sample size
1.0.0
This page is devoted to the calculation of the number of patients required for several studies such as an randomized clinical trial (RCT) for causal inference or a cohort for constructing or validating a predictive tool. R codes are proposed. We can click here to access to the related user-friendly calculators. Some reminders:
Superiority RCT: Use to demonstrate that the experimental treatment is more effective than standard therapy.
Non-inferiority RCT: Use to demonstrate that the experimental treatment is as effective as standard therapy.
Sequential RCT: Intermediate analyses for early stopping the study.
Consider the following RCT with two parallel groups with a 1:1 randomization ratio. The expected mean is 66 units in patients in the experimental arm versus 72 units in the control arm. In order to demonstrate such a difference of 6 units, with a standard deviation of 23, a 5% two-sided type I error rate and a power of 80%, the minimum sample size per arm equals 231 (i.e., a total of 462 patients).
library(epiR)
epi.sscompc(treat = 66, control = 72, sigma = 23, n = NA, power = 0.8,
r = 1, sided.test = 2, conf.level = 1-0.05)
#> $n.total
#> [1] 462
#> $n.treat
#> [1] 231
#> $n.control
#> [1] 231
#> $power
#> [1] 0.8
#> $delta
#> [1] 6Input parameters:
Consider the following RCT with two parallel groups with a 1:1 randomization ratio and 2 planned intermediate analyses for efficacy by using the O'Brien-Fleming method for considering the inflation of the type I error rate). The expected mean is 66 units in patients in the experimental arm versus 72 units in the control arm. In order to demonstrate such a difference of 6 units, with a standard deviation of 23, a 5% two-sided type I error rate and a power of 80%, the final analysis should be carried out on 472 patients (236 patients per group). The first and second intermediate analyses would be performed on 158 and 316 patients respectively, i.e. 33% and 66% of the maximum number of included patients if their is no decision of stopping the study.
library("rpact")
design <- getDesignGroupSequential(
typeOfDesign = "OF", informationRates = c(1/3, 2/3, 1),
alpha = 0.05, beta = 1-0.8, sided = 2)
designPlan <- getSampleSizeMeans(design, alternative = 6, stDev = 23,
allocationRatioPlanned = 1)
summary(designPlan)
#> Stage 1 2 3
#> Planned information rate 33.3% 66.7% 100%
#> Cumulative alpha spent 0.0005 0.0143 0.0500
#> Stage levels (two-sided) 0.0005 0.0141 0.0451
#> Efficacy boundary (z-value scale) 3.471 2.454 2.004
#> Lower efficacy boundary (t) -13.012 -6.405 -4.258
#> Upper efficacy boundary (t) 13.012 6.405 4.258
#> Cumulative power 0.0329 0.4424 0.8000
#> Number of subjects 157.1 314.2 471.3
#> Expected number of subjects under H1 396.7
#> Exit probability for efficacy (under H0) 0.0005 0.0138
#> Exit probability for efficacy (under H1) 0.0329 0.4095 Input parameters:
Consider the following RCT with two parallel groups with a 1:1 randomization ratio. The expected mean is 66 units in patients in the control arm and no difference compared to the experimental arm. Assuming an absolute non-inferiority margin of 7 points, a standard deviation of 23, the minimum sample size per arm equals 134 (i.e., a total of 268 patients) to achieve a 5% one-sided type I error rate and a power of 80%
library(epiR)
epi.ssninfc(treat = 66, control = 66, sd= 23, delta = 7,
power = 0.8, alpha = 0.05, r = 1, n = NA)
#> $n.total
#> [1] 268
#> $n.treat
#> [1] 134
#> $n.control
#> [1] 134
#> $delta
#> [1] 7
#> $power
#> [1] 0.8Input parameters:
Consider the following RCT with two parallel groups with a 1:1 randomization ratio. The expected proportion of events is 35% in the experimental arm compared to 28% in the control arm. In order to demonstrate such a difference of 7%, with a two-sided type I error rate of 5% and a power of 80%, the minimum sample size per arm equals 691 (i.e., a total of 1382 patients).
