Worksheet 1 MAS367
Sample Sizes, Allocation, Bias and Randomization
The Beck depression inventory score is often used as the outcome measure in trialsthat assess treatments for depression. It has standard deviation 8 (no units). Supposethat in a trial comparing two treatments there is interest in detecting a difference of 3.5between the mean Beck depression inventory scores in the two groups.
a) If the groups are of equal size and the power is 90% at the 5% level of significance, how many patients need to be recruited in total? b) How would this number change if the power were reduced to 80%? c) I can recruit 100 patients: what is the largest power I can have to detect a mean d) If the value used for the standard deviation for the Beck depression inventory score should have been 10% larger, what would have been the effect on the numbers ina) and b)? 2.
Patients who contract bronchiolitis, a respiratory disease of infants, often suffer fromwheezing for some time after they have been ostensibly cured. In a trial to see if aninhaled steroid, budesonide, can alleviate this after-effect, patients were randomized toreceive budesonide or a placebo. The outcome was the presence of wheeze one yearafter the diagnosis of bronchiolitis.
a) It is thought that the proportion of children with wheeze at one year post-diagnosis is 0.65. There is interest in reducing this to 0.25. How many infants would haveto be randomized to each of two equal-sized groups in order to have 80% powerto detect this difference at the 5% significance level? b) Suppose that the initial estimate of 0.65 is too high, and it should be 0.55. How many patients would be needed to detect a change to 0.15 (i.e. the same change inproportion) with 80% power? c) If the initial estimate were too low and 0.75 was a more reasonable choice, how many patients would be needed to detect a change to 0.35 (again the same change,0.4, in proportions) with 80% power? Comment on your answer.
The method for determining the sample size for a trial with binary outcome usesthe approximation: arcsin( p ) ~ N (arcsin( π 1 where r successes have been observed in n trials and p=r/n: the populationprobability of success is π. Although no longer a standard method, this result canbe used as a basis for the analysis of binary data. If two groups are to becompared, with observed proportions of successes from n1 and n2 trials of,respectively, p1 and p2, then show that a statistic testing the null hypothesis that thetwo population proportions are equal is: arcsin( p ) − arcsin( p ) To which distribution would you compare this statistic in order to perform the test?Using the result arcsin( p ) ≈ arcsin( π ) + ( p π ) and relate this to one of the more common ways of testing this hypothesis.
4. Ten patients are randomly allocated to treatments A and B in a clinical trial.
Unequal allocation is used, with Pr(Allocate to A) = 0.6. What is the probabilitythat two or fewer patients are allocated to treatment B? 5. Suppose you were designing the trial outlined in question 1 and had decided to opt for 80% power. However it was thought appropriate that one treatment groupshould be twice the size of the other. What would be the sizes of the two groups?Suppose that allocation was to be made using random permuted blocks of length 6.
How many blocks are there? What is the disadvantage of using a single block size?How many blocks of length 9 are there?


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