MATH38072 Page 2 of 11 A1. A randomised controlled trial of an electronic glucose monitor (E) against standard finger-prick glucose monitoring (F) in patients with Type 1 diabetes has been designed to use deterministic minimisation based on two baseline factors. The minimisation factors are trial centre (3 levels: Manchester; Birmingham, Cambridge) and HbA1c level (2 levels: 7.5%-9.0%; >9.0% to 11.0). Note: HbA1c is glycated haemoglobin, a measure of average glucose level over the past 2-3 months. The number of participants with each baseline factor level after 20 participants have been allocated to the intervention groups are given in the table below. Baseline factor Trial Centre HbA1c level Manchester Birmingham Cambridge 7.5% - 9.0% >9.0% - 11.0% Treatment E F E F E F E F E F Number of participants 4 2 3 3 3 5 5 6 5 4 (i) How many participants have been allocated to each treatment group? [1 mark] (ii) The characteristics of the next participant (participant 21) entered into the trial is: Trial Centre: Manchester HbA1c level: 7.5%-9.0%. Determine the intervention group to which this participant has to be allocated. [2 marks] (iii) What type of bias has an increased risk when deterministic minimisation is used rather than stratified randomisation? How would the use of stochastic minimisation reduce this risk of bias? [2 marks] MATH38072 Page 3 of 11 (iv) If stochastic minimisation is used with allocation probabilities 0.7 and 0.3, use the random number 0.60131 (from a uniform distribution on [0,1]) and a suitably-defined rule to allocate participant 21 to a treatment group. [2 marks] [Total 7 marks] A2. Van de Port (2012) reported testing for differences between treatment groups for 35 baseline variables, 4 of which had p-values less than 0.05. In the Discussion section of their paper they report some limitations of their trial, which included the text: “Firstly, although the trial was well powered, including 250 patients with few drop-outs, we found significant baseline differences in favour of the circuit training group for a few secondary outcomes. All analyses, however, were adjusted for these covariates at baseline.” (i) Explain why performing significance tests for baseline differences between treatment groups is not recommended in randomised controlled trials. [2 marks] (ii) When should the between-groups comparison of continuous outcome measures in randomized controlled trials be adjusted for baseline variables? Is it appropriate to use significance testing to select the baseline covariates to include in an adjusted analysis with the aim of reducing potential bias? (Justify your answer.) [3 marks] [Total 5 marks] MATH38072 Page 4 of 11 A3. A randomised controlled trial is carried out to compare two doses of a new vaccine for the prevention of H1N1 influenza. The effectiveness of the vaccine is tested by measuring the immune response in the blood measured 9 days following vaccination (as this measures successful vaccination and protection against influenza). The results are summarised in the table below: the numbers in the table are the respective counts for the different combinations of Dose (15mg; 30mg) and Immune response (Yes; No), with the totals randomised to each of the dose levels. Dose 15mg 30mg Immune response Yes 200 220 No 50 30 Total 250 250 (i) Estimate the odds ratio for the effectiveness of a 30mg dose as compared to a 15mg dose. [1 mark] (ii) Calculate the 95% confidence interval of this odds ratio. [5 marks] (iii) Interpret the results of this trial. [2 marks] [Total 8 marks] A4. In a meta-analysis of k trials, suppose that ̂ is an estimate of the treatment effect for the ith trial and let [̂] be its sampling variance. The minimum variance estimate is defined by: ̂ = ∑ ̂ =1 ∑ =1 where = 1 [̂] ⁄ , and [̂] = 1 ∑ 1 [̂] =1 . MATH38072 Page 5 of 11 The table below summarises the outcome of three trials of Neuromuscular Electrical Stimulation (NMES) versus no treatment for the condition of patellofemoral pain (PFP). The outcome measures was a Visual Analogue Scale (VAS) Pain score (range 0-10, with 0 representing no pain and 10 representing worst imaginable pain) during normal activities, assessed at the end of treatment. The treatment effect for each study ( , i =1, 2, 3) is the difference in mean VAS pain for the two treatments (NMES group - No treatment group). Study (Date of publication) Difference ̂ [̂] Akarcali(2002) 1.12 0.3969 Bily (2008) 0.55 1.2769 Tunay (2003) 1.90 0.1296 (i) Compute the minimum variance estimate of the overall treatment effect, ̂, and determine its 95% confidence interval, stating any assumptions that you make. [6 marks] (ii) What can you conclude from the meta-analysis