Overview of survival endpoint design

Power and sample size with nEvents()

Based on the above, the power for the logrank test is approximated by

\[P[Z\le z]=\Phi(z -\sqrt n\theta)=\Phi(z- \sqrt{nr}/(1+r)\log\nu).\] Thus, assuming \(n=100\) events and \(\delta = \log\nu=-\log(.7)\), and \(r=1\) (equal randomization) we approximate power for the logrank test when \(\alpha=0.025\) as

n <- 100
hr <- .7
delta <- log(hr)
alpha <- .025
r <- 1
pnorm(qnorm(alpha) - sqrt(n * r) / (1 + r) * delta)
#> [1] 0.4299155

We can compute this with gsDesign::nEvents() as:

nEvents(n = n, alpha = alpha, hr = hr, r = r)
#> [1] 0.4299155

We solve for the number of events \(n\) to see how many events are required to obtain a desired power

\[1-\beta=P(Z\ge \Phi^{-1}(1-\alpha))\] with

\[n = \left(\frac{\Phi^{-1} (1-\alpha)+\Phi^{-1}(1-\beta)}{\theta}\right)^2 =\frac{(1+r)^2}{r(\log\nu)^2}\left({\Phi^{-1} (1-\alpha)+\Phi^{-1}(1-\beta)}\right)^2.\] Thus, the approximate number of events required to power for HR=0.7 with \(\alpha=0.025\) one-sided and power \(1-\beta=0.9\) is

beta <- 0.1
(1 + r)^2 / r / log(hr)^2 * ((qnorm(1 - alpha) + qnorm(1 - beta)))^2
#> [1] 330.3779

which, rounding up, matches (with tabular output):

nEvents(hr = hr, alpha = alpha, beta = beta, r = 1, tbl = TRUE) %>%
  kable()

The notation delta in the above table changes the sign for the standardized treatment effect \(\theta\) in the above:

theta <- delta * sqrt(r) / (1 + r)
theta
#> [1] -0.1783375

The se in the table is the estimated standard error for the log hazard ratio \(\delta=\log\hat\nu\)

(1 + r) / sqrt(331 * r)
#> [1] 0.1099299

Group sequential design

We can create a group sequential design for the above problem either with \(\theta\) or with the fixed design sample size. The parameter delta in gsDesign() corresponds to standardized effect size with sign changed \(-\theta\) in notation used above and by Jennison and Turnbull (2000), while the natural parameter, \(\log(\hbox{HR})\) is in the parameter delta1 passed to gsDesign(). The name of the effect size is specified in deltaname and the parameter logdelta = TRUE indicates that delta input needs to be exponentiated to obtain HR in the output below. This example code can be useful in practice. We begin by passing the number of events for a fixed design in the parameter n.fix (continuous, not rounded) to adapt to a group sequential design.

Schoenfeld <- gsDesign(
  k = 2,
  n.fix = nEvents(hr = hr, alpha = alpha, beta = beta, r = 1),
  delta1 = log(hr)
)
Schoenfeld %>%
  gsBoundSummary(deltaname = "HR", logdelta = TRUE) %>%
  kable(row.names = FALSE)

Information based design

Exactly the same result can be obtained with the following, passing the standardized effect size theta from above to the parameter delta in gsDesign().

Schoenfeld <- gsDesign(k = 2, delta = -theta, delta1 = log(hr))

We noted above that the asymptotic variance for \(\hat\theta\) is \(1/n\) which corresponds to statistical information \(\mathcal I=n\) for the parameter \(\theta\). Thus, the value

Schoenfeld$n.I
#> [1] 172.2757 344.5514

corresponds both to the number of events and the statistical information for the standardized effect size \(\theta\) required to power the trial at the desired level. Note that if you plug in the natural parameter \(\delta= -\log\nu > 0\), then \(n.I\) returns the statistical information for the log hazard ratio.

gsDesign(k = 2, delta = -log(hr))$n.I
#> [1] 43.06893 86.13786

The reader may wish to look above to derive the exact relationship between events and statistical information for \(\delta\).

Approximating boundary characteristics

Another application of the Schoenfeld (1981) method is to approximate boundary characteristics of a design. As noted in the introduction above, we will consider zn2hr(), gsHR() and gsBoundSummary() to approximate the treatment effect required to cross design bounds. zn2hr() is complemented by the functions hrn2z() and hrz2n(). We begin with the basic approximation used across all of these functions in this section and follow with a sub-section with example code to reproduce some of what is in the table above.

