Mixed Continuous and Binary Co-Primary Endpoints • twoCoprimary

Overview

This vignette demonstrates sample size calculation for clinical trials with two co-primary endpoints where one is continuous and one is binary. The methodology is based on Sozu et al. (2012).

Important note on notation: In Sozu et al. (2012), the allocation ratio is defined as $\kappa = n_{2}/n_{1}$ (control/treatment), which is the inverse of the notation used in other papers in this package where $r = n_{1}/n_{2}$ (treatment/control). Therefore, $\kappa = 1/r$ . In this vignette, we follow the notation from the original paper using $\kappa$ to maintain consistency with the published formulas.

library(twoCoprimary)
library(dplyr)
library(tidyr)
library(knitr)

Background

Clinical Context

Mixed continuous and binary co-primary endpoints are common in:

Rheumatoid arthritis trials: The mean change from baseline in the modified total Sharp score (mTSS) + An ACR50 response (yes/no)
Chronic kidney disease trials: The mean peak percentage change from baseline serum creatinine over the 72-h period after contrast media administration + The contrast-induced nephropathy (yes/no)

Why Mixed Endpoints?

Combining continuous and binary endpoints provides:

Comprehensive assessment: Magnitude of change (continuous) + clinical meaningfulness (binary)
Regulatory relevance: Many guidelines require both types
Clinical interpretability: Binary endpoints are easier to communicate to patients

Statistical Framework

Model and Assumptions

Consider a two-arm superiority trial with sample sizes $n_{1}$ (treatment) and $n_{2}$ (control). The allocation ratio supplied to the package is $r = n_{1}/n_{2}$ , and, as noted above, the formulas below are written in terms of $\kappa = n_{2}/n_{1} = 1/r$ .

For subject $i$ in group $j$ ( $j = 1$ : treatment, $j = 2$ : control), we observe two outcomes:

Outcome 1 (Continuous) ( $k = 1$ ): $X_{i,j,1} \sim \text{N}(\mu_{j}, \sigma^2)$

where $\mu_{j}$ is the population mean in group $j$ and $\sigma^{2}$ is the common variance across groups.

Outcome 2 (Binary) ( $k = 2$ ): $X_{i,j,2} \in \{0, 1\}$

where $X_{i,j,2} = 1$ if subject $i$ in group $j$ responds successfully, and 0 otherwise. Let $p_{j}$ denotes the true success probability in group $j$ for endpoint $2$ .

Correlation Structure: Biserial Correlation

The correlation between a continuous outcome and a binary outcome requires special consideration. Following Sozu et al. (2012), we assume that both outcomes have latent bivariate normal distributions.

Key concept: The binary outcome $X_{i,j,2}$ is assumed to arise from dichotomizing a latent continuous variable $X_{i,j,2}^{*}$ (i.e., $X_{i,j,2}^{*} \sim \text{N}(\mu_{j}^{\ast}, \sigma^{2\ast})$ ):

$X_{i,j,2} = \begin{cases} 1 & \text{if } X_{i,j,2}^* \geq g_{j} \\ 0 & \text{if } X_{i,j,2}^* < g_{j} \end{cases}$

where $g_{j}$ is a threshold (cut-off point) such that $\Pr(X_{i,j,2} = 1) = p_{j} = \Pr(X_{i,j,2}^{*} \geq g_{j})$ .

Biserial correlation: Assuming that $(X_{i,j,1}, X_{i,j,2}^{*})$ follow a bivariate normal distribution, the biserial correlation $\rho$ measures the correlation between the continuous outcome and the latent continuous variable underlying the binary outcome.

For the detailed formula relating the biserial correlation to the correlation between test statistics, see Sozu et al. (2012), equation (1) in the Supporting Information.

