trial.params: Calculate Parameters for a Simple Model of a Clinical Trial
In tobyjohnson/gtx: Genetics ToolboX

Description Usage Arguments Details Value Author(s) Examples

Calculates parameters for a simple model of a clinical trial, with a constant rate recruitment period and then a followup period, with constant event rate per recruited live subject.

1	trial.params(hazardRate = NA, surviveEnd = NA, timeTotal = NA, timeRecruit = NA)

`hazardRate`	Instantaneous rate of events
`surviveEnd`	Proportion of subjects surviving at end of trial
`timeTotal`	Total duration of the trial
`timeRecruit`	Time of subject recruitment

This function assumes simplified model of a clinical trial (albeit less simplified than other models). Events occur at a fixed instantaneous rate (hazardRate; h) per recruited live subject and per unit time, and the expected surviving fraction at the end of the trial is surviveEnd (sEnd). The total duration of the trial (from start of recruitment until the time at which sEnd is evaulated, is timeTotal (tTotal). Recruitment is assumed to occur at a constant rate over a period timeRecruit (tRecruit). The two extreme possibilities are tRecruit=0, where all subjects are recruited instantaneously at the start of the trial, and tRecruit=tTotal, where recruitment is continuous over the entire duration of the trial.

For specified values for any three of the four parameters, this function calculates the value for the missing paramater.

The constant instantaneous event rate h implies that for an exponential lifetime distribution, the mean lifetime is 1/h and the median lifetime is qexp(0.5)/h ~= 0.693/h.

Note that with an exponential lifetime model, some parameter combinations cannot be solved numerically, e.g. calculating h when tTotal=tRecruit and sEnd very close to zero.

A structure with missing values calculated.

Toby Johnson Toby.x.Johnson@gsk.com

trial.params(hazardRate = c(0.1, 0.3, 0.5, NA),
            surviveEnd = c(NA, 0.4, 0.6, 0.8),
            timeTotal = c(4, NA, 3, 2),
            timeRecruit = c(2, 1.6, NA, 2))
## check missing return value is indeed not achievable:
trial.params(hazardRate = 0.5, surviveEnd = NA,
            timeTotal = 3, timeRecruit = c(0, 3))
## cannot solve all these cases numerically:
trial.params(NA, .00001, 1, c(.9, .95, .99, 1))