Author: John Aponte
Date: 2025-12-02

# Sample size with Uncertainty

#### John Aponte

#### 2025-12-02

## Introduction

This example show in details how to construct a new factory of `DISTRIBUTION` objects using a convolution of parameters to propagate the uncertainty to the final calculation.

The estimation of sample size require the input of different parameters in a way that when the parameters are correct, the power to have a significant result is correct. However many times there are some uncertainty on the parameters and this is not reflected in the power.

The `convdist` package could help on explicitly express uncertainty on those parameters and estimate the distribution of sample sizes that would satisfy the power requirements of the assay.

We explore how to estimate the sample size for binomial outcome with uncertainty and explain the details of how to construct a `DISTRIBUTION` factory function.

## Sample size for a proportion

For an individually randomized trial, the number of subjects per group could be defined as \[n = (z_{\alpha/2} + z_\beta)^2 [\pi_0(1 – \pi_0) + \pi_1(1 – \pi1)]/(\pi_0 – \pi_1)^2\]

In this formula, \(z_{\alpha/2}\) and \(z_\beta\) are standard normal distribution values corresponding to upper tail probabilities of \(z_{\alpha/2}\) and \(z_\beta\) respectively. This choice of sample size provides a power of 100(1 – \(\beta\))% to obtaining a significant difference (\(P < \alpha\) on a two-sided test), assuming that the true (population) proportion in presence or absence of the interventions are \(\pi_1\) and \(\pi_0\) respectively

\(\pi_1\) can be express on terms of \(\pi_0\) as \(\pi_1 = \pi_0 RR\) where \(RR\) is the relative risk.

In the real world, there could be uncertainty on the exact values of \(\pi_0\) and \(RR\) as they may come from previous studies or surveys. What would be the distribution of \(n\) if the uncertainty is taking into account?

## A new `DISTRIBUTION` factory

We can construct a `DISTRIBUTION` object which would produce random numbers taking into account the a probability distribution for \(\pi_0\) and \(RR\)

See the code of `new_UNIFORM` function as an example of how is a factory of `DISTRIBUTION` objects:

    library(convdistr)
    new_UNIFORM
    #> function (p_min, p_max, p_dimnames = "rvar")
    #> {
    #>     stopifnot(p_min <= p_max)
    #>     .oval = (p_max + p_min)/2
    #>     names(.oval) <- p_dimnames
    #>     structure(list(distribution = "UNIFORM", seed = sample(1:2^15,
    #>         1), oval = .oval, rfunc = restrict_environment(function(n) {
    #>         matrix(runif(n, p_min, p_max), ncol = 1, dimnames = list(1:n,
    #>             p_dimnames))
    #>     }, p_min = p_min, p_max = p_max, p_dimnames = p_dimnames)),
    #>         class = c("UNIFORM", "DISTRIBUTION"))
    #> }
    #> <bytecode: 0xaaab0ff03e18>
    #> <environment: namespace:convdistr>

The function should return an structure which consist of a list of four named vector and a class attribute that should include “DISTRIBUTION” The components of the list are:

distribution : A character with the name of the distribution implemented. It is a good practice to use the same name also in the class attribute

seed : A numerical seed that is use to get a repeatable sample in the `summary` function Use a `sample` function to ensure each new object will have a different seed

oval : The observed value. It is the value expected. It is used as a number for the mathematical operations of the distributions as if they were a simple scalar

rfunc(n) : A function that generate random numbers from the distribution. Its only parameter `n` is the number of drawns of the distribution. It returns a matrix with as many rows as `n`, and as many columns as the dimensions of the distributions, The `rfunc` should be defined within a `restrict_environment` function to ensure only the variables required to the function are included in the environment of the function. This will prevent reference leaking which could make the objects big and slow when saved in a file.

