Chapter 8 Probability distributions
8.1 R as a set of statistical tables
One convenient use of R is to provide a comprehensive set of statistical tables. Functions are provided to evaluate the cumulative distribution function P(X <= x), the probability density function and the quantile function (given q, the smallest x such that P(X <= x) > q), and to simulate from the distribution.
Distribution R name additional arguments beta
shape1, shape2, ncp
df1, df2, ncp
m, n, k
Prefix the name given here by ‘d’ for the density, ‘p’ for the CDF, ‘q’ for the quantile function and ‘r’ for simulation (random deviates). The first argument is
rxxx (except for
rwilcox, for which it is
nn). In not quite all cases is the non-centrality parameter
ncp currently available: see the on-line help for details.
qxxx functions all have logical arguments
log.p and the
dxxx ones have
log. This allows, e.g., getting the cumulative (or “integrated”) hazard function, H(t) = - log(1 - F(t)), by
- pxxx(t, ..., lower.tail = FALSE, log.p = TRUE)
or more accurate log-likelihoods (by
dxxx(…, log = TRUE)), directly.
In addition there are functions
qtukey for the distribution of the studentized range of samples from a normal distribution, and
rmultinom for the multinomial distribution. Further distributions are available in contributed packages, notably SuppDists.
Here are some examples
> ## -tailed p-value for t distribution > 2*pt(-2.43, df = 13) > ## upper 1% point for an F(2, 7) distribution > qf(0.01, 2, 7, lower.tail = FALSE)
See the on-line help on
RNG for how random-number generation is done in R.
8.2 Examining the distribution of a set of data
Given a (univariate) set of data we can examine its distribution in a large number of ways. The simplest is to examine the numbers. Two slightly different summaries are given by
fivenum and a display of the numbers by
stem (a “stem and leaf” plot).
> attach(faithful) > summary(eruptions) Min. 1st Qu. Median Mean 3rd Qu. Max. 1.600 2.163 4.000 3.488 4.454 5.100 > fivenum(eruptions)  1.6000 2.1585 4.0000 4.4585 5.1000 > stem(eruptions) The decimal point is 1 digit(s) to the left of the | 16 | 070355555588 18 | 000022233333335577777777888822335777888 20 | 00002223378800035778 22 | 0002335578023578 24 | 00228 26 | 23 28 | 080 30 | 7 32 | 2337 34 | 250077 36 | 0000823577 38 | 2333335582225577 40 | 0000003357788888002233555577778 42 | 03335555778800233333555577778 44 | 02222335557780000000023333357778888 46 | 0000233357700000023578 48 | 00000022335800333 50 | 0370
> hist(eruptions) ## make the bins smaller, make a plot of density > hist(eruptions, seq(1.6, 5.2, 0.2), prob=TRUE) > lines(density(eruptions, bw=0.1)) > rug(eruptions) # show the actual data points
More elegant density plots can be made by
density, and we added a line produced by
density in this example. The bandwidth
bw was chosen by trial-and-error as the default gives too much smoothing (it usually does for “interesting” densities). (Better automated methods of bandwidth choice are available, and in this example
bw = “SJ” gives a good result.)
> plot(ecdf(eruptions), do.points=FALSE, verticals=TRUE)
This distribution is obviously far from any standard distribution. How about the right-hand mode, say eruptions of longer than 3 minutes? Let us fit a normal distribution and overlay the fitted CDF.
> long <- eruptions[eruptions > 3] > plot(ecdf(long), do.points=FALSE, verticals=TRUE) > x <- seq(3, 5.4, 0.01) > lines(x, pnorm(x, mean=mean(long), sd=sqrt(var(long))), lty=3)
par(pty="s") # arrange for a square figure region qqnorm(long); qqline(long)
which shows a reasonable fit but a shorter right tail than one would expect from a normal distribution. Let us compare this with some simulated data from a t distribution
x <- rt(250, df = 5) qqnorm(x); qqline(x)
which will usually (if it is a random sample) show longer tails than expected for a normal. We can make a Q-Q plot against the generating distribution by
qqplot(qt(ppoints(250), df = 5), x, xlab = "Q-Q plot for t dsn") qqline(x)
> shapiro.test(long) Shapiro-Wilk normality test data: long W = 0.9793, p-value = 0.01052
> ks.test(long, "pnorm", mean = mean(long), sd = sqrt(var(long))) One-sample Kolmogorov-Smirnov test data: long D = 0.0661, p-value = 0.4284 alternative hypothesis: two.sided
(Note that the distribution theory is not valid here as we have estimated the parameters of the normal distribution from the same sample.)
8.3 One- and two-sample tests
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.
> 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 (in contrast to the similar S-PLUS
t.test function). 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)