K-Sample Test Procedure Survival

Procedure

Step-by-step procedure for comparing survival experience across $K$ independent groups.

1. Formulate Hypotheses

$H_0:h_1(t)=h_2(t)=\dots =h_K(t)\forall t\le \tau$

$H_1:\text{not all }{h}_j(t)\text{ are equal for some }t\le \tau$

2. Choose Weight Function

Test	$W(t_i)$	When to Use
Log-Rank	$1$	Proportional hazards assumed; balanced weighting
Gehan / Breslow	$Y_i$	Early differences matter more
Tarone-Ware	$\sqrt{Y_i}$	Compromise between log-rank and Gehan

3. Compute Test Components

For each group $j=1,\dots ,K$:

$Z_j(\tau )={\sum }_{i=1}^{D}W(t_i)\left[d_{ij}-Y_{ij}\frac{d_i}{Y_i}\right]$

where:

• $t_i$ : ordered distinct event times (across all groups pooled)
• $d_{ij}$ : events in group $j$ at $t_i$
• $Y_{ij}$ : at risk in group $j$ just before $t_i$
• $d_i={\sum }_j{d}_{ij}$ : total events at $t_i$
• $Y_i={\sum }_j{Y}_{ij}$ : total at risk at $t_i$

Under $H_0$, each group’s expected events: $E[d_{ij}]=Y_{ij}\cdot d_i/Y_i$

4. Compute Variance-Covariance Matrix

For $j=1,\dots ,K$: ${\hat{\sigma }}_{jj}={\sum }_{i=1}^{D}W(t_i{)}^2\frac{Y_{ij}}{Y_i}\left(1-\frac{Y_{ij}}{Y_i}\right)\left(\frac{Y_i-d_i}{Y_i-1}\right)d_i$

For $g\ne j$: ${\hat{\sigma }}_{jg}=-{\sum }_{i=1}^{D}W(t_i{)}^2\frac{Y_{ij}{Y}_{ig}}{Y_i^2}\left(\frac{Y_i-d_i}{Y_i-1}\right)d_i$

5. Compute Test Statistic

${\chi }^2=(Z_1,\dots ,Z_{K-1}){\hat{\Sigma }}^{-1}(Z_1,\dots ,Z_{K-1}{)}^T$

Note: Only $K-1$ components are needed because ${\sum }_{j=1}^K{Z}_j=0$.

For $K=2$ (two-sample case): $Z=\frac{{\sum }_{i=1}^{D}W(t_i)\left[d_{i1}-Y_{i1}\frac{d_i}{Y_i}\right]}{\sqrt{{\sum }_{i=1}^{D}W(t_i{)}^2\frac{Y_{i1}}{Y_i}\left(1-\frac{Y_{i1}}{Y_i}\right)\left(\frac{Y_i-d_i}{Y_i-1}\right)d_i}}\sim N(0,1)$

6. Decision Rule

Case	Statistic	Distribution	Reject $H_0$ if
$K>2$	${\chi }^2$	${\chi }_{K-1}^2$	${\chi }^2>{\chi }_{K-1,\alpha }^2$
$K=2$ (two-sided)	$Z$	$N(0,1)$	$\Vert Z\Vert >z_{\alpha /2}$
$K=2$ (one-sided)	$Z$	$N(0,1)$	$Z>z_{\alpha }$ or $Z<-z_{\alpha }$

7. Interpret Results

The interpretation depends on the nature of the event $T$:

Event Type	”Better” means	Survival Curve	Hazard
Time-to-relapse (death, recurrence)	Longer time	Higher $S(t)$	Lower $h(t)$
Time-to-recovery (healing, remission)	Shorter time	Lower $S(t)$	Higher $h(t)$

Interpretation

For time-to-relapse: the group with the higher survival curve has better prognosis. For time-to-recovery: the group with the lower survival curve recovers faster. Always check which direction is “better” before interpreting test results.

R Implementation

library(survival)
 
# K-sample log-rank test
fit <- survdiff(Surv(time, status) ~ group)
fit
# Output: N, Observed, Expected, (O-E)^2/E, (O-E)^2/V, Chisq, df, p-value

Example: Cancer Stages (4 Groups)

Case: Larynx cancer patients grouped by disease stage (I, II, III, IV). Question: Does survival differ across stages?

fit <- survdiff(Surv(time, status) ~ as.factor(stage), data = dat)
fit  # χ² with 3 df

Interpretation flow:

1. Omnibus test: survdiff gives overall ${\chi }^2$ with $df=K-1=3$. If significant → at least two stages differ.
2. Direction check: KM curves plotted by stage show Stage I (highest) → Stage IV (lowest), suggesting monotonic ordering.
3. Follow-up: If ordering exists, proceed to Trend Test to test whether hazard increases monotonically with stage.
4. Partial comparisons: Test specific pairs (e.g., Stage II vs I) to identify which stages actually differ.

Interpretation

The omnibus k-sample test tells whether groups differ. The trend test tells in what direction. Always do the omnibus test first, then follow up with a trend test if groups have a natural ordering.

Related

• Log-Rank Test
• Trend Test
• Stratified Test
• Hypothesis Testing Cheatsheet