Hypothesis Testing

We test superiority of treatment over control for both endpoints:

For continuous endpoint (1): $\text{H}_{01}: \mu_{1} - \mu_{2} \leq 0 \text{ vs. } \text{H}_{11}: \mu_{1} - \mu_{2} > 0$

For binary endpoint (2): $\text{H}_{02}: p_{1} - p_{2} \leq 0 \text{ vs. } \text{H}_{12}: p_{1} - p_{2} > 0$

Co-primary endpoints (intersection-union test): $\text{H}_0 = \text{H}_{01} \cup \text{H}_{02} \text{ vs. } \text{H}_1 = \text{H}_{11} \cap \text{H}_{12}$

Reject $\text{H}_{0}$ at level $\alpha$ if and only if both $\text{H}_{01}$ and $\text{H}_{02}$ are rejected at level $\alpha$ .

Test Statistics

Continuous endpoint (Equation 2 in Sozu et al., 2012):

$Z_{1} = \frac{\bar{X}_{1} - \bar{X}_{2}}{\sigma\sqrt{\frac{1+\kappa}{n_{1}}}}$

where $\bar{X}_{1}$ and $\bar{X}_{2}$ are the sample means. When $\sigma$ is unknown, use the pooled sample standard deviation.

Binary endpoint - Asymptotic Normal (AN) method (Equation 3 in Sozu et al., 2012):

$Z_{2} = \frac{\hat{p}_{1} - \hat{p}_{2}}{\sqrt{\left(\frac{1}{n_{1}} + \frac{1}{n_{2}}\right) \hat{p}(1 - \hat{p})}}$

where:

$\hat{p}_{1}$ and $\hat{p}_{2}$ are the sample proportions
$\hat{p} = \frac{n_{1}\hat{p}_{1} + n_{2}\hat{p}_{2}}{n_{1} + n_{2}}$ is the pooled proportion

Other test methods: Sozu et al. (2012) also present:

ANc: Asymptotic normal with continuity correction (equation 5)
AS: Arcsine transformation without continuity correction (equation 6)
ASc: Arcsine transformation with continuity correction (equation 7)
Fisher: Fisher’s exact test (simulation-based)

See the paper for detailed formulas. The twoCoprimary package implements all five methods.

Fisher’s exact test is different in kind from the other four. Under the null hypothesis the number of responders in group 1 is hypergeometric given the total, which gives an exact conditional $p$ -value for the binary endpoint, but the joint distribution of that $p$ -value and the $t$ statistic of the continuous endpoint has no closed form. Sozu et al. (2012) therefore obtain the overall power by Monte Carlo integration over the latent variables. Two consequences follow for users. The result varies between calls unless a seed is set with set.seed(), and the sequential search compares a simulated power against the target, so the returned sample size can move by a subject or two between runs. Raising nMC reduces both effects at a proportional cost.

Joint Distribution and Power Calculation

Under $\text{H}_{1}$ , the test statistics $(Z_{1}, Z_{2})$ asymptotically follow a bivariate normal distribution. The overall power is given by (Equation 4 in Sozu et al., 2012):

$1 - \beta = \Pr\left[\bigcap_{k=1}^{2} \{Z_{k} > z_{1-\alpha}\}\right] \approx \Pr\left[\bigcap_{k=1}^{2} \left\{Z_{k}^{*} > c_{k}^{*}\right\}\right]$

where $Z_{k}^{*}$ is $Z_{k}$ centered at its mean under the alternative and scaled to unit variance, so that it is standard normal, and $c_{k}^{*}$ is $z_{1-\alpha}$ carried through the same centering and scaling. The two endpoints are standardized by their own quantities, the mean difference for $k = 1$ and the risk difference for $k = 2$ :

For continuous endpoint ( $k = 1$ ): $c_{1}^{*} = z_{1-\alpha} - \frac{\delta_{1}}{\sigma} \sqrt{\frac{\kappa n_{1}}{1+\kappa}}$

where $\delta_{1} = \mu_{1} - \mu_{2}$ is the effect size for the endpoint 1.

