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)
#> [1] 2 39 8478 9395 10780 18320 21564 36562
#> [9] 40298 50000 176602 2268583 3404930
```

We can verify this with R as well:

To calculate a two-sided p-value, we need to find

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))
#> [1] 0.1459961
```

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:

Finally, sometimes we are interest in one sided sign tests. For the test

the p-value is given by

\[ P(B > 3) = 1 - P(B \le 2) \]

which we calculate with

```
1 - cdf(X, b - 1)
#> [1] 0.9807129
```

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)
#> [1] 0.07299805
```

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)
#>
#> Exact binomial test
#>
#> data: 3 and 12
#> number of successes = 3, number of trials = 12, p-value = 0.146
#> alternative hypothesis: true probability of success is not equal to 0.5
#> 95 percent confidence interval:
#> 0.05486064 0.57185846
#> sample estimates:
#> probability of success
#> 0.25
```

For \(H_0 : m \le 3\)

```
binom.test(3, n = 12, p = 0.5, alternative = "greater")
#>
#> Exact binomial test
#>
#> data: 3 and 12
#> number of successes = 3, number of trials = 12, p-value = 0.9807
#> alternative hypothesis: true probability of success is greater than 0.5
#> 95 percent confidence interval:
#> 0.07187026 1.00000000
#> sample estimates:
#> probability of success
#> 0.25
```

For \(H_0 : m \ge 3\)

```
binom.test(3, n = 12, p = 0.5, alternative = "less")
#>
#> Exact binomial test
#>
#> data: 3 and 12
#> number of successes = 3, number of trials = 12, p-value = 0.073
#> alternative hypothesis: true probability of success is less than 0.5
#> 95 percent confidence interval:
#> 0.0000000 0.5273266
#> sample estimates:
#> probability of success
#> 0.25
```

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