---
title: "The Chi-Square test"
output: html_document
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```

The chi-square test is a method for assessing a model when the data are binned.
It is often use to test if a sample is representative of a given population.

## Basic example

Here's a potentially important question: Is a given pool of potential jurors in a county racially representative of that county?

### Specific data

Here's some specific data representing 275 jurors in a small county. Jurors identified their racial group, as shown in the table below. We would like to determine if these jurors are racially representative of the population. 

| Race                           |     | White | Black | Hispanic | Other |     | Total |
|:-------------------------------|:----|:-----:|:-----:|:--------:|:-----:|:----|:------|
| Representation in juries  |     |  205  |   26  |    25    |   19  |     | 275   |
| Percentages for registered voters |     |  0.72 |  0.07 |   0.12   |  0.09 |     | 1.00  |
| Expected count            |     |  198 | 19.25 | 33 | 24.75 | | 275

### Using `chisq.test`

R's `chisq.test` is built for exactly this situation and it's pretty easy to use:

```{r}
chisq.test(c(205, 26, 25, 19), p=c(0.72,0.07,0.12,0.09))
```

There's a lot going on in the background here but, ultimately, we are interested in that $p$-value. If we are looking for a 95% confidence level, then we are unable to reject the null hypothesis here, in spite of the deviation from expected counts that we see in the data.

### Some formulae

The $p$-value is computed using the $\chi^2$ statistic, which we find as follows: 

We suppose that we are to evaluate whether there is convincing evidence that a set of observed counts $O_1$, $O_2$, ..., $O_k$ in $k$ categories are unusually different from what might be expected under a null hypothesis. Call the \emph{expected counts} that are based on the null hypothesis $E_1$, $E_2$, ..., $E_k$. If each expected count is at least 5 and the null hypothesis is true, then the test statistic below follows a chi-square distribution with $k-1$ degrees of freedom:
$$
\chi^2 = \frac{(O_1 - E_1)^2}{E_1} + \frac{(O_2 - E_2)^2}{E_2} + \cdots + \frac{(O_k - E_k)^2}{E_k}
$$
We then evaluate the area under the tail of the $\chi^2$-distribution with $k-1$ degrees of freedom. In the example above, the $\chi^2$-statistic is

```{r}
ch_sq = ((205-198)^2/198 + (26-19.25)^2/19.25 +(25-33)^2/33 + (19-24.75)^2/24.75)
ch_sq
```
and the probability can be computed via

```{r}
1-pchisq(ch_sq,3)
```

Geometrically, this represents the area under the curve below and to the right of 5.88

```{r, echo=F}
f = function(x) dchisq(x,3)
plot(f,0,10, yaxs='i', xlab="", ylab="", lwd=3, yaxt='n', ylim=c(0,0.25))
x <- seq(5.88,10, by=0.01)
y <- f(x)
polygon(c(x,rev(x)), c(rep(0, length(y)), rev(y)), col='lightgray')
```

Note that we can use a table like the one in the back of our text to assess the null-hypothesis.

## A two-way example

The previous example was one-way. That is, the percentages were known and we just wanted to know if the jury pool matched those percentages. Sometimes, we have two categorical variables and we want to know if they are independent or not. This is also called the Chi-Square *test for homogeneity*. One example from [the R-Tutorial](http://www.r-tutor.com/elementary-statistics/goodness-fit/chi-squared-test-independence) examines whether exercise and smoking are independent of one another using the following data:

| Smokes/Exercises    |     | Frequently | Some | None  |
|:-------------------------------|:----|:-----:|:-----:|:--------|
| Never   |     |  435 | 420 | 90    |
| Occassionally |     |  60 | 20 | 15 |
| Regularly            |     |  45 | 35 |  5 |
| Heavily |   | 35 | 15 |  5 |

The rows indicate how much the participant smokes and the columns indicate how much they exercise. Our null hypothesis is that these are independent; our alternative hypothesis is contrary.

We can enter a small table like this into R as follows:

```{r}
m = matrix(c(435,420,90, 60,20,15, 45,35,5, 35,15,5), 4,3, byrow=T)
tbl = as.table(m)
tbl
```

Note that the row and column labels are different but that's not important for the computation, which we perform like so:

```{r}
chisq.test(tbl)
```

It looks like we reject the null hypothesis of independence. 

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