Lecture 11

class: center, middle, inverse, title-slide

# Lecture 11
### JMG
### MATH 204

---

# Inference for Categorical Data

In this lecture, we will examine two new inferential techniques:

- Testing for goodness of fit using chi-square, which is applied to a categorical variable with more than two levels. This is commonly used in two circumstances:

- Given a sample of cases that can be classified into several groups, determine if the sample is representative of the general population. 
  
--

- Evaluate whether data resemble a particular distribution, such as a normal distribution or a geometric distribution. 
  
--

- Testing for independence in two-way tables.

---

# Learning Objectives

- In this lecture, we cover some inferential techniques for categorical data. After this lecture you should be able to

- Identify one-way and two-way table problