Cancer Survival Analysis

One-way ANOVA

---

# reading data into R
cancer.df <- read.csv(paste("cancer-survival.csv"))
# attaching data columns of the dataframe
attach(cancer.df)
# dimension of the dataframe
dim(cancer.df)

[1] 64  2

This data set tracks the number of days cancer patients survived after getting cancer treatment, as a function of the type of cancer they were sufferring from i.e. as a function of the body organ which had cancer.

# average survival time by organ
library(data.table)
dt <- data.table(cancer.df)
dt[, list(Count = .N,
        mean = round(mean(Survival), 3), 
        sd = round(mean(Survival), 3),
        median = round(median(Survival), 3),
        min = min(Survival),
        max = max(Survival)), 
   by = list(Organ)]

      Organ Count     mean       sd median min  max
1:  Stomach    13  286.000  286.000    124  25 1112
2: Bronchus    17  211.588  211.588    155  20  859
3:    Colon    17  457.412  457.412    372  20 1843
4:    Ovary     6  884.333  884.333    406  89 2970
5:   Breast    11 1395.909 1395.909   1166  24 3808

# descriptive statistics by each organ
library(psych)
describeBy(Survival, Organ)

 Descriptive statistics by group 
group: Breast
   vars  n    mean      sd median trimmed    mad min  max range skew
X1    1 11 1395.91 1238.97   1166 1280.33 662.72  24 3808  3784 0.81
   kurtosis     se
X1     -0.7 373.56
-------------------------------------------------------- 
group: Bronchus
   vars  n   mean     sd median trimmed    mad min max range skew kurtosis
X1    1 17 211.59 209.86    155   181.2 133.43  20 859   839 1.75     2.66
     se
X1 50.9
-------------------------------------------------------- 
group: Colon
   vars  n   mean     sd median trimmed    mad min  max range skew
X1    1 17 457.41 427.17    372   394.2 244.63  20 1843  1823 1.96
   kurtosis    se
X1     3.76 103.6
-------------------------------------------------------- 
group: Ovary
   vars n   mean      sd median trimmed    mad min  max range skew
X1    1 6 884.33 1098.58    406  884.33 386.96  89 2970  2881 1.01
   kurtosis     se
X1    -0.75 448.49
-------------------------------------------------------- 
group: Stomach
   vars  n mean     sd median trimmed    mad min  max range skew kurtosis
X1    1 13  286 346.31    124  234.64 121.57  25 1112  1087 1.27     0.25
      se
X1 96.05

# number of observations in organ
addmargins(table(Organ))

Organ
  Breast Bronchus    Colon    Ovary  Stomach      Sum 
      11       17       17        6       13       64 

# box plot of organ
boxplot(Survival ~ Organ, data = cancer.df,
        main = "Boxplot of Organ",
        xlab = "Organ", ylab = "Survival (Days)")

# mean plot by organ
library(gplots)
plotmeans(Survival ~ Organ, data = cancer.df,
          xlab = "Organ", ylab = "Survival (Days)",
          digits=2, col = "black", ccol = "blue", barwidth = 2,
          legends = TRUE, mean.labels = TRUE, frame = TRUE)

# check for normality in each group
with(cancer.df, tapply(Survival, Organ, shapiro.test))

$Breast

    Shapiro-Wilk normality test

data:  X[[i]]
W = 0.86857, p-value = 0.07431


$Bronchus

    Shapiro-Wilk normality test

data:  X[[i]]
W = 0.76596, p-value = 0.0007186


$Colon

    Shapiro-Wilk normality test

data:  X[[i]]
W = 0.76056, p-value = 0.0006134


$Ovary

    Shapiro-Wilk normality test

data:  X[[i]]
W = 0.76688, p-value = 0.029


$Stomach

    Shapiro-Wilk normality test

data:  X[[i]]
W = 0.75473, p-value = 0.002075

Null hypothesis (\( H_0 \)): The data is normally distributed.

• we reject the null hypothesis for Breast

• We fail to reject the null hypothesis for other organs like Bronchus, Colon, Ovary, Stomach.

In other words, we can not assume the normality in each group.

# Check for homogeneity of variance
library(car)
leveneTest(Survival ~ Organ, data = cancer.df)

Levene's Test for Homogeneity of Variance (center = median)
      Df F value   Pr(>F)   
group  4  4.4524 0.003271 **
      59                    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Null hypothesis (\( H_0 \)): The variance of cancer Survival is homogenous across diffrent types of cancers (Organ).

We reject the null hypothesis. There is heterogeneity in variance of cancer survival.

We use log transformation to the dependent variable (Survival) and again check for the normality and homogenity of the variance.

