ratio for treatment B relative to
treatment A and provide a 95% confidence interval for this ratio. What does this
suggest to you about the effect of the treatments on the risk of relapse?
Solutions
The estimated ratio of the hazard rate for treatment B to that for treatment A is
exp(−1.25) = 0.287. An approximate 95% CI for the hazard ratio is
exp[ ˆβ±1.96 ese( ˆβ)] = (exp(−2.97),exp(0.47)) = (0.05,1.60).
As this interval contains the value 1 there is no evidence of any significant
difference in hazard rates between the two treatments.
Problem 9
In heart transplant surgery, the following factors are thought to influence survival
times after operation:
(a) sex of the patient;
(b) age of the patient at operation; and
(c) heart-matching score.
statisticsassignmentexperts.com
This last variable is an attempt to measure how well the transplanted heart
matches the characteristics of the patient’s own heart and is scored on a scale of 0
(very poor) to 3 (very good).
Patients within a hospital trust have been randomly allocated to one of four
treatment courses after surgery, one of these being the standard regime and the
others all being minor modifications of the standard. There is particular interest
in whether any of these three modifications provide a general improvement in
survival time over the standard.
Write down the hazard function for a proportional hazards model incorporating
all these factors, and explain briefly how you would investigate the question of
interest.
Solutions
The model is h(t, z) = h0(t) exp(z⊤β),
With
The indicator variables for treatment should have the standard treatment as a
baseline, and each of the three indicator variables corresponding to one of the
three modifications.
statisticsassignmentexperts.com
Here β⊤= (β1, β2, β3, β4, β5, η21, η31, η41), where (η21, η31, η41) are the three
treatment effects relative to the standard treatment. Thus it is of interest to test
first whether η21 =η31 =η41 = 0 assuming all three covariates (sex, age and heart
status) are in the model.
If this proves significant, one could carry out a set of pairwise comparisons of the
form
ˆη21 ±N(0,1; 1 −0.025/3) ese(ˆη21)
ˆη31 ±N(0,1; 1 −0.025/3) ese(ˆη31)
ˆη41 ±N(0,1; 1 −0.025/3) ese(ˆη41)
to see if any of the three treatment effects are significantly different from zero
and therefore different from the standard treatment. Then one could compare the
new treatment effects via comparisons of η21 −η31 etc.
statisticsassignmentexperts.com

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Linear Modeling Survival Analysis Statistics Assignment Help

  • 1. For any Assignment related queries, call us at : - +1 678 648 4277 visit : - https://www.statisticsassignmentexperts.com/, or Email : - info@statisticsassignmentexperts.com
  • 2. statisticsassignmentexperts.com Problem 1 Given a hazard function h(t) = c, derive the corresponding survival function S(t) and probability density function f(t). Solutions That is 7 ~ Exponential(c) here. Problem 2 Censoring is a missing data problem that is specific to survival analysis. Explain what is meant by a right censored observation that occurred at week 18.
  • 3. Solutions A right-censored observation at 18 weeks means that the event of interest (e.g. death, recurrence of an illness) had not occurred by 18 weeks and no further information is known. An observation can be censored due to several reasons, e.g. subject lost during follow-up, event of interest did not happen during the follow-up period or the subject died from a different cause unrelated to the event of interest. Problem 3 Define the Kaplan-Meier estimator of the population survival function. Consider a sample of survival data with event times: 6+ 66679 10+ 10 11 + 13 16 17+ 19 + 20+ 22 where a indicates a censored observation. Compute the Kaplan-Meier estimator for these data. Solutions Consider a sample consisting of nobservation times (t1, . . . , tn), which contain distinct event times. statisticsassignmentexperts.com
  • 4. Arrange the mdistinct ordered event/survival times in the sample by t(1) < t(2) < . . . < t(m)where m≤n. Further let r(i)represent the number of individuals who are alive just before t(i)and s(i)=r(i)−d(i) represent the number of individuals who survive at least beyond time t(i). Then the Kaplan-Meier estimator is given by Then for the data at hand the Kaplan-Meier estimator can be summarised by the table below: Table 1: Summary table of survival analysis. Compute the numbers that are missing in the table above. statisticsassignmentexperts.com
  • 5. Solutions • At time 2 we have r(i)= 20 and s(i)= 20 −3 = 17, so we have S(2) = 0.9524 ×17/20 = 0.8095. • At time 17 we have r(i)= 8 and s(i)= 8 −1 = 7, so we have S(17) = 0.4906 ×7/8 = 0.4293. • At time 9 the standard error should be which here equates to 0.6071 ×0.1790 = 0.1087. • Finally, at time t= 13 the ∗is Problem 5 Two samples of five patients on two different treatments (A and B) were followed until relapse of their disease. The survival times were recorded to the nearest month and were as follows: Time 3 6 9 + 11 14 7 11 16 + 17 21+ Treatment A A A A A B B B B B statisticsassignmentexperts.com
  • 6. Using the log-rank test, decide if there is sufficient evidence to suggest a significant difference in the(population) survival functions for the two treatments. Solutions For the log-rank test we have Table 3: Summary table of log-rank test. statisticsassignmentexperts.com
