Free examples and use-cases:   rpact vignettes
rpact: Confirmatory Adaptive Clinical Trial Design and Analysis

# Summary

This R Markdown document provides an example for planning a trial with a survival endpoint using rpact thereby illustrating the different ways of entering recruitment schemes. It also demonstrates the use of the survival simulation function.

# 1 Types of calculations

library(rpact)
packageVersion("rpact") # version should be version 2.0.5 or later
## [1] '3.3.2'

A survival trial is planned to be performed with one interim stage and using an O’Brien & Fleming type $$\alpha$$-spending approach at $$\alpha = 0.025$$. The interim is planned to be performed after half of the necessary events were observed. It is assumed that the median survival time is 18 months in the treatment group, and 12 months in the control. We assume that the drop-out rate is 5% after 1 year and the drop-out time is exponentially distributed.

## 1.1 Accrual and follow-up time given

The patients should be recruited within 12 months assuming uniform accrual. We assume an additional follow-up time of 12 months, i.e., the study should be conducted within 2 years. We also assume the survival time to be exponentially distributed.

In this simplest example, accrual and follow-up time needs to be specified and the necessary number of events and patients (total and per month) in order to reach power 90% with the assumed median survival times will be calculated.

The effect size is defined in terms of lambda1 and lambda2 (you can also specify lambda2 and hazardRatio) and the function getLambdaByMedian() is used (for lambda2 the direct calculation is illustrated):

dGS <- getDesignGroupSequential(kMax = 2, typeOfDesign = "asOF", beta = 0.1)

x1 <- getSampleSizeSurvival(dGS,
lambda1 = getLambdaByMedian(18), lambda2 = log(2) / 12,
dropoutRate1 = 0.05, dropoutRate2 = 0.05, dropoutTime = 12,
accrualTime = 12, followUpTime = 12
)
kable(summary(x1))

Sample size calculation for a survival endpoint

Sequential analysis with a maximum of 2 looks (group sequential design), overall significance level 2.5% (one-sided). The sample size was calculated for a two-sample logrank test, H0: hazard ratio = 1, H1: treatment lambda(1) = 0.039, control lambda(2) = 0.058, accrual time = 12, accrual intensity = 38.9, dropout rate(1) = 0.05, dropout rate(2) = 0.05, dropout time = 12, power 90%.

Stage 1 2
Information rate 50% 100%
Efficacy boundary (z-value scale) 2.963 1.969
Overall power 0.2525 0.9000
Expected number of subjects 467.3
Number of subjects 467.3 467.3
Cumulative number of events 128.3 256.5
Expected study duration 21.3
Cumulative alpha spent 0.0015 0.0250
One-sided local significance level 0.0015 0.0245
Efficacy boundary (t) 0.593 0.782
Exit probability for efficacy (under H0) 0.0015
Exit probability for efficacy (under H1) 0.2525

Legend:

• (t): treatment effect scale

We conclude that a maximum number of 257 events needs to be observed to reach 90% power. Under the assumed accrual and follow-up time (and the median survival times) this number of events are expected to be observed if 468 subjects are observed at the end of the trial. This corresponds to about 39 ( = accrualIntensity) patient per month to be accrued.

The interim analysis is performed after 128.3 events (which in practice needs to be rounded up or down). This number is expected to be observed at analysis time 13.14.

Note that in this case you can also enter the accrual time as accrualTime = c(0,12) and so accrualTime = 12 is just a shortcut (see below).

## 1.2 Accrual time and maximum number of patients given

Assume that accrual stops after 16 months with 25 patients per month, i.e., after 400 patients were recruited. We now can estimate the necessary follow-up time such that 257 events will be observed at the end of the trial.

There are two ways of defining the situation of given maximum number of subjects: You can either specify accrualTime and maxNumberOfSubjects directly; or you specify accrualTime and accrualIntensity.

In both cases, at given accrual time and number of patients, the follow-up time is calculated.