regarding the performance of NMES? [2 marks] (iii) It is known that the minimal clinically important difference for VAS pain is in the range 1.5 to 2.0. By comparing these values with the 95% confidence interval that you computed in (i), provide interpretation of the confidence interval in the context of clinical importance. [2 marks] [Total 10 marks] MATH38072 Page 6 of 11 A5. Consider a randomized controlled trial. Suppose the patient population can be divided into three latent sub-groups as follows:  Compliers: patients who will comply with the allocated treatment;  Always control treatment: patients who will receive control treatment regardless of allocation;  Always new treatment: patients who will receive the new treatment regardless of allocation. This assumes that there are no defiers, namely patients who will always receive the opposite of the treatment to which they are randomized. Assuming that the proportion and characteristics of compliers, always control treatment, always new treatment is the same in both arms and that randomization can only affect the outcome through the receipt of treatment, show that: (i) An intention-to-treat estimate of the treatment effect is biased towards the null hypothesis of no treatment effect. [5 marks] (ii) A per-protocol estimate of the treatment effect may be biased either towards or away from the null hypothesis of no treatment effect. [5 marks] [Total 10 marks] MATH38072 Page 7 of 11 B1. For an AB/BA crossover trial, a standard model for a continuous outcome ijy for the i th participant in the jth period is: 1 = µ +  + 1 for a participant in sequence AB in period 1 2 = µ +  +  +  + 2 for a participant in sequence AB in period 2 1 = µ +  +  + 1 for a participant in sequence BA in period 1 2 = µ +  +  + 2 for a participant in sequence BA in period 2 where µ is the mean for the sequence BA in period 1,  is the treatment effect of A compared to B, and  is the period effect, with  and  being independent random variables with ~[0, 2 ] and ~[0, 2]. Defining = 2 − 1, let ̅, ̅, µ and µ be the sample and population means of for sequences AB and BA respectively. (i) Show that ̂ = ̅−̅ 2 is an unbiased estimator of , that is [̂] = [ ̅−̅ 2 ] =  . Hence also show that a test of the null hypothesis 0:  =  is the same as a test of the null hypothesis of no treatment effect 0:  = 0. [4 marks] (ii) Show that the variance of ̂ is given by [̂ ] = [ ̅−̅ 2 ] =  2 2 ( 1 + 1 ) . (You may assume the general result (̅) = 2 where () = 2, and is the sample size.) [5 marks] MATH38072 Page 8 of 11 (iii) Given that the variance of the estimator based on the first period observations from the two groups, which is equivalent to a parallel-design trial treatment effect estimator, is given by: ( 2 +  2) ( 1 + 1 ) , what does the result given in (iii) suggest is an advantage of using an AB/BA crossover design rather than a parallel-design trial? State the condition when this advantage is at its greatest. [3 marks] (iv) An AB/BA crossover trial was designed to compare two treatments for hypertension (high blood pressure): Drug A and Drug B. Thirty-eight participants, all suffering from hypertension, were randomised: 18 to the sequence Drug A then Drug B, and 20 to the sequence Drug B then Drug A. A measurement of the primary outcome measure, diastolic blood pressure (in mm Hg), was taken at the end of each treatment period and are summarised in the table below. Test the hypothesis 0:  = 0 versus 1:  ≠ 0 using a 5% significance level. Also compute and briefly interpret the 95% confidence interval for . Sequence Period 1 Period 2 Period 2 – Period 1 N Mean S.D. Mean S.D. Mean S.D. AB 84.50 7.01 83.39 6.57 -1.11 2.47 18 BA 84.10 8.94 86.65 8.12 2.55 2.26 20 [8 marks] [Total 20 marks] MATH38072 Page 9 of 11 B2. A randomised controlled trial is planned to compare a treatment (A) with the current standard therapy (B). For an outcome measure Y, let ̅̅ ̅, ̅̅ ̅, ,  be the sample and population means of Y for each treatment. Let s and  be the common within-group sample and population standard deviations of Y. Assume that the null hypothesis of no treatment effect will be tested by the statistic = ̅̅ ̅̅ −̅̅ ̅̅  , with  = √1 ⁄ + 1 ⁄ , where and are the number of participants allocated to each treatment. (i) Write down an expression for the power (1-) of the test for a two-tailed alternative hypothesis to detect a treatment effect , stating the assumptions you make. [4 marks] (ii) Assuming that participants are allocated in the ratio of 1: k such that = . , show that the total sample size required to give a power of (1-) for a two-tailed - sized test is: = ( + 1)2 2 2 ( 2⁄ + ) 2 . [5 marks] (iii) Show that the minimum sample