To determine whether patients are responsive to a particular chemotherapceutic a

Question

To determine whether patients are responsive to a particular chemotherapceutic agent, colonies of cells are grown from each patient and then treated with the agent in question. To compare two agents, two samples are prepared simultaneously, from each patient, one sample treated with each drug, and the number of colonies in the culture counted. The data listed below show the results of such an experiment. On each line, the colony counts resulting from treatment 1 and treatment 2 are given for a particular patient. Do the two treatments have the same effect on colony formation? Use two techniques, one parametric and one nonparametric, to answer this question. (20) The data are stored on file cellgr.jmp on Blackboard. 252 253 227 260 181 344 167 248 83 98 35 69 20 85 18 62 12 22 613 54 23 18 23 11 120 73 2 1

Explanation / Answer

As each patient is given two treatments, so the data is correlated. So we have test of equality of mean or medians over the treatment in the paired dataset. We will do in R.

Code tp fetch the data in R.

trt1<-c(252,227,181,167,83,35,20,18,12,6,5,23,23,12,7,2)
trt2<-c(253,260,344,248,98,69,85,62,22,13,4,18,11,0,3,1)

Non-Paramteric Test:

As the data is paired, to test whether the median of the both the treatment is equal or not we will conduct wilcoxon sign rank test.

here is the code for it:

wilcox.test(trt1,trt2,
alternative ="two.sided",
mu = 0, paired = TRUE,correct=TRUE,exact=FALSE,
conf.int = FALSE, conf.level = 0.95)

Output:

Wilcoxon signed rank test with continuity correction

data: trt1 and trt2
V = 30, p-value = 0.05229
alternative hypothesis: true location shift is not equal to 0

You see in the code that conf.level is given ass 95%, so significance level is 5%, now we sse that the p-value of the test is 0.05229 whuch is greater than 0.05.

So at 5% level we can conclude that the effect of both the treatment are equal.

Parametric Test:

Code:

t.test(trt1,trt2,paired=TRUE,mu=0,conf.level=0.95)

Output:

Paired t-test

data: trt1 and trt2
t = -2.2869, df = 15, p-value = 0.03716
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-50.474634 -1.775366
sample estimates:
mean of the differences
-26.125

Now here p-value is less than 0.05 so we can conclude that there is a significant difference in means between the two treatments.

So we came to contradictory results based on paramteric and non parametric test.

But before jumping to the conclusion let's validate the assumption of the paired t-test in parametric assumption.

the 1st one is assumption of normality. As sample size is small, so we can't apply CLT, so the data must follow normal distribution for paired t-test.

for that validation we will do shapiro wilk test in R. where null hypothesis is normality.

code;

shapiro.test(trt1)
shapiro.test(trt2)

Output:

> shapiro.test(trt1)

Shapiro-Wilk normality test

data: trt1
W = 0.725, p-value = 0.0003224

> shapiro.test(trt2)

Shapiro-Wilk normality test

data: trt2
W = 0.7762, p-value = 0.001338

>

Look at the p-value of the two shapiro wilks test for the two dataset. It's too smaller than 0.05. So the data is not anyway normal. So results from paired t-test can be misleading, where as nonparametric results are more logical.

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