Quection

Using the AML data set, calculate the KM estimate for survival and draw the KM curves for both groups on the same graph. Compare the two curves using Log Rank Test.

1). Introduction

Below we can see AML dataset. It contains a clinical trial in patients with Acute Myelogenous Leukemia (AML) comparing two groups of patients having maintenance chemotherapy vs no maintenance chemotherapy.

2). Data

library(readxl)
library(knitr)

aml <- read_excel("AML.xlsx")
kable(aml)
Time Status Group
9 1 Maintained
13 1 Maintained
13 0 Maintained
18 1 Maintained
23 1 Maintained
28 0 Maintained
31 1 Maintained
34 1 Maintained
45 0 Maintained
48 1 Maintained
161 0 Maintained
5 1 NonMaintained
5 1 NonMaintained
8 1 NonMaintained
8 1 NonMaintained
12 1 NonMaintained
16 0 NonMaintained
23 1 NonMaintained
27 1 NonMaintained
30 0 NonMaintained
33 1 NonMaintained
43 1 NonMaintained
45 1 NonMaintained 
attach(aml)

In these data sets, Status = 1 means event occur(die) and Status = 0 means lost to follow(censored) data.

library(survival)
## 
## Attaching package: 'survival'
## The following object is masked _by_ '.GlobalEnv':
## 
##     aml
aml.surv <- Surv(time = Time, event = Status)
aml.surv
##  [1]   9   13   13+  18   23   28+  31   34   45+  48  161+   5    5    8    8 
## [16]  12   16+  23   27   30+  33   43   45

3). Life table

aml.fit <- survfit(Surv(time = Time, event = Status) ~ Group, data = aml)
summary(aml.fit)
## Call: survfit(formula = Surv(time = Time, event = Status) ~ Group, 
##     data = aml)
## 
##                 Group=Maintained 
##  time n.risk n.event survival std.err lower 95% CI upper 95% CI
##     9     11       1    0.909  0.0867       0.7541        1.000
##    13     10       1    0.818  0.1163       0.6192        1.000
##    18      8       1    0.716  0.1397       0.4884        1.000
##    23      7       1    0.614  0.1526       0.3769        0.999
##    31      5       1    0.491  0.1642       0.2549        0.946
##    34      4       1    0.368  0.1627       0.1549        0.875
##    48      2       1    0.184  0.1535       0.0359        0.944
## 
##                 Group=NonMaintained 
##  time n.risk n.event survival std.err lower 95% CI upper 95% CI
##     5     12       2    0.833   0.108       0.6470        1.000
##     8     10       2    0.667   0.136       0.4468        0.995
##    12      8       1    0.583   0.142       0.3616        0.941
##    23      6       1    0.486   0.148       0.2675        0.883
##    27      5       1    0.389   0.147       0.1854        0.816
##    33      3       1    0.259   0.144       0.0871        0.771
##    43      2       1    0.130   0.117       0.0222        0.756
##    45      1       1    0.000     NaN           NA           NA

4). KM Curves

plot(aml.fit, mark.time = T, 
     col = c("red","blue"), lwd = 1:1, lty = 1:1, cex = 2,main="KM Curves",
     xlab = "Time(t in weeks)", ylab = "Survival Probability = S(t)")
legend("topright", legend = c("Maintained Chemotherapy", "NonMaintained Chemotherapy"),
       fill = c("red","blue"), lty = 1:1)

aml.fit
## Call: survfit(formula = Surv(time = Time, event = Status) ~ Group, 
##     data = aml)
## 
##                      n events median 0.95LCL 0.95UCL
## Group=Maintained    11      7     31      18      NA
## Group=NonMaintained 12     10     23       8      NA

5). Log Rank Test (Comparing Groups)

To compare survival between gropus we can use the log rank test. Hypotheis,

\(H_0\) = The two survival curves are identicall (i.e. \(S_{1t} = S_{2t}\)) vs \(H_1\) = The two survival curves are not identicall (i.e. \(S_{1t} \neq S_{2t}\))

survdiff(Surv(time = Time, event = Status) ~ Group, data = aml)
## Call:
## survdiff(formula = Surv(time = Time, event = Status) ~ Group, 
##     data = aml)
## 
##                      N Observed Expected (O-E)^2/E (O-E)^2/V
## Group=Maintained    11        7    10.13     0.969      2.61
## Group=NonMaintained 12       10     6.87     1.430      2.61
## 
##  Chisq= 2.6  on 1 degrees of freedom, p= 0.1

Is the \(\alpha = 0.05\) since \(p-value > \alpha\) we failed to reject \(H_{0}\)(null hypothesis).

In fact calculated Chi-Squre value is 2.6. And table value of Chi-Squre \(\alpha=0.05\) df \((k-1) = 2-1 = 1\) where \(k\) is number of groups. So table value is 3.34.

Therefore calculated value is less than critical value. Then we failed to reject \(H_0\).

That means we have enough evidence to tell that the two Survival curves are identical for 5% significance level.