This is the code and the output for the first case:

x2 <- getSampleSizeSurvival(dGS,
lambda1 = log(2) / 18, lambda2 = log(2) / 12,
dropoutRate1 = 0.05, dropoutRate2 = 0.05, dropoutTime = 12,
accrualTime = c(0, 16), accrualIntensity = 25
)

ceiling(x2$maxNumberOfSubjects) ## [1] 400 x2$followUpTime
## [1] 15.96226
x2$analysisTime ## [,1] ## [1,] 16.82864 ## [2,] 31.96226 This is the code and the output for the second case: x2 <- getSampleSizeSurvival(dGS, lambda1 = log(2) / 18, lambda2 = log(2) / 12, dropoutRate1 = 0.05, dropoutRate2 = 0.05, dropoutTime = 12, accrualTime = c(0, 16), maxNumberOfSubjects = 400 ) x2$accrualIntensity
## [1] 25
x2$followUpTime ## [1] 15.96226 x2$analysisTime
##          [,1]
## [1,] 16.82864
## [2,] 31.96226

Note that for these two cases entering the accrual time in the form of c(0,16) is mandatory, i.e., you cannot enter just the end of accrual which was the shortcut before.

Alternatively, you might enter the accrualTime as a list in the form

atList <- list(
"0 - <16" = 25
)

and then enter accrualTime = atList in getSampleSizeSurvival():

x2 <- getSampleSizeSurvival(dGS,
lambda1 = log(2) / 18, lambda2 = log(2) / 12,
dropoutRate1 = 0.05, dropoutRate2 = 0.05, dropoutTime = 12,
accrualTime = atList
)

x2$accrualIntensity ## [1] 25 x2$followUpTime
## [1] 15.96226
x2$analysisTime ## [,1] ## [1,] 16.82864 ## [2,] 31.96226 ## 1.3 Follow-up time and absolute intensity given Assume now that 25 patients can be recruited each month and that there is uniform accrual. We now want to estimate the necessary accrual time if the planned follow-up time is given and equal to 12 as before. Here the end of accrual and the number of patients is calculated at given follow-up time and absolute accrual intensity: x3 <- getSampleSizeSurvival(dGS, hazardRatio = 2 / 3, lambda2 = log(2) / 12, dropoutRate1 = 0.05, dropoutRate2 = 0.05, dropoutTime = 12, accrualTime = 0, accrualIntensity = 25, followUpTime = 12 ) ceiling(x3$maxNumberOfSubjects)
## [1] 435
x3$accrualTime ## [1] 17.38334 x3$analysisTime
##          [,1]
## [1,] 16.79806
## [2,] 29.38334

## 1.4 Staggered patient entry

This case illustrates how the accrual of subjects can be entered if there is piecewise accrual over predefined intervals where it is assumed that accrual within the intervals is uniform.

For example, assume that in the first 3 months 15 patients, in the second 3 months 20 patients, and after 6 months 25 patients per month can be accrued.

If the follow-up time at given number of patients is to be calculated (Section 1.2), it is simply to enter accrualTime = c(0, 3, 6, 16) and accrualIntensity = c(15, 20, 25) as follows (maxNumberOfSubjects will be calculated):

x4 <- getSampleSizeSurvival(dGS,
lambda1 = log(2) / 18, lambda2 = log(2) / 12,
dropoutRate1 = 0.05, dropoutRate2 = 0.05, dropoutTime = 12,
accrualTime = c(0, 3, 6, 16), accrualIntensity = c(15, 20, 25)
)

ceiling(x4$maxNumberOfSubjects) ## [1] 355 x4$followUpTime
## [1] 23.60973
x4$analysisTime ## [,1] ## [1,] 18.83368 ## [2,] 39.60973 Alternatively, you can enter atList <- list( "0 - < 3" = 15, "3 - < 6" = 20, "6 - <= 16" = 25 ) and run x4 <- getSampleSizeSurvival(dGS, lambda1 = log(2) / 18, lambda2 = log(2) / 12, dropoutRate1 = 0.05, dropoutRate2 = 0.05, dropoutTime = 12, accrualTime = atList ) ceiling(x4$maxNumberOfSubjects)
x4$followUpTime x4$analysisTime

If maxNumberOfSubjects i