For binary endpoint ( $k = 2$ , AN method) (Equation 5 in Sozu et al., 2012): $c_{2}^{*} = \frac{\sqrt{\frac{(p_{1}+\kappa p_{2})\{(1-p_{1})+\kappa(1-p_{2})\}}{1+\kappa}} z_{1-\alpha} - \sqrt{\kappa n_{1}}(p_{1}-p_{2})}{\sqrt{\kappa p_{1}(1-p_{1})+p_{2}(1-p_{2})}}$

The vector $(Z_{1}^{*}, Z_{2}^{*})^\text{T}$ is approximately distributed as a standardized bivariate normal distribution $\text{N}_{2}(\mathbf{0}, \gamma)$ , where $\gamma$ is the correlation between the test statistics.

Correlation between test statistics: For the mixed continuous and binary case, the correlation $\gamma$ between $Z_{1}$ and $Z_{2}$ depends on the biserial correlation $\rho$ between outcomes. The explicit formula involves the standard normal density function and success probabilities. See equation (1) in the Supporting Information of Sozu et al. (2012) for details:

$\text{Corr}(X_{i,j,1}, X_{i,j,2}) = \frac{\rho_{j} \xi_{j}}{\sqrt{p_{j}(1-p_{j})}}$

where $\xi_{j} = \frac{1}{\sqrt{2\pi}} \exp\left\{-\frac{(g_{j} - \mu_{j}^{\ast})^2}{2\sigma_{j}^{2\ast}}\right\}$ .

Sample Size Determination

The sample size is determined by solving the power equation numerically. For a given allocation ratio $r$ , target power $1 - \beta$ , and significance level $\alpha$ , we find the smallest $n_{2}$ such that the overall power equals or exceeds $1 - \beta$ .

Computational approach:

Calculate initial sample size based on single-endpoint formulas
Compute the correlation $\gamma$ between test statistics using the biserial correlation
Calculate joint power using the bivariate normal distribution
Iterate until target power is achieved

Fisher’s exact test: the binary outcome is generated as the dichotomization of the latent continuous variable at a group specific threshold $g_{j}$ chosen so that $\Pr(X_{i,j,2}^{\ast} \geq g_{j}) = p_{j}$ , jointly with the observed continuous outcome. Each replicate yields a $t$ statistic and a hypergeometric $p$ -value, and the overall power is the proportion of replicates in which both reject. The sample size calculation uses the same sequential search, starting from the AN sample size.

Replicating Sozu et al. (2012) Table 2

Table 2 from Sozu et al. (2012) shows sample sizes for the PREMIER study scenario with different standard deviations and correlations.

# Recreate Sozu et al. (2012) Table 2
library(dplyr)
library(tidyr)

param_grid_mixed_cb_ss <- expand.grid(
  delta = 4.4,
  sd = c(19, 20, 21, 22),
  p1 = 0.59,
  p2 = 0.46
)

result_mixed_cb_ss <- design_table(
  param_grid = param_grid_mixed_cb_ss,
  rho_values = c(0, 0.3, 0.5, 0.8),
  r = 1,
  alpha = 0.025,
  beta = 0.2,
  endpoint_type = "mixed_cont_binary",
  Test = "AN"
) %>% 
  mutate_at(vars(starts_with("rho_")), ~ . / 2)

kable(result_mixed_cb_ss,
      caption = "Table 2: Sample Size per Group (n) for PREMIER Study Scenario (delta1 = 4.4, p1 = 0.59, p2 = 0.46, α = 0.025, 1-β = 0.80)",
      digits = 2,
      col.names = c("delta", "σ", "p1", "p2", "ρ=0.0", "ρ=0.3", "ρ=0.5", "ρ=0.8"))

Table 2: Sample Size per Group (n) for PREMIER Study Scenario (delta1 = 4.4, p1 = 0.59, p2 = 0.46, α = 0.025, 1-β = 0.80)
delta	σ	p1	p2	ρ=0.0	ρ=0.3	ρ=0.5	ρ=0.8
4.4	19	0.59	0.46	346	340	334	323
4.4	20	0.59	0.46	369	363	358	347
4.4	21	0.59	0.46	394	389	384	374
4.4	22	0.59	0.46	422	417	413	404

Interpretation: This table shows that as the standard deviation increases, the required sample size increases. The correlation has a modest effect, reducing the sample size by 4 to 7% at $\rho = 0.8$ across the four standard deviations shown.