We return to the following equation from above:

\[Z\approx Z_W\approx \frac{\sqrt {nr}}{1+r}\hat\delta=\frac{\ln(\hat\nu)\sqrt{nr}}{1+r}.\] By fixing \(Z=z, n\) we can solve for \(\hat\nu\) from the above:

\[\hat{\nu} = \exp(z(1+r)/\sqrt{rn}).\] By fixing \(\hat\nu\) and \(z\), we can solve for the corresponding number of events required: \[ n = (z(1+r)/\log(\hat{\nu}))^2/r.\]

Examples

For our first example, we note that the event counts in Schoenfeld are actually continuous numbers that are rounded up in the table:

Schoenfeld$n.I
#> [1] 172.2757 344.5514

We reproduce the approximate hazard ratios required to cross efficacy bounds using the Schoenfeld approximations above:

gsHR(
  z = Schoenfeld$upper$bound, # Z-values at bound
  i = 1:2, # Analysis number
  x = Schoenfeld, # Group sequential design from above
  ratio = r # Experimental/control randomization ratio
)
#> [1] 0.6576844 0.8077846

For the following examples, we assume \(r=1\).

r <- 1

1. Assuming a Cox model estimate \(\hat\nu\) and a corresponding event count, approximately what Z-value (p-value) does this correspond to? We use the first equation above:

hr <- .73 # Observed hr
events <- 125 # Events in analysis

z <- log(hr) * sqrt(events * r) / (1 + r)
c(z, pnorm(z)) # Z- and p-value
#> [1] -1.75928655  0.03926443

We replicate the Z-value with

hrn2z(hr = hr, n = events, ratio = r)
#> [1] -1.759287

2. Assuming an efficacy bound Z-value and event count, approximately what hazard ratio must be observed to cross the bound? We use the second equation above:

z <- qnorm(.025)
events <- 120
exp(z * (1 + r) / sqrt(r * events))
#> [1] 0.6991858

We can reproduce this with zn2hr() by switching the sign of z above; note that the default is ratio = 1 for all of these functions and often is not specified:

zn2hr(z = -z, n = events, ratio = r)
#> [1] 0.6991858

3. Finally, if we want an observed hazard ratio \(\hat\nu = .8\) to represent a positive result, how many events would be need to observe to achieve a 1-sided p-value of 0.025? assuming 2:1 randomization? We use the third equation above:

r <- 2
hr <- .8
z <- qnorm(.025)
events <- (z * (1 + r) / log(hr))^2 / r
events
#> [1] 347.1683

This is replicated with

hrz2n(hr = hr, z = z, ratio = r)
#> [1] 347.1683

Model assumptions

Some detail in specification comes With the flexibility allowed by the Lachin and Foulkes (1986) method. The model assumes

• A fixed enrollment period with piecewise constant enrollment rates
• A fixed minimum follow-up period
• Piecewise exponential failure rates for the control group
• A single, constant hazard ratio for the experimental group
• Piecewise exponential loss-to-follow-up rates
• A stratified population
• A fixed randomization ratio of experimental to control group assignment

Other than the proportional hazards assumption, this allows a great deal of flexibility in trial design assumptions. While Lachin and Foulkes (1986) adjusts the piecewise constant enrollment rates proportionately to derive a sample size, gsDesign$nSurv() also enables the approach of Kim and Tsiatis (1990) which fixes enrollment rates and extends the final enrollment rate duration to power the trial; the minimum follow-up period is still assumed this approach. We do not enable the drop-in option proposed in Lachin and Foulkes (1986).

The two practical differences the Lachin and Foulkes (1986) method has from the Schoenfeld (1981) method are:

1. By assuming enrollment, failure and dropout rates the method delivers sample size \(N\) as well as events required.
2. The variance for the log hazard ratio \(\hat\delta\) is computed differently and both a null (\(\sigma^2_0\)) and alternate hypothesis (\(\sigma^2_1\)) variance are incorporated through the formula \[N = \left(\frac{\Phi^{-1}(1-\alpha)\sigma_0 + \Phi^{-1}(1-\beta)\sigma_1}{\delta}\right).\] The null hypothesis is derived by averaging the alternate hypothesis rates, weighting according to the proportion randomized in each group.

Fixed design

We will use the same hazard ratio 0.7 as for the Schoenfeld (1981) sample size calculations above. We assume further that the trial will enroll for a constant rate for 12 months, have a control group median of 8 months (exponential failure rate \(\lambda = \log(2)/8\)), a dropout rate of 0.001 per month, and 16 months of minimum follow-up. As before, we assume a randomization ratio \(r=1\), one-sided Type I error \(\alpha=0.025\), 90% power which is equivalent to Type II error \(\beta=0.1\).