  • 7. Problem 6 In a study of peritonitis infection after kidney transplants, the time (from the end of the first episode) until occurrence of a second episode of peritonitis was recorded, as well as whether or not a patient had a low or high immune response. The data are Low immune response subjects: 1+ 1+ 1+ 1 2+ 2 5 8 9 9 12+ 12 12 14+ 15 17 20 23+ 23 26 30+ 39 40+. High immune response subjects: 1+ 2 3 3 4+ 6 7+ 14 17 17 23 26 40+ 48+ 48+ 49 89+ 101+ 109+ 117+. statisticsassignmentexperts.com
  • 8. Figure 1: Kaplan-Meier survival curves plotted separately for high and low immune response subjects in a study of peritonitis infection after kidney transplants. Comment on the plot before you carry out any formal statistical tests. Do you anticipate to find a difference in the survival times between the two immune responses? Fill in the asterisks, *, in the table below and carry out the corresponding log- rank test of whether or not the population survival distributions are identical for the two types of immune response. Table 4: Summary table of log-rank test with some values missing. statisticsassignmentexperts.com
  • 9. Solutions The two survival curves in the plot cross each other. It is therefore reasonable to expect no significant difference in the distribution of survival times between the two immune response groups. That is only an inital impression and needs to be followed up with a formal statistical analysis. For the log-rank test we have Table 5: Summary table of log-rank test. statisticsassignmentexperts.com
  • 10. The log-rank test statistic is which is not significant when compared to a χ2(1). One can therefore only conclude that there is no strong evidence against the two survival distributions being equal in the light of these data, i.e there is no significant difference in the distribution of time to second episode of peritonitis between the two immune response status populations. Thus the exploratory plot and formal statistical analysis agree. statisticsassignmentexperts.com
  • 11. Problem 7 Write down the general form of the proportional hazards model and specify the assumptions required for such a model to hold. Derive the survival function for this model. Solutions The PH model is h(t, z) = h0(t) exp(z⊤β). The key assumptions are that the effects of time, t, and explanatory variables, z, are multiplicative in the above model, i.e. the ratio of the hazards between two different values of zdoes not depend upon t. One is also assuming that the logarithm of the hazard ratio (relative to the baseline hazard, h0(t)) is linear and additive in z. The survival function is given by statisticsassignmentexperts.com
  • 12. where H0(t) = {t 0h0(u)du is the baseline cumulative hazard function. Problem 8 A Cox proportional hazards model is fit in Rto the data from Problem 6 with treatment as its single explanatory variable. Treatment is coded as 0 for treatment A and 1 for treatment B. By default, R chooses treatment A as baseline. This results in the following output: statisticsassignmentexperts.com
  • 13. Using the above output, estimate the hazard ratio for treatment B relative to treatment A and provide a 95% confidence interval for this ratio. What does this suggest to you about the effect of the treatments on the risk of relapse? Solutions The estimated ratio of the hazard rate for treatment B to that for treatment A is exp(−1.25) = 0.287. An approximate 95% CI for the hazard ratio is exp[ ˆβ±1.96 ese( ˆβ)] = (exp(−2.97),exp(0.47)) = (0.05,1.60). As this interval contains the value 1 there is no evidence of any significant difference in hazard rates between the two treatments. Problem 9 In heart transplant surgery, the following factors are thought to influence survival times after operation: (a) sex of the patient; (b) age of the patient at operation; and (c) heart-matching score. statisticsassignmentexperts.com
  • 14. This last variable is an attempt to measure how well the transplanted heart matches the characteristics of the patient’s own heart and is scored on a scale of 0 (very poor) to 3 (very good). Patients within a hospital trust have been randomly allocated to one of four treatment courses after surgery, one of these being the standard regime and the others all being minor modifications of the standard. There is particular interest in whether any of these three modifications provide a general improvement in survival time over the standard. Write down the hazard function for a proportional hazards model incorporating all these factors, and explain briefly how you would investigate the question of interest. Solutions The model is h(t, z) = h0(t) exp(z⊤β), With The indicator variables for treatment should have the standard treatment as a baseline, and each of the three indicator variables corresponding to one of the three modifications. statisticsassignmentexperts.com
  • 15. Here β⊤= (β1, β2, β3, β4, β5, η21, η31, η41), where (η21, η31, η41) are the three treatment effects relative to the standard treatment. Thus it is of interest to test first whether η21 =η31 =η41 = 0 assuming all three covariates (sex, age and heart status) are in the model. If this proves significant, one could carry out a set of pairwise comparisons of the form ˆη21 ±N(0,1; 1 −0.025/3) ese(ˆη21) ˆη31 ±N(0,1; 1 −0.025/3) ese(ˆη31) ˆη41 ±N(0,1; 1 −0.025/3) ese(ˆη41) to see if any of the three treatment effects are significantly different from zero and therefore different from the standard treatment. Then one could compare the new treatment effects via comparisons of η21 −η31 etc. statisticsassignmentexperts.com