Replicating Sozu et al. (2012) Supporting Information Table 5

Table 5 from the Supporting Information shows sample sizes for scenarios with higher success probabilities and different test methods.

# Recreate Supporting Information Table 5
param_grid_mixed_cb_ss2 <- tibble(
  delta = c(0.235, 0.397, 0.521, 0.190, 0.335, 0.457),
  sd = 1,
  p1 = c(rep(0.99, 3), rep(0.95, 3)),
  p2 = c(seq(0.95, 0.85, length.out = 3), seq(0.90, 0.80, length.out = 3))
)

result_mixed_cb_ss2 <- do.call(
  bind_rows,
  lapply(c("ANc", "ASc"), function(test) {
    design_table(
      param_grid = param_grid_mixed_cb_ss2,
      rho_values = c(0, 0.3, 0.5, 0.8),
      r = 1,
      alpha = 0.025,
      beta = 0.2,
      endpoint_type = "mixed_cont_binary",
      Test = test
    ) %>% 
      mutate_at(vars(starts_with("rho_")), ~ . / 2) %>% 
      mutate(Test = test)
  })
) %>% 
  arrange(desc(p1), delta) %>% 
  select(delta, sd, p1, p2, Test, everything())

# Display for ANc
result_anc <- result_mixed_cb_ss2 %>% 
  filter(Test == "ANc") %>% 
  select(-Test)

kable(result_anc,
      caption = "Table 5 (Part A): Sample Size per Group (n) with Continuity Correction (ANc) (σ = 1, α = 0.025, 1-β = 0.80)^a^",
      digits = 3,
      col.names = c("delta", "σ", "p1", "p2", "ρ=0.0", "ρ=0.3", "ρ=0.5", "ρ=0.8"))

Table 5 (Part A): Sample Size per Group (n) with Continuity Correction (ANc) (σ = 1, α = 0.025, 1-β = 0.80)^a
delta	σ	p1	p2	ρ=0.0	ρ=0.3	ρ=0.5	ρ=0.8
0.235	1	0.99	0.95	400	397	395	391
0.397	1	0.99	0.90	143	142	141	139
0.521	1	0.99	0.85	84	83	82	81
0.190	1	0.95	0.90	592	585	579	569
0.335	1	0.95	0.85	195	192	190	187
0.457	1	0.95	0.80	106	105	104	102


# Display for ASc
result_asc <- result_mixed_cb_ss2 %>% 
  filter(Test == "ASc") %>% 
  select(-Test)

kable(result_asc,
      caption = "Table 5 (Part B): Sample Size per Group (n) with Arcsine and Continuity Correction (ASc) (σ = 1, α = 0.025, 1-β = 0.80)^a^",
      digits = 3,
      col.names = c("delta", "σ", "p1", "p2", "ρ=0.0", "ρ=0.3", "ρ=0.5", "ρ=0.8"))

Table 5 (Part B): Sample Size per Group (n) with Arcsine and Continuity Correction (ASc) (σ = 1, α = 0.025, 1-β = 0.80)^a
delta	σ	p1	p2	ρ=0.0	ρ=0.3	ρ=0.5	ρ=0.8
0.235	1	0.99	0.95	376	373	371	367
0.397	1	0.99	0.90	129	128	127	125
0.521	1	0.99	0.85	74	74	73	72
0.190	1	0.95	0.90	585	578	572	562
0.335	1	0.95	0.85	189	187	185	182
0.457	1	0.95	0.80	102	101	100	98

^a Four of the forty-eight values differ from Supporting Information Table 5 of Sozu et al. (2012) by one subject. The standardized effect size of the continuous endpoint is printed there to three decimal places, and for each of the four a value inside the rounding window of the printed figure returns the published sample size, so the difference is what the third decimal place is worth rather than a difference in the calculation.

