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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