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)
To test for the equality of the means of the two examples, we can useanunpairedt-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
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 usethe classicalt-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, andqqplotwill perform a Q-Qplot of the two samples. The Kolmogorov-Smirnov test is of the maximalvertical distance between the two ecdf’s, assuming a common continuousdistribution:
> 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)
