---
title: "One sample sign tests"
output: rmarkdown::html_vignette
vignette: >
  %\VignetteIndexEntry{One sample sign tests}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---

```{r, include = FALSE}
knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>",
  message = FALSE,
  warning = FALSE
)
```

## Problem setup

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:

```{r}
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).

## Null hypothesis and test statistic

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:

```{r}
sort(incomes)
```

We can verify this with R as well:

```{r}
b <- sum(incomes > 50000)
b

n <- sum(incomes != 50000)
n
```

## Calculating p-values

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:

```{r}
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

```{r}
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

```{r}
cdf(X, b)
```

## Using the binom.test() function

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

```{r}
binom.test(3, n = 12, p = 0.5)
```

For $H_0 : m \le 3$

```{r}
binom.test(3, n = 12, p = 0.5, alternative = "greater")
```

For $H_0 : m \ge 3$

```{r}
binom.test(3, n = 12, p = 0.5, alternative = "less")
```

All of these results agree with our manual computations, which is reassuring.