Part C: Fisher’s exact test

Table 5 of the Supporting Information also reports Fisher’s exact test. The row with the largest standardized effect is reproduced here; the remaining rows behave the same way but need sample sizes up to 584 per group, which makes a simulated search slow enough to be unsuitable for a vignette. The test suite checks the achieved power at two of the published Fisher sample sizes, and the verification scripts kept in the package’s repository check all 24.

set.seed(20260829)

rho_values <- c(0, 0.3, 0.5, 0.8)
published_n <- c(76, 75, 75, 74)

fisher_row <- lapply(seq_along(rho_values), function(i) {
  ss <- ss2MixedContinuousBinary(
    delta = 0.521, sd = 1,
    p1 = 0.99, p2 = 0.85,
    rho = rho_values[i], r = 1,
    alpha = 0.025, beta = 0.2,
    Test = "Fisher", nMC = 10000
  )
  data.frame(rho = rho_values[i], n_per_group = ss$n1, published = published_n[i])
})

kable(do.call(rbind, fisher_row),
      caption = "Table 5 (Part C): Sample Size per Group (n) with Fisher's Exact Test (delta1 = 0.521, p1 = 0.99, p2 = 0.85, sigma = 1, alpha = 0.025, 1-beta = 0.80)",
      col.names = c("rho", "n per group", "Sozu et al. (2012)"))

Table 5 (Part C): Sample Size per Group (n) with Fisher’s Exact Test (delta1 = 0.521, p1 = 0.99, p2 = 0.85, sigma = 1, alpha = 0.025, 1-beta = 0.80)
rho	n per group	Sozu et al. (2012)
0.0	76	76
0.3	77	75
0.5	75	75
0.8	74	74

Agreement within a subject or two is the most that a simulated power can give at this number of replications; raising nMC tightens it at a proportional cost.

Key findings:

ANc and ASc give similar sample sizes
Fisher’s exact test agrees closely with ASc, as Sozu et al. (2012) report
Correlation effect is modest for these scenarios
Higher success probabilities ( $p_{1}$ ) generally require larger sample sizes when the effect size is small

Basic Usage Examples

Example 1: Balanced Design

Calculate sample size for a balanced design with moderate effect sizes:

# Balanced design: nT = nC (i.e., r = 1, which corresponds to kappa = 1)
result_balanced <- ss2MixedContinuousBinary(
  delta = 0.5,           # Standardized effect for continuous endpoint
  sd = 1,                # Standard deviation
  p1 = 0.7,              # Success prob in treatment group
  p2 = 0.5,              # Success prob in control group
  rho = 0.5,             # Biserial correlation
  r = 1,                 # Balanced allocation (r = nT/nC = 1)
  alpha = 0.025,
  beta = 0.2,
  Test = "AN"
)

print(result_balanced)
#> 
#> Sample size calculation for mixed continuous and binary co-primary endpoints
#> 
#>              n1 = 102
#>              n2 = 102
#>               N = 204
#>           delta = 0.5
#>              sd = 1
#>               p = 0.7, 0.5
#>             rho = 0.5
#>      allocation = 1
#>           alpha = 0.025
#>            beta = 0.2
#>            Test = AN

Note: In the function, $r = n_{1}/n_{2}$ . Thus $r = 1$ corresponds to balanced allocation ( $n_{1} = n_{2}$ ), which is equivalent to $\kappa = 1$ in Sozu et al. (2012).

