Lab 9: nonparametric inference

Course activity

STAT218

The objective of this lab is to learn how to implement nonparamteric alternatives to $t$ and $F$ tests:

sign test
signed rank test
rank sum test
Kruskal-Wallis test

Examples utilize data from lecture and past assignments.

load('data/census.RData')
load('data/temps.RData')

One-sample inference

Sign test

The sign test is a nonparametric inference procedure for inference on a population median $m$ . Suppose we wish to test:

$H_{0} : m = m_{0} H_{A} : m \neq m_{0}$

If the median is in fact $m_{0}$ , then we’d expect this value to evenly divide the observations. We can determine the sign of each observation as follows:

$sign (x_{i}) = {\begin{cases} + if x_{i} \geq m_{0} \\ - if x_{i} < m_{0} \end{cases}$ We then tabulate counts of the negative signs:

$X = # {x_{i} < m_{0}}$ And test whether the proportion of negative signs is 0.5. The test can be directional or two-sided.

Here’s an example of testing whether median DDT in kale samples is 3:

# ddt data
ddt <- MASS::DDT

# how many observations are less than 3?
ddt.x <- sum(ddt < 3)
ddt.n <- length(ddt)

# sign test
binom.test(x = ddt.x, n = ddt.n, p = 0.5, alternative = 'two.sided')


    Exact binomial test

data:  ddt.x and ddt.n
number of successes = 2, number of trials = 15, p-value = 0.007385
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
 0.01657591 0.40460270
sample estimates:
probability of success 
             0.1333333

The $p$ -value is used to determine the result of the test; in this class, we will ignore the confidence interval and other output.

The data provide evidence that the median DDT in kale is not 3ppm (sign test, *p = 0.0074).

Your turn

Using the census data, test whether median total personal income is 25K using the sign test. Interpret the result in context.

Extra: Which alternative should you use in binom.test to test whether median income is less than 25K?

Solution

# incomes from census data
incomes <- census$total_personal_income

# how many observations are less than 25K?
incomes.x <- sum(incomes < 25000)
incomes.n <- length(incomes)

# sign test
binom.test(x = incomes.x, n = incomes.n, p = 0.5, alternative = 'two.sided')


    Exact binomial test

data:  incomes.x and incomes.n
number of successes = 228, number of trials = 377, p-value = 5.557e-05
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
 0.5534322 0.6544581
sample estimates:
probability of success 
             0.6047745

The data provide evidence that median income is not 25K.

Signed rank test

The signed rank test provides inference on the “center” of a population distribution, assuming the distribution is symmetric. For example, a two-sided test would pertain to the hypothesis and alternative:

$H_{0} : c = c_{0} H_{A} : c \neq c_{0}$

Under the symmetry assumption, one can interpret the center as a median.

The implementation in R is almost identical to the $t$ test; instead, simply use wilcox.test(...). For example, to test whether median male body temperature is 98.7 degrees Farenheit:

# male body temperatures
temps.split <- split(temps$body.temp, temps$sex)
temps.m <- temps.split$male

# signed rank test
wilcox.test(temps.m, mu = 98.7, alternative = 'two.sided')


    Wilcoxon signed rank test with continuity correction

data:  temps.m
V = 39.5, p-value = 0.01505
alternative hypothesis: true location is not equal to 98.7

We’d interpret the result as follows:

The data provide evidence that median male body temperature is not 98.7 degrees Farenheit (signed rank test, p = 0.0151).

Your turn

Test whether median female body temperature is 98.7 degrees Farenheit and interpret the result in context.

Solution

# female body temperatures
temps.f <- temps.split$female

# signed rank test
wilcox.test(temps.f, mu = 98.7, alternative = 'two.sided')


    Wilcoxon signed rank test with continuity correction

data:  temps.f
V = 79, p-value = 0.7935
alternative hypothesis: true location is not equal to 98.7

The data do not provide evidence that median female body temperature differs from 98.7 degrees Farenheit.

It is important to remember that this method relies on assuming the underlying data distribution is symmetric.

Two-sample inference

Consider now comparing the centers of two groups. For this we can use the rank sum test, which for a two-sided alternative would assess the hypotheses:

$H_{0} : c_{1} = c_{2} H_{A} : c_{1} \neq c_{2}$

This is the analogue of the two-sample $t$ test for means, and similarly can have directional one-sided alternatives.

The implementation in R is straightforward. To test whether male and female body temperatures differ:

# rank sum test
wilcox.test(body.temp ~ sex, data = temps)


    Wilcoxon rank sum test with continuity correction

data:  body.temp by sex
W = 249.5, p-value = 0.09682
alternative hypothesis: true location shift is not equal to 0

We’d interpret the result as follows:

The data do not provide evidence that body temperatures differ by sex (rank sum test, p = 0.0968).

Your turn

Using the census data, test whether incomes are higher among men than among women. Interpret your result in context.

Solution

# rank sum test
wilcox.test(total_personal_income ~ sex, data = census, alternative = 'less')


    Wilcoxon rank sum test with continuity correction

data:  total_personal_income by sex
W = 11889, p-value = 1.363e-08
alternative hypothesis: true location shift is less than 0

The data provide evidence that incomes difer by sex (rank sum test, p < 0.0001).

By including conf.int and conf.level arguments, we can obtain an estimate and confidence interval for the magnitude of the location shift:

# rank sum test
wilcox.test(body.temp ~ sex, data = temps, conf.int = T, conf.level = 0.95)


    Wilcoxon rank sum test with continuity correction

data:  body.temp by sex
W = 249.5, p-value = 0.09682
alternative hypothesis: true location shift is not equal to 0
95 percent confidence interval:
 -0.1999996  1.1000358
sample estimates:
difference in location 
             0.5000276

This is interpreted as follows:

Body temperatures are estimated to be 0.5 degrees Farenheit higher among women.

As discussed in lecture, the rank sum test is designed to detect differences in location – meaning that the alternative, though written in terms of centers, is really that the values from one group tend to be uniformly larger/smaller than those from the other group. It’s important to keep this in mind.