Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 207 / 3 / a / ii

Table of contents
- Solution ii

Solution

For

k = 4

, the optimum is

q = 2

. The supplied two-sided sample size formula gives

n_{k} = \frac{1 + 2}{2} (\frac{2}{2})^{2} (0.84 + 1.96)^{2} = \frac{3}{2} (7.84) = 11.76,

so round upward to

12

patients in each new-treatment arm and take

n_{0} = 24

. The analyzed total is therefore

n_{tot} = 4 (12) + 24 = 72.

Allowing for ten percent attrition requires

⌈ \frac{72}{0.9} ⌉ = 80

patients to be recruited.

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version