# change Survival to log of Survival
cancer.df$LogSurvival <- log(cancer.df$Survival)
# first few rows of the dataframe
head(cancer.df)

  Survival   Organ LogSurvival
1      124 Stomach    4.820282
2       42 Stomach    3.737670
3       25 Stomach    3.218876
4       45 Stomach    3.806662
5      412 Stomach    6.021023
6       51 Stomach    3.931826

# check for normality in each group after log transformation
with(cancer.df, tapply(LogSurvival, Organ, shapiro.test))

$Breast

    Shapiro-Wilk normality test

data:  X[[i]]
W = 0.802, p-value = 0.009995


$Bronchus

    Shapiro-Wilk normality test

data:  X[[i]]
W = 0.98047, p-value = 0.9613


$Colon

    Shapiro-Wilk normality test

data:  X[[i]]
W = 0.92636, p-value = 0.1891


$Ovary

    Shapiro-Wilk normality test

data:  X[[i]]
W = 0.983, p-value = 0.9655


$Stomach

    Shapiro-Wilk normality test

data:  X[[i]]
W = 0.92837, p-value = 0.3245

# Check for homogeneity of variance after log transformation
library(car)
leveneTest(LogSurvival ~ Organ, data = cancer.df)

Levene's Test for Homogeneity of Variance (center = median)
      Df F value Pr(>F)
group  4  0.6685 0.6164
      59               

Now we can see that after log transformation, problem of normality and equality of variance has been resolved, as suggested by test output.

Hence, we have normality in each group as well as homogeneity in variance in each group.

# mean plot by organ
library(gplots)
plotmeans(Survival ~ Organ, data = cancer.df,
          xlab = "Organ", ylab = "Survival (Days)",
          digits=2, col = "black", ccol = "blue", barwidth = 2,
          legends = TRUE, mean.labels = TRUE, frame = TRUE)

# one-way ANOVA 
oneWayfit <- aov(Survival ~ Organ, data = cancer.df)
# summary of the ANOVA model
summary(oneWayfit)

            Df   Sum Sq Mean Sq F value   Pr(>F)    
Organ        4 11535761 2883940   6.433 0.000229 ***
Residuals   59 26448144  448274                     
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

# one-way ANOVA 
oneWayTransfit <- aov(LogSurvival ~ Organ, data = cancer.df)
# summary of the ANOVA model
summary(oneWayTransfit)

            Df Sum Sq Mean Sq F value  Pr(>F)   
Organ        4  24.49   6.122   4.286 0.00412 **
Residuals   59  84.27   1.428                   
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

As the p-value < 0.05, we can conclude that there are significant differences in survival rates between the cancer organ groups.

In one-way ANOVA test, a significant p-value indicates that some of the group means are different, but we don't know which pairs of groups are different.

It's possible to perform multiple pairwise-comparison, to determine if the mean difference between specific pairs of group are statistically significant.

Hence, we use Tukey Honestly Significant Differences (HSD) test to check pairwise-comparison.

# Tukey comparison test
TukeyHSD(oneWayTransfit)

  Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = LogSurvival ~ Organ, data = cancer.df)

$Organ
                        diff       lwr        upr     p adj
Bronchus-Breast  -1.60543320 -2.906741 -0.3041254 0.0083352
Colon-Breast     -0.80948110 -2.110789  0.4918267 0.4119156
Ovary-Breast     -0.40798703 -2.114754  1.2987803 0.9615409
Stomach-Breast   -1.59068365 -2.968399 -0.2129685 0.0158132
Colon-Bronchus    0.79595210 -0.357534  1.9494382 0.3072938
Ovary-Bronchus    1.19744617 -0.399483  2.7943753 0.2296079
Stomach-Bronchus  0.01474955 -1.224293  1.2537924 0.9999997
Ovary-Colon       0.40149407 -1.195435  1.9984232 0.9540004
Stomach-Colon    -0.78120255 -2.020245  0.4578403 0.3981146
Stomach-Ovary    -1.18269662 -2.842480  0.4770864 0.2763506

# Tukey pair-wise comparisons plot
plot(TukeyHSD(oneWayfit))

The classical one-way ANOVA test requires an assumption of equal variances for all groups. In our example, the homogeneity of variance assumption turned out to be fine: the Levene test is not significant.

But, if homogeneity of variance is violated?

This test is used when variances are heterogeneous.

# anova test when variances are not same
oneway.test(Survival ~ Organ, data = cancer.df)

    One-way analysis of means (not assuming equal variances)

data:  Survival and Organ
F = 3.5152, num df = 4.000, denom df = 19.862, p-value = 0.02514

pairwise.t.test(Survival, Organ, data = cancer.df,
                p.adjust.method = "BH", pool.sd = FALSE)

    Pairwise comparisons using t tests with non-pooled SD 

data:  Survival and Organ 

         Breast Bronchus Colon Ovary
Bronchus 0.073  -        -     -    
Colon    0.110  0.110    -     -    
Ovary    0.443  0.349    0.443 -    
Stomach  0.073  0.502    0.349 0.349

P value adjustment method: BH 

This is a non-parametric test. This test can be used when the normality assumption is violated.

# Kruskal-Wallis rank sum test
kruskal.test(Survival ~ Organ, data = cancer.df)

    Kruskal-Wallis rank sum test

data:  Survival by Organ
Kruskal-Wallis chi-squared = 14.954, df = 4, p-value = 0.004798

ANOVA Assumptions Before-After Transformation