The factory of object for the sample size of proportions could be defined as:

    #' Sample size for proportions, with uncertainty
    #'
    #' Create an new DISTRIBUTION object that produce
    #' random drawns of the estimated sample size for
    #' two proportions
    #' @param p0 a DISTRIBUTION object that drawns for proportions in control group
    #' @param logrr a DISTRIBUTION object that drawns log(RR) of the intervention
    #' @param alpha significant value
    #' @param beta  1-power
    new_SAMPLE_SIZE <- function(p0,logrr, alpha= 0.05, beta=0.2){
      #checkings
      stopifnot(inherits(p0,"DISTRIBUTION"))
      stopifnot(inherits(logrr,"DISTRIBUTION"))
      stopifnot(0 < alpha & alpha < 1)
      stopifnot(0 < beta & beta < 1)

     # function of alpha and beta
      f_alpha_beta <- (qnorm(alpha/2,lower.tail = F) + qnorm(beta, lower.tail = F)) ^ 2

      # The expected value of the distribution with default name for the dimension
      # based on the oval value of the individual distributions
      v0 <- p0$oval
      v1 <- p0$oval*exp(logrr$oval)

      # expected value of the distribution
      .oval <- f_alpha_beta * (v0 * (1 - v0) + v1 * (1 - v1)) / (v0 - v1) ^ 2
      names(.oval) <- "rvar"

      #random function within a restricted environment where only
      #the specified variables can be accesed within the function
      .rfunc <- restrict_environment(
         function(n) {
           d_p0 = rfunc_p0(n)
           d_p1 = d_p0 * exp(rfunc_logrr(n))
           fab * (d_p0 * (1 - d_p0) + d_p1 * (1 - d_p1)) / (d_p0 - d_p1) ^ 2
         },
         rfunc_p0 = p0$rfunc,
         rfunc_logrr = logrr$rfunc,
         fab = f_alpha_beta
      )
      # Create the object with 4 slots
      structure(
        list(
          distribution = "SAMPLE_SIZE",
          seed = sample(1:2 ^ 15, 1),
          oval = .oval,
          rfunc = .rfunc
        ),
        class = c("SAMPLE_SIZE","DISTRIBUTION")
      )

    }

## Using the Sample Size object

As example, suppose that the incidence of anaemia in children under 5 have been estimated from a survey as 35% with 95%CI (30%,40%), and a prevention treatment that in other settings have shown to decrease anaemia by 50% with 95%CI(8% to 73%).

We can define a `BETA` distribution based on the CI as

    d_p0 = new_BETA_lci(0.35,0.30,0.40)

And a `NORMAL` distribution of the `log(RR)` as

    rr = log(1 - 50/100)
    sd = (log((1 - 8 / 100)) - log((1 - 73 / 100))) / 4
    d_logrr = new_NORMAL(rr,sd)

So the input distributions are

distribution | varname | oval | nsample | mean_ | sd_ | lci_ | median_ | uci_
--- | --- | --- | --- | --- | --- | --- | --- | ---
BETA | rvar | 0.35 | 10000 | 0.35 | 0.03 | 0.3 | 0.35 | 0.40
NORMAL | rvar | -0.69 | 10000 | -0.70 | 0.31 | -1.3 | -0.70 | -0.08

And base on this we can calculate the distribution of samples with those conditions

    d_sample <- new_SAMPLE_SIZE(d_p0, d_logrr)

distribution | varname | oval | nsample | mean_ | sd_ | lci_ | median_ | uci_
--- | --- | --- | --- | --- | --- | --- | --- | ---
SAMPLE_SIZE | rvar | 95 | 10000 | 28610 | 1630112 | 37 | 94 | 2327

The oval value of the distribution give us the result we obtain from the formula. With the conditions specified, a sample size of `95` is required. However, the distribution is quite skewed as you can see the mean and the median value are far, and the standard deviation very big. This is because the sample size require for combinations of low efficacy and low prevalence could be very large and there is no restriction on how low could be the efficacy or the prevalence, which although have low frequency, could happen in the simulations.

    rdrawn <- rfunc(d_sample, 10000)
    f_ecdf <- ecdf(rdrawn)
    n_80 = trunc(uniroot(function(x){f_ecdf(x) - 0.8}, interval = c(10,10000))$root)

The empirical cumulative distribution values of a random sample of the distribution show us that with `209` subjects per arm, 80% of the possible scenarios are covered

## Enhance the Sample Size object

The function created above show us a distribution of sample size, but we can’t inspect condition on which a particular drawn is obtained, i.e. how low an prevalent anaemia value could be compensated with a high efficacy value? To do that we could create a second object that have drawns in several dimensions, the prevalence, the RR and the sample size so we can make a graph with the three values.

    #' Sample size for proportions, with uncertainty (V2)
    #'
    #' Create an new DISTRIBUTION object that produce
    #' random drawns of the estimated sample size for
    #' two proportions as well as the samples from the parameters
    #' @param p0 a DISTRIBUTION object that drawns for proportions in control group
    #' @param logrr a DISTRIBUTION object that drawns log(RR) of the intervention
    #' @param alpha significant value
    #' @param beta  1-power
    new_SAMPLE_SIZE2 <- function(p0,logrr, alpha= 0.05, beta=0.2){
      #checkings
      stopifnot(inherits(p0,"DISTRIBUTION"))
      stopifnot(inherits(logrr,"DISTRIBUTION"))
      stopifnot(0 < alpha & alpha < 1)
      stopifnot(0 < beta & beta < 1)

     # function of alpha and beta
      f_alpha_beta <- (qnorm(alpha/2,lower.tail = F) + qnorm(beta, lower.tail = F)) ^ 2

      # The expected value of the distribution with default name for the dimension
      # based on the oval value of the individual distributions
      v0 <- p0$oval
      v1 <- p0$oval*exp(logrr$oval)

      # expected value of the distribution
      ss <- f_alpha_beta * (v0 * (1 - v0) + v1 * (1 - v1)) / (v0 - v1) ^ 2
      .oval = c(p0$oval, exp(logrr$oval), ss)
      names(.oval) <- c('prevalence', 'rr', 'sample_size')

      #random function within a restricted environment where only
      #the specified variables can be accesed within the function
      .rfunc <- restrict_environment(
         function(n) {
           d_p0 = rfunc_p0(n)
           d_rr = exp(rfunc_logrr(n))
           d_p1 = d_p0 * d_rr
           d_ss = fab * (d_p0 * (1 - d_p0) + d_p1 * (1 - d_p1)) / (d_p0 - d_p1) ^ 2
           matrix(
             c(d_p0, d_rr, d_ss),
             ncol = 3,
             dimnames = list(1:n, c('prevalence','rr','sample_size')))
         },
         rfunc_p0 = p0$rfunc,
         rfunc_logrr = logrr$rfunc,
         fab = f_alpha_beta
      )
      # Create the object with 4 slots
      structure(
        list(
          distribution = "SAMPLE_SIZE2",
          seed = sample(1:2 ^ 15, 1),
          oval = .oval,
          rfunc = .rfunc
        ),
        class = c("SAMPLE_SIZE2","DISTRIBUTION")
      )

    }

    d_sample2 <- new_SAMPLE_SIZE2(d_p0, d_logrr)

distribution | varname | oval | nsample | mean_ | sd_ | lci_ | median_ | uci_
--- | --- | --- | --- | --- | --- | --- | --- | ---
SAMPLE_SIZE2 | prevalence | 0.35 | 10000 | 0.35 | 0.03 | 0.30 | 0.35 | 0.40
SAMPLE_SIZE2 | rr | 0.50 | 10000 | 0.53 | 0.17 | 0.27 | 0.50 | 0.92
SAMPLE_SIZE2 | sample_size | 95.31 | 10000 | 7871669.99 | 783772715.29 | 37.37 | 96.45 | 2534.42

    ss <- rfunc(d_sample2, 10000)
    df <- data.frame(ss)
    df$category = cut(df$sample_size, c(0,150,Inf),include.lowest = T, ordered_result = T)
    head(df)
    #>   prevalence        rr sample_size  category
    #> 1  0.3522787 0.4720035    83.21674   [0,150]
    #> 2  0.3466862 0.8010541   704.65712 (150,Inf]
    #> 3  0.3411333 0.4405050    75.93916   [0,150]
    #> 4  0.3186313 0.7592984   534.42234 (150,Inf]
    #> 5  0.3451038 0.6612591   230.96343 (150,Inf]
    #> 6  0.3315541 0.3018421    45.65738   [0,150]

The following graph shows the combination of options that would lead to a sample size lower than 150 subjects

We see that under this particular set of conditions, it is possible to have more than 80% power in all prevalence range provided the relative risk is really lower than 50%.