Example 2: Effect of Correlation

Demonstrate how biserial correlation affects sample size:

# Fixed effect sizes
delta <- 0.5
p1 <- 0.7
p2 <- 0.5

# Range of correlations
rho_values <- c(0, 0.3, 0.5, 0.8)

ss_by_rho <- sapply(rho_values, function(rho) {
  result <- ss2MixedContinuousBinary(
    delta = delta,
    sd = 1,
    p1 = p1,
    p2 = p2,
    rho = rho,
    r = 1,
    alpha = 0.025,
    beta = 0.2,
    Test = "AN"
  )
  result$n2
})

result_df <- data.frame(
  rho = rho_values,
  n_per_group = ss_by_rho,
  N_total = ss_by_rho * 2,
  reduction_pct = round((1 - ss_by_rho / ss_by_rho[1]) * 100, 1)
)

kable(result_df,
      caption = "Effect of Biserial Correlation on Sample Size",
      col.names = c("ρ", "n per group", "N total", "Reduction (%)"))

Effect of Biserial Correlation on Sample Size
ρ	n per group	N total	Reduction (%)
0.0	104	208	0.0
0.3	103	206	1.0
0.5	102	204	1.9
0.8	99	198	4.8


# Plot
plot(rho_values, ss_by_rho, 
     type = "b", pch = 19,
     xlab = "Biserial Correlation (ρ)", 
     ylab = "Sample size per group (n)",
     main = "Effect of Correlation on Sample Size",
     ylim = c(90, max(ss_by_rho) * 1.1))
grid()

Interpretation: Higher positive correlation reduces the required sample size. At $\rho = 0.8$ the reduction is about 5% relative to $\rho = 0$ in this scenario.

Example 3: Comparison of Test Methods

Compare different test methods for the binary endpoint:

# Fixed design parameters
delta <- 0.5
p1 <- 0.7
p2 <- 0.5
rho <- 0.5

test_methods <- c("AN", "ANc", "AS", "ASc")

test_comparison <- lapply(test_methods, function(test_method) {
  result <- ss2MixedContinuousBinary(
    delta = delta,
    sd = 1,
    p1 = p1,
    p2 = p2,
    rho = rho,
    r = 1,
    alpha = 0.025,
    beta = 0.2,
    Test = test_method
  )
  
  data.frame(
    Test_method = test_method,
    n_per_group = result$n2,
    N_total = result$N
  )
})

test_comparison_table <- bind_rows(test_comparison)

kable(test_comparison_table,
      caption = "Comparison of Test Methods for Binary Endpoint",
      digits = 0,
      col.names = c("Test Method", "n per group", "N total"))

Comparison of Test Methods for Binary Endpoint
Test Method	n per group	N total
AN	102	204
ANc	109	218
AS	101	202
ASc	109	218

Key findings:

AS (arcsine): Smallest sample size
AN (no continuity correction): One subject larger than AS here
ANc and ASc (with continuity correction): About 7% larger than AN
ASc (arcsine with CC): Similar to ANc

Example 4: Unbalanced Allocation

Calculate sample size with 2:1 allocation ratio:

# Balanced design (r = 1, equivalent to kappa = 1)
result_balanced <- ss2MixedContinuousBinary(
  delta = 0.5,
  sd = 1,
  p1 = 0.7,
  p2 = 0.5,
  rho = 0.5,
  r = 1,
  alpha = 0.025,
  beta = 0.2,
  Test = "AN"
)

