Question

To determine whether patients are responsive to a particular chemotherapeutic 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 the 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. This is a paired data problem.

Summarize all results giving hypotheses, significance level, tests performed, statistics, and conclusions.

Treatment 1	Treatment 2
252	253
227	260
181	344
167	248
83	98
35	69
20	85
18	62
12	22
6	13
5	4
23	18
23	11
12	0
7	3
2	1

Answer 1

ANSWER::

We will do both non-parametric and paramteric test in R.

1st note that this a paired dataset, the values in tretament-1 are correlated to treatment-2 as both the measurements are taken from the same set of individuals.

Here is the to take 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-parametric test:

We will test whether there is any shift in location of the two treatments by wilcoxon sign rank test.

Code:

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

>

From the output we see that p-value = 0.5229 > specified significance.level = 0.05, so we fail to reject H0 to conclude that there is no sufficient evidence to suggest any difference in the effect of two treatments.

Paramteric Test:

We will conduct paired t-test in R .

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

>

As the p-value of the test = 0.037 which is less than 0.05, so we will reject H0 based on paramteric test to conclude that there is sufficient evidence to suggest that there is a siginifficant difference in the mean effect of the two treatments.

before moving on, let's see why the two conclusion based on paramteric and non-parametric test are not equal.

The main assumption of paire t-test is data follows normal. but we not tested that. we will do shapiro wilks test for normality

code:

diff<-trt1-trt2
shapiro.test(diff)

Output:

Shapiro-Wilk normality test

data: diff
W = 0.7711, p-value = 0.001153

>

As the p-value is much less than 0.05 ,so we have the sufficient evidence that the data is not from normal distribution. So paired t-test is not valid here, and the results and conclusion may be misleading. Whereas nonparametric results are more reliable.

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