r <- 1 # Experimental/control randomization ratio
alpha <- 0.025 # 1-sided Type I error
beta <- 0.1 # Type II error (1 - power)
hr <- 0.7 # Hazard ratio (experimental / control)
controlMedian <- 8
dropoutRate <- 0.001 # Exponential dropout rate per time unit
enrollDuration <- 12
minfup <- 16 # Minimum follow-up
Nlf <- nSurv(
  lambdaC = log(2) / controlMedian,
  hr = hr,
  eta = dropoutRate,
  T = enrollDuration + minfup, # Trial duration
  minfup = minfup,
  ratio = r,
  alpha = alpha,
  beta = beta
)
cat(paste("Sample size: ", ceiling(Nlf$n), "Events: ", ceiling(Nlf$d), "\n"))
#> Sample size:  422 Events:  330

Recall that the Schoenfeld (1981) method recommended 331 events. The two methods tend to yield very similar event count recommendations, but not the same. Other methods will also differ slightly; see Lachin and Foulkes (1986). Sample size recommendations can vary more between methods.

We can get the same result with the nSurvival() routine since only a single enrollment, failure and dropout rate is proposed for this example.

lambda1 <- log(2) / controlMedian
nSurvival(
  lambda1 = lambda1,
  lambda2 = lambda1 * hr,
  Ts = enrollDuration + minfup,
  Tr = enrollDuration,
  eta = dropoutRate,
  ratio = r,
  alpha = alpha,
  beta = beta
)
#> Fixed design, two-arm trial with time-to-event
#> outcome (Lachin and Foulkes, 1986).
#> Study duration (fixed):          Ts=28
#> Accrual duration (fixed):        Tr=12
#> Uniform accrual:              entry="unif"
#> Control median:      log(2)/lambda1=8
#> Experimental median: log(2)/lambda2=11.4
#> Censoring median:        log(2)/eta=693.1
#> Control failure rate:       lambda1=0.087
#> Experimental failure rate:  lambda2=0.061
#> Censoring rate:                 eta=0.001
#> Power:                 100*(1-beta)=90%
#> Type I error (1-sided):   100*alpha=2.5%
#> Equal randomization:          ratio=1
#> Sample size based on hazard ratio=0.7 (type="rr")
#> Sample size (computed):           n=422
#> Events required (computed): nEvents=330

Group sequential design

k <- 2 # Total number of analyses
lfgs <- gsSurv(
  k = 2,
  lambdaC = log(2) / controlMedian,
  hr = hr,
  eta = dropoutRate,
  T = enrollDuration + minfup, # Trial duration
  minfup = minfup,
  ratio = r,
  alpha = alpha,
  beta = beta
)
lfgs %>%
  gsBoundSummary() %>%
  kable(row.names = FALSE)

Although we did not use the Schoenfeld (1981) for sample size, it is still used for the approximate HR at bound calculation above:

events <- lfgs$n.I
z <- lfgs$upper$bound
zn2hr(z = z, n = events) # Schoenfeld approximation to HR
#> [1] 0.6571386 0.8074431

Plotting

There are various plots available. The approximate hazard ratios required to cross bounds again use the Schoenfeld (1981) approximation. For a ggplot2 version of this plot, use the default base = FALSE.

plot(lfgs, pl = "hr", dgt = 4, base = TRUE)

Event accrual

The variance calculations for the Lachin and Foulkes method are mostly determined by expected event accrual under the null and alternate hypotheses. The null hypothesis characterized above is seemingly designed so that event accrual will be similar under each hypothesis. You can see the expected events accrued at each analysis under the alternate hypothesis with:

tibble::tibble(
  Analysis = 1:2,
  `Control events` = lfgs$eDC,
  `Experimental events` = lfgs$eDE
) %>%
  kable()

It is worth noting that if events accrue at the same rate in both the null and alternate hypothesis, then the expected duration of time to achieve the targeted events would be shortened. Keep in mind that there can be many reasons events will accrue at a different rate than in the design plan.

The expected event accrual of events over time for a design can be computed as follows:

Month <- seq(0.025, enrollDuration + minfup, .025)
plot(
  c(0, Month),
  c(0, sapply(Month, function(x) {
    nEventsIA(tIA = x, x = lfgs)
  })),
  type = "l", xlab = "Month", ylab = "Expected events",
  main = "Expected event accrual over time"
)

On the other hand, if you want to know the expected time to accrue 25% of the final events and what the expected enrollment accrual is at that time, you compute using:

b <- tEventsIA(x = lfgs, timing = 0.25)
cat(paste(
  " Time: ", b$T,
  "\n Expected enrollment:", b$eNC + b$eNE,
  "\n Expected control events:", b$eDC,
  "\n Expected experimental events:", b$eDE, "\n"
))
#>  Time:  8.88072854965729 
#>  Expected enrollment: 325.066468979827 
#>  Expected control events: 49.0303966242417 
#>  Expected experimental events: 36.7672374839294

For expected accrual of events without a design returned by gsDesign::gsSurv(), see the help file for gsDesign::eEvents().