# Unbalanced design (r = 2, i.e., nT = 2*nC, equivalent to kappa = 0.5)
result_unbalanced <- ss2MixedContinuousBinary(
  delta = 0.5,
  sd = 1,
  p1 = 0.7,
  p2 = 0.5,
  rho = 0.5,
  r = 2,
  alpha = 0.025,
  beta = 0.2,
  Test = "AN"
)

comparison_allocation <- data.frame(
  Design = c("Balanced (1:1)", "Unbalanced (2:1)"),
  n_treatment = c(result_balanced$n1, result_unbalanced$n1),
  n_control = c(result_balanced$n2, result_unbalanced$n2),
  N_total = c(result_balanced$N, result_unbalanced$N),
  kappa = c(1, 0.5)
)

kable(comparison_allocation,
      caption = "Comparison: Balanced vs Unbalanced Allocation",
      col.names = c("Design", "nT", "nC", "N total", "κ"))

Comparison: Balanced vs Unbalanced Allocation
Design	nT	nC	N total	κ
Balanced (1:1)	102	102	204	1.0
Unbalanced (2:1)	152	76	228	0.5


cat("\nIncrease in total sample size:", 
    round((result_unbalanced$N - result_balanced$N) / result_balanced$N * 100, 1), "%\n")
#> 
#> Increase in total sample size: 11.8 %

Note: In the function, $r = n_{1}/n_{2}$ , so $r = 2$ means $n_{1} = 2 \times n_{2}$ , which corresponds to $\kappa = n_{2}/n_{1} = 0.5$ in Sozu et al. (2012) notation.

Power Verification

Verify that calculated sample sizes achieve target power:

# Use result from Example 1
power_result <- power2MixedContinuousBinary(
  n1 = result_balanced$n1,
  n2 = result_balanced$n2,
  delta = 0.5,
  sd = 1,
  p1 = 0.7,
  p2 = 0.5,
  rho = 0.5,
  alpha = 0.025,
  Test = "AN"
)

cat("Target power: 0.80\n")
#> Target power: 0.80
cat("Achieved power (Continuous endpoint):", round(as.numeric(power_result$power1), 4), "\n")
#> Achieved power (Continuous endpoint):
cat("Achieved power (Binary endpoint):", round(as.numeric(power_result$power2), 4), "\n")
#> Achieved power (Binary endpoint):
cat("Achieved power (Co-primary):", round(as.numeric(power_result$powerCoprimary), 4), "\n")
#> Achieved power (Co-primary): 0.8044

Practical Recommendations

Design Considerations

Estimating biserial correlation: Use pilot data or historical studies; be conservative if uncertain. Biserial correlation is more challenging to estimate than Pearson correlation.
Latent variable assumption: Ensure the binary endpoint conceptually has an underlying continuous scale (e.g., “improved” means crossing a threshold on a continuous improvement scale).
Test method selection:
- AN: Most common, smallest sample size
- ANc: Adds continuity correction for conservatism
- AS: Uses arcsine transformation for variance stabilization
- ASc: Combines arcsine transformation with continuity correction
- Fisher: Provides exact inference for the binary endpoint, at the cost of a simulated overall power
Balanced allocation: Generally most efficient ( $\kappa = 1$ , i.e., $r = 1$ ) unless practical constraints require otherwise.
Sensitivity analysis: Calculate for range of plausible correlations and effect sizes.

When to Use This Method

Use mixed continuous-binary methods when:

One endpoint is naturally continuous (e.g., change in test score)
Other endpoint is naturally binary (e.g., clinical response yes/no)
Both endpoints are clinically meaningful co-primary endpoints
Sample sizes are moderate to large ( $N > 50$ )

Challenges and Considerations

Correlation estimation: Biserial correlation involves a latent variable and is harder to estimate than Pearson correlation
Threshold specification: The dichotomization threshold affects correlation; ensure it’s clinically meaningful
Asymmetric power: Mixed endpoints often have unequal power for the two endpoints; the endpoint with lower power dominates sample size
Asymptotic approximation: Methods rely on asymptotic normality; may not be accurate for very small samples ( $N < 50$ )

References

Sozu, T., Sugimoto, T., & Hamasaki, T. (2012). Sample size determination in clinical trials with multiple co-primary endpoints including mixed continuous and binary variables. Biometrical Journal, 54(5), 716-729.