knitr::opts_chunk$set( collapse = TRUE, comment = "#>", message = FALSE, warning = FALSE )
Sometimes you want to do a Z-test or a T-test, but for some reason these tests are not appropriate. Your data may be skewed, or from a distribution with outliers, or non-normal in some other important way. In these circumstances a sign test is appropriate.
For example, suppose you wander around Times Square and ask strangers for their salaries. Incomes are typically very skewed, and you might get a sample like:
[ 8478, 21564, 36562, 176602, 9395, 18320, 50000, 2, 40298, 39, 10780, 2268583, 3404930 ]
If we look at a QQ plot, we see there are massive outliers:
incomes <- c(8478, 21564, 36562, 176602, 9395, 18320, 50000, 2, 40298, 39, 10780, 2268583, 3404930) qqnorm(incomes) qqline(incomes)
Luckily, the sign test only requires independent samples for valid inference (as a consequence, it has been low power).
The sign test allows us to test whether the median of a distribution equals some hypothesized value. Let's test whether our data is consistent with median of 50,000, which is close-ish to the median income in the U.S. if memory serves. That is
[ H_0: m = 50,000 \qquad H_A: \mu \neq 50,000 ]
where $m$ stands for the population median. The test statistic is then
[ B = \sum_{i=1}^n 1_{(50, 000, \infty)} (x_i) \sim \mathrm{Binomial}(N, 0.5) ]
Here $B$ is the number of data points observed that are strictly greater than the median, and $N$ is sample size after exact ties with the median have been removed. Forgetting to remove exact ties is a very frequent mistake when students do this test in classes I TA.
If we sort the data we can see that $B = 3$ and $N = 12$ in our case:
sort(incomes)
We can verify this with R as well:
b <- sum(incomes > 50000) b n <- sum(incomes != 50000) n
To calculate a two-sided p-value, we need to find
[ \begin{align} 2 \cdot \min(P(B \ge 3), P(B \le 3)) = 2 \cdot \min(1 - P(B \le 2), P(B \le 3)) \end{align} ]
To do this we need to c.d.f. of a binomial random variable:
library(distributions3) X <- Binomial(n, 0.5) 2 * min(cdf(X, b), 1 - cdf(X, b - 1))
In practice computing the c.d.f. of binomial random variables is rather tedious and there aren't great shortcuts for small samples. If you got a question like this on an exam, you'd want to use the binomial p.m.f. repeatedly, like this:
[ \begin{align} P(B \le 3) &= P(B = 0) + P(B = 1) + P(B = 2) + P(B = 3) \ &= \binom{12}{0} 0.5^0 0.5^12 + \binom{12}{1} 0.5^1 0.5^11 + \binom{12}{2} 0.5^2 0.5^10 + \binom{12}{3} 0.5^3 0.5^9 \end{align} ]
Finally, sometimes we are interest in one sided sign tests. For the test
[ \begin{align} H_0: m \le 3 \qquad H_A: m > 3 \end{align} ]
the p-value is given by
[ P(B > 3) = 1 - P(B \le 2) ]
which we calculate with
1 - cdf(X, b - 1)
For the test
[ H_0: m \ge 3 \qquad H_A: m < 3 ]
the p-value is given by
[ P(B < 3) ]
which we calculate with
cdf(X, b)
To verify results we can use the binom.test()
from base R. The x
argument gets the value of $B$, n
the value of $N$, and p = 0.5
for a test of the median.
That is, for $H_0 : m = 3$ we would use
binom.test(3, n = 12, p = 0.5)
For $H_0 : m \le 3$
binom.test(3, n = 12, p = 0.5, alternative = "greater")
For $H_0 : m \ge 3$
binom.test(3, n = 12, p = 0.5, alternative = "less")
All of these results agree with our manual computations, which is reassuring.
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