library(epiR)
epi.sscohortc(irexp1 = 0.35, irexp0 = 0.28, power = 0.80, r = 1,
sided.test = 2, conf.level = 1-0.05)
#> $n.total
#> [1] 1382
#> $n.exp1
#> [1] 691
#> $n.exp0
#> [1] 691
#> $power
#> [1] 0.8
#> $irr
#> [1] 1.25
#> $or
#> [1] 1.384615Input parameters:
Consider the following RCT with two parallel groups with a 1:1 randomization ratio and 2 planned intermediate analyses for efficacy by using the O'Brien-Fleming method for considering the inflation of the type I error rate. The expected proportion of event is 11% in patients in the experimental arm versus 15% units in the control arm. In order to demonstrate such a difference of 4%, with a 5% two-sided type I error rate and a power of 80%, the final analysis should be carried out on 2,256 patients (1,128 patients per group). The first and second intermediate analyses would be performed on 752 and 1,504 patients respectively, i.e. 33% and 66% of the maximum number of included patients if their is no decision of stopping the study.
library("rpact")
design <- getDesignGroupSequential(typeOfDesign = "OF",
informationRates = c(1/3, 2/3, 1), alpha = 0.05,
beta = 1-0.8, sided = 2)
designPlan <- getSampleSizeRates(design, pi1 = 0.11, pi2 = 0.15,
allocationRatioPlanned = 1)
summary(designPlan)
#> Stage 1 2 3
#> Planned information rate 33.3% 66.7% 100%
#> Cumulative alpha spent 0.0005 0.0143 0.0500
#> Stage levels (two-sided) 0.0005 0.0141 0.0451
#> Efficacy boundary (z-value scale) 3.471 2.454 2.004
#> Lower efficacy boundary (t) -0.079 -0.042 -0.029
#> Upper efficacy boundary (t) 0.101 0.048 0.031
#> Cumulative power 0.0329 0.4424 0.8000
#> Number of subjects 751.8 1503.7 2255.5
#> Expected number of subjects under H1 1898.1
#> Exit probability for efficacy (under H0) 0.0005 0.0138
#> Exit probability for efficacy (under H1) 0.0329 0.4095 Input parameters:
Consider the following RCT with two parallel groups with a 1:1 randomization ratio. The expected percentage of events is 35% in patients in the control arm and no difference compared to the experimental arm. Assuming an absolute non-inferiority margin of 5%, the minimum sample size per arm equals 1,126 (i.e., a total of 2,252 patients) to achieve a 5% one-sided type I error rate and a power of 80%.
epi.ssninfb(treat = 0.35, control = 0.35, delta = 0.05,
n = NA, r = 1, power = 0.8, alpha = 0.05)
#> $n.total
#> [1] 2252
#> $n.treat
#> [1] 1126
#> $n.control
#> [1] 1126
#> $delta
#> [1] 0.05
#> $power
#> [1] 0.8Parameters :
For developing a model/alghorithm based on 34 predictors as candidates with an expected R2 of at least 0.25 and an expected shrinkage of 0.9 (equation 11 in Riley et al. Statistics in Medicine. 2019;38:1276–1296), the minimal sample size is 1045.
34/((0.9-1)*log(1-0.25/0.9))
#> [1] 1044.796Consider O/E the ratio between the number of observed events versus expected ones. To achieve a precision defined as a length of the (1-α)% confidence interval of this ratio equals to 0.2, if the expected proportions is 50%, the required sample size is 386 (Riley et al. Minimum sample size for external validation of a clinical prediction model with a binary outcome. Statistics in Medicine. 2021;19:4230-4251).
se <- function(width, alpha) # The standard error associated with the 1-alpha confidence interval
{
fun <- function(x) { exp( qnorm(1-alpha/2, mean=0, sd=1) * x ) - exp(-1* qnorm(1-alpha/2, mean=0, sd=1) * x ) - width }
return(uniroot(fun, lower = 0.001, upper = 100)$root)
}
size.calib <- function(p, width, alpha) # the minimum sample size to achieve this precision
{
(1-p) / ((p * se(width=width, alpha=alpha)**2 ))
}
size.calib(p=0.5, width=0.2, alpha=0.05)
#> [1] 385.4265Input parameters:
