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Question 1: 25 marks
A clinician conducted a study to investigate a new chemotherapy in the prevention of recurrent tumours in patients who have bladder cancer. Sixty-nine patients were allocated to one of three groups: group one received standard treatment, group two received a low dose of the new chemotherapy and group three received a high dose of the new chemotherapy. The outcome of interest was the number of new tumours. Results are presented in the attached EXCEL sheet, each cell shows the number of new tumours for a patient. You can assume that all patients had the same follow-up period.
Since the outcome is a count, a possible model for these data is a Poisson model. For each hypothesis test in parts a and b, clearly state your null hypothesis, show all steps in your derivation, calculate the p-value and interpret the results.
Question 2: 10 marks
We wish to assess a new intervention aimed at reducing the serum uric acid levels in patients with gout. To do this a randomized control trial is to be the conducted, where patients are randomized to either the new intervention or the control arm, which receives standard care. At the initial planning stage, the chief investigator stated that it will be feasible to recruit 100 patients in total for the study (50 in each arm). From a pilot study the pooled standard deviation was estimated to be 3 mg/100 mL. The planned analysis at the end of the study is to compare the mean serum uric acid levels between the two groups using a two-sample t-test and 5% significance level.
Let the true mean serum uric acid level in the control group and intervention be notated and , respectively. Hence the true difference in means can be notated . The estimator of is denoted which can be assumed to have a normal distribution.
i.e. Var( ̅ ̅ = where s is the standard deviation and n is the sample size per group
b. Plot a power function for the difference between the two groups ranging from -2 to 2 mg/100 mL. Assume that the hypothesis test will be two-sided. In a table, display the power for differences: -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, 2 mg/100 mL. Note, you may wish to use a finer grid to get a smooth curve on your plot. Explain or show how you calculated the power. 4 marks
c. One of your collaborators suggested conducting a one-sided test (at 5% significance level) instead of a two-sided test. On the same plot as the two-sided test, show the power function for a one-sided test. What is the advantage of a one-sided test (if any)? Which test would you use? Explain your answer. 3 marks
Question 3: 15 marks
When a patient receives an organ transplant, there is a risk that the graft will fail at some point. A clinician wishes to estimate the proportion of kidney transplant patients who experience graft failure within the first month of receiving their transplant. We shall denote this parameter by θ.
a. The clinician has a prior belief that the risk of having a graft failure in the first month is about 30%, and upon discussion with the clinician it was agreed that a Beta prior with α = 3 and β = 7 should be used.
Data from ten patients are collected, out of which 2 patient had a graft failure in the first month. Calculate the posterior distribution for θ. 2 mark
b. Data from another twenty patients are collected, out of which 3 had a graft failure in the first month. Using the posterior distribution from part a as your new prior, calculate a new posterior distribution for θ from the twenty new observations. 1 mark
c. Plot the prior and posterior distributions from part a. Also plot the posterior distribution from part b. How do the distributions change as data are added? Explain why this occurs. 3 marks
d. Suppose that we had waited to analyse the data until we had collected data from all thirty patients. Using the original Beta prior with α = 3 and β = 7. Calculate the posterior distribution for θ from the thirty observations (i.e. 5 graft failures out of 30). Compare this posterior distribution to the posterior distribution in part b. What are the implications of this? 2 marks
Using the posterior distribution calculated in part b:
e. What is the posterior probability that more than 25% of patients will have graft failure in the first month? 1 mark
f. Calculate a 95% credible interval for θ. Interpret this interval. 3 marks
g. Calculate a frequentist 95% CI for θ from the thirty observations. How does this interval differ from the interval calculated in part f? Explain why. 3 marks
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For this problem, there are three means of occurrence of tumours that we need to compare. The comparison will be done in pairs. The first comparison will be between group 1 (control group) vs. the group 2 (low dose of chemotherapy). And the second one will be between the group 1 vs. group 3 (high dose of chemotherapy). Before going to the comparison, the basic principle of likelihood ratio test is described below.
The likelihood function of the sample is now a joint probability function of all xi's and I’s, and is given by
L(θ1, θ2) = ∏ θ1xi e– θ1 /xi! ∏ θ2yi e– θ2 /you! (as the occurrence of tumours being a rare event follows Poisson distribution)
Under H0 : θ1=θ2 = θ, hence the likelihood function becomes a function of single parameter θ
L(θ)= 1/xi! yi! ∏ θ∑x i+∑ yi e– 2nθ
l(θ) = ln L(θ)= ln(1/ xi! yi! )+ (∑x i+∑ yi) ln θ -2n θ
dl(θ)/dθ= (∑x i+∑ yi)/ θ -2n
Now maximizing l(θ), we get
dl(θ)/dθ = 0
or, ∑x i+∑ yi -2n θ = 0
or, θ = 1/2n(∑x i+∑ yi) = ½(x̄ + ӯ)
L(θ1, θ2) is maximized when both θ1 and θ2 are replaced by their maximum likelihood estimates. Thus,
Likelihood ratio = Λ = θn(x̄ + ӯ)/ (x̄)nx̄ + (ӯ)nӯ
Now, the first comparison as indicated in the introduction. Let, the mean of the first group is θ1 and the same for the second group is θ2. Then,
H0 : θ1=θ2
The mean for the first group is calculated to be 42/23= 1.83 and the same for the second group is 25/23= 1.09. The total population is n=46. And, θ= ½(1.83+1.09)= 1.46.
Now, test statistic for likelihood ratio test = -2lnΛ= -2*[46*(1.83+1.09) * ln1.46 – 46*1.83*ln1.83 – 46*1.09*ln1.09] = -2*(50.83-50.87-4.32) = 8.72.
Now, the approximated distribution of likelihood ratio test statistic is a Χ21. Now, at 95% significance, the critical value is 3.84. The value of the test statistic is > 3.84. Hence, we can reject the null hypothesis successfully. The p-value calculated from Χ21 distribution is 0.003. The conclusion can be that the mean of tumours of group 1 and group 2 are not equal and there is a very low chance of committing type I error, equal to 0.3%. This also indicates that a low dose of chemotherapy actually helps in reducing tumours.
Coming to the second comparison. Let, the mean of group 3 is θ3 and the same for group 1 is already used as θ1. Then,
H0 : θ1=θ3
The mean for the third group is 23/23=1, the mean for the first group is 1.83. Total population n=46. And, θ= ½(1.83+1)= 1.42.
Test statistic for likelihood ratio test = -2lnΛ= -2*[46*(1.83+1) * ln1.42 – 46*1.83*ln1.83 – 46*1*ln1] = -2*(45.65-50.87-0) = 10.44.
Similarly, the value for test statistic is >>3.84. Hence, again we can reject the null hypothesis. The p value, in this case, is 0.001. Therefore, the mean of tumours of group1 and group 3 are also not equal with a 0.01% chance of committing type I error. This also shows that a high dose of chemotherapy seems to have a negative effect on the growth of new tumours.