size is obtained by using an equal allocation ratio. [5 marks] (iv) In a randomised controlled trial it is planned to allocate 240 participants to two treatments. The pooled within-group standard deviation is thought to be 12 units. Estimate the power of a study to detect a treatment effect of 5 units for two-tailed 0.05-sized test, if an allocation ratio of 1:2 (treatment: control) is used. [4 marks] (v) In a randomised controlled trial, it is decided to allocate twice as many participants to the control group as to the treatment group, using blocked randomisation with a block size of 3. Briefly describe how to create a randomisation list to implement this randomisation scheme by using random numbers with a simple allocation rule. [2 marks] [Total 20 marks] MATH38072 Page 10 of 11 B3. In a parallel-group non-inferiority trial, a new treatment T is being compared with a control treatment C using a normally distributed outcome measure Y. Assume that large values of Y represent a worse outcome for the participant. Let ̅̅ ̅,  and and ̅̅ ̅,  and be the sample mean, population mean and the sample size for the new treatment and for the control treatment, respectively. Suppose  is the population standard deviation of both treatments. The treatment effect is defined as  =  − . (i) Provide an example, with justification, of a situation where a non-inferiority trial rather than a superiority trial should be used. [2 marks] (ii) Explain why a significance test of the hypothesis 0:  = 0 versus 1:  > 0 would not be appropriate in a non-inferiority trial. [3 marks] (iii) For a non-inferiority trial, the null hypothesis 0:  ≥  is rejected if the upper limit for a 100(1 − )% one-sided confidence interval for  is less than the limit of non- inferiority  , for which the probability, under 0, is given by: Pr[Reject 0|  ]= ((   − ) −   ) , where  is the cumulative distribution function of the standard normal distribution and  = √1 + 1 ⁄⁄ . Given this, show that Pr[Reject 0| ] has a maximum under 0 when  =  . Hence, show that this procedure has a Type I error probability less than or equal to  . [7 marks] MATH38072 Page 11 of 11 (iv) A randomised controlled non-inferiority trial was carried out to test whether Nurse-led Telephone follow-up (Telephone Group) was non-inferior to the current standard care of Consultant-led Hospital follow-up (Hospital Group) for women following treatment for cancer of the endometrium (the inner lining of the uterus). The primary outcome was measured on a continuous scale using a questionnaire-based measure of anxiety, with higher scores representing greater anxiety. Follow-up data were collected on 111 participants from the Telephone Group and 106 from the Hospital Group. The computer output from the statistical analysis is given below. Two-sample t test with equal variances ------------------------------------------------------------------------------ Group | Obs Mean Std. Err. Std. Dev. [90% Conf. Interval] ---------+-------------------------------------------------------------------- telephon | 111 33.00901 1.045759 11.01775 31.27428 34.74374 hospital | 106 35.5283 1.265849 13.03271 33.42763 37.62897 ---------+-------------------------------------------------------------------- combined | 217 34.23963 .8201926 12.08219 32.88472 35.59454 ---------+-------------------------------------------------------------------- diff | -2.519293 1.635633 -5.221312 .1827265 ------------------------------------------------------------------------------ diff = mean(telephon) - mean(hospital) t = -1.5403 Ho: diff = 0 degrees of freedom = 215 Ha: diff < 0 Ha: diff != 0 Ha: diff > 0 Pr(T < t) = 0.0625 Pr(|T| > |t|) = 0.1250 Pr(T > t) = 0.9375 A difference of less than 3.5 points on the anxiety scale was considered by researchers to be an unimportant difference between treatments. Using the computer output given above, test whether the Nurse-Led Telephone follow-up was non-inferior to the Consultant-Led Clinic Hospital-based follow-up for the primary outcome measure. Interpret your findings in the context of the trial. Can you also conclude superiority of the Nurse-led Telephone follow-up on the primary outcome measure? [8 marks] [Total 20 marks] END OF EXAMINATION PAPER































































































































































































































































































































































































































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