Author: Jan Meis
Date: 2023-09-26

# Introduction to adestr

#### Jan Meis

#### 2023-09-26

# Introduction

This package implements methods to evaluate the performance characteristics of various point and interval estimators for adaptive two-stage designs with prespecified sample-size recalculation rules. Further, it allows for evaluation of these estimators on real datasets, and it implements methods to calculate p-values.

Currently, it works for designs objects which were produced by the R-package `adoptr`, which calculates optimal design parameters for adaptive two-stage designs. You can learn about adoptr here: kkmann.github.io/adoptr/.

# Fitting a design with adoptr

In order to showcase the capabilities of this package, we need a trial design first. We refer to the example from the adoptr documentation for this. You can read more about optimal adaptive designs fitted via the adoptr package here: kkmann.github.io/adoptr/articles/adoptr_jss.html.

For the sake of this introduction, a pre-computed version of the first example from kkmann.github.io/adoptr/articles/adoptr.html is provided with this package via the `get_example_design` function.

    library(adestr)
    get_example_design(two_armed = TRUE)
    #> TwoStageDesign<n1=56;0.8<=x1<=2.3:n2=18-80>

# Example

## Evaluating the mean squared of an estimator

Now that we have created an optimal adaptive design, we can investigate the performance characteristics of various estimators for the mean in that design. To this end, the `evaluate_estimator` function can be used.

    evaluate_estimator(
     score = MSE(),
     estimator = SampleMean(),
     data_distribution = Normal(two_armed = TRUE),
     design = get_example_design(two_armed = TRUE),
     mu = 0.3,
     sigma = 1
    )
    #> Design:                              TwoStageDesign<n1=56;0.8<=x1<=2.3:n2=18-80>
    #> Data Distribution:                                             Normal<two-armed>
    #> Estimator:                                                           Sample mean
    #> Assumed sigma:                                                                 1
    #> Assumed mu:                                                                  0.3
    #> Results:
    #>  Expectation:                                                          0.3056727
    #>  Bias:                                                               0.005672677
    #>  Variance:                                                            0.03777784
    #>  MSE:                                                                 0.03781002

The mean squared error of the sample mean depends on the true underlying value of the paramter \(\mu\), which of course is unknown. Therefore, it may be advisable to use the `evaluate_estimator` function on an array of values for \(\mu\) to investigate the distributional properties of an estimator.

In the following, the MSE of the sample mean vs. a weighted sample mean with fixed weights will be plotted.

    mse_mle <- evaluate_estimator(
      score = MSE(),
      estimator = SampleMean(),
      data_distribution = Normal(two_armed = TRUE),
      design = get_example_design(two_armed = TRUE),
      mu = seq(-0.75, 1.32, .03),
      sigma = 1
    )
    mse_weighted_sample_means <- evaluate_estimator(
      score = MSE(),
      estimator = WeightedSampleMean(w1 = .8),
      data_distribution = Normal(two_armed = TRUE),
      design = get_example_design(two_armed = TRUE),
      mu = seq(-0.75, 1.32, .03),
      sigma = 1
    )
    plot(c(mse_mle, mse_weighted_sample_means))

## Analyzing datasets

Next, let us look at how to the package can be used to calculate estimates after data has been collected.

The first stage data of a trial might look like this:

    set.seed(321)
    dat <- data.frame(
     endpoint = c(rnorm(56, .3, 1), rnorm(56, 0, 1)),
     group = factor(rep(c("trt", "ctl"),
                        c(56,56)), levels = c("trt", "ctl")),
     stage = rep(1, 112)
    )
    head(dat)
    #>      endpoint group stage
    #> 1  2.00490322   trt     1
    #> 2 -0.41203856   trt     1
    #> 3  0.02201509   trt     1
    #> 4  0.18035098   trt     1
    #> 5  0.17603938   trt     1
    #> 6  0.56818377   trt     1

    analyze(data = dat,
            statistics = list(),
            data_distribution = Normal(two_armed = TRUE),
            design = get_example_design(two_armed = TRUE),
            sigma = 1)
    #> Design:                              TwoStageDesign<n1=56;0.8<=x1<=2.3:n2=18-80>
    #> Data Distribution:                                             Normal<two-armed>
    #> Observed number of stages:                                                     1
    #> Observed n1 (group 1)                                                         56
    #> Observed n1 (group 2)                                                         56
    #> Observed n1 (total)                                                          112
    #> Z1                                                                          1.75
    #> Interim decision:                                       continue to second stage
    #> Calculated n2(Z1) (per group)                                           46.99923
    #> Calculated c2(Z1)                                                           1.14

The results suggest recruiting 23 more patients per group for the second stage.

We will simulate 47 more patients per group:

    dat <- rbind(dat,
                 data.frame(
                   endpoint = c(rnorm(47, .3, 1), rnorm(47, 0, 1)),
                   group = factor(rep(c("trt", "ctl"),
                                      c(47, 47)), levels = c("trt", "ctl")),
                   stage = rep(2, 94)
                 ))

Now, we can use the `analyze` function to evaluate a selection of estimators on the complete dataset:

    analyze(
     data = dat,
     statistics = c(
       SampleMean(),
       BiasReduced(),
       PseudoRaoBlackwell(),
       MedianUnbiasedStagewiseCombinationFunctionOrdering(),
       StagewiseCombinationFunctionOrderingCI(),
       StagewiseCombinationFunctionOrderingPValue()
       ),
     data_distribution = Normal(two_armed = TRUE),
     sigma = 1,
     design = get_example_design(two_armed = TRUE)
    )
    #> Design:                              TwoStageDesign<n1=56;0.8<=x1<=2.3:n2=18-80>
    #> Data Distribution:                                             Normal<two-armed>
    #> Observed number of stages:                                                     2
    #> Observed n1 (group 1)                                                         56
    #> Observed n1 (group 2)                                                         56
    #> Observed n1 (total)                                                          112
    #> Z1                                                                          1.75
    #> Interim decision:                                       continue to second stage
    #> Calculated n2(Z1) (per group)                                           46.99923
    #> Calculated c2(Z1)                                                           1.14
    #> Observed n2 (group 1)                                                         47
    #> Observed n2 (group 2)                                                         47
    #> Observed n2 (in total)                                                        94
    #> Z2                                                                          2.71
    #> Final test decision:                                                 reject null
    #>
    #> Stage 2 results:
    #>  Sample mean:                                                           0.434684
    #>  Bias reduced MLE (iterations=1):                                      0.4221532
    #>  Pseudo Rao-Blackwellized:                                             0.2658506
    #>  Median unbiased (SWCF ordering):                                      0.3047428
    #>  SWCF ordering CI:                                       [0.04435513, 0.5484439]
    #>  SWCF ordering p-value:                                               0.01097266

The estimates presented here are for the difference in means of the two normal distributions. Keep in mind that a difference of \(\mu=0.3\) was used in the simulation.

Note that while the median unbiased estimator performs well in this particular example, this is not universally true.