So far we have compared a single sample to a normal distribution. A much more common operation is to compare aspects of two samples. Note that in R, all "classical" tests including the ones used below are in package stats which is normally loaded.

Consider the following sets of data on the latent heat of the fusion of ice (cal/gm) from Rice (1995, p.490) 

Method A: 79.98 80.04 80.02 80.04 80.03 80.03 80.04 79.97
          80.05 80.03 80.02 80.00 80.02
Method B: 80.02 79.94 79.98 79.97 79.97 80.03 79.95 79.97

Boxplots provide a simple graphical comparison of the two samples. 

A <- scan()
79.98 80.04 80.02 80.04 80.03 80.03 80.04 79.97
80.05 80.03 80.02 80.00 80.02

B <- scan()
80.02 79.94 79.98 79.97 79.97 80.03 79.95 79.97

boxplot(A, B)

which indicates that the first group tends to give higher results than the second.


To test for the equality of the means of the two examples, we can use an unpaired t-test by 

> t.test(A, B)

         Welch Two Sample t-test

data:  A and B
t = 3.2499, df = 12.027, p-value = 0.00694
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 0.01385526 0.07018320
sample estimates:
mean of x mean of y
 80.02077  79.97875

which does indicate a significant difference, assuming normality. By default the R function does not assume equality of variances in the two samples. We can use the F test to test for equality in the variances, provided that the two samples are from normal populations. 

> var.test(A, B)

         F test to compare two variances

data:  A and B
F = 0.5837, num df = 12, denom df =  7, p-value = 0.3938
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.1251097 2.1052687
sample estimates:
ratio of variances
         0.5837405

which shows no evidence of a significant difference, and so we can use the classical t-test that assumes equality of the variances. 

> t.test(A, B, var.equal=TRUE)

         Two Sample t-test

data:  A and B
t = 3.4722, df = 19, p-value = 0.002551
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 0.01669058 0.06734788
sample estimates:
mean of x mean of y
 80.02077  79.97875

All these tests assume normality of the two samples. The two-sample Wilcoxon (or Mann-Whitney) test only assumes a common continuous distribution under the null hypothesis. 

> wilcox.test(A, B)

         Wilcoxon rank sum test with continuity correction

data:  A and B
W = 89, p-value = 0.007497
alternative hypothesis: true location shift is not equal to 0

Warning message:
Cannot compute exact p-value with ties in: wilcox.test(A, B)

Note the warning: there are several ties in each sample, which suggests strongly that these data are from a discrete distribution (probably due to rounding).

There are several ways to compare graphically the two samples. We have already seen a pair of boxplots. The following 

> plot(ecdf(A), do.points=FALSE, verticals=TRUE, xlim=range(A, B))
> plot(ecdf(B), do.points=FALSE, verticals=TRUE, add=TRUE)

will show the two empirical CDFs, and qqplot will perform a Q-Q plot of the two samples. The Kolmogorov-Smirnov test is of the maximal vertical distance between the two ecdf's, assuming a common continuous distribution: 

> ks.test(A, B)

         Two-sample Kolmogorov-Smirnov test

data:  A and B
D = 0.5962, p-value = 0.05919
alternative hypothesis: two-sided

Warning message:
cannot compute correct p-values with ties in: ks.test(A, B)

Source: R Core Team, https://cran.r-project.org/doc/manuals/r-release/R-intro.html#One_002d-and-two_002dsample-tests
Creative Commons License This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 License.

Last modified: Monday, 9 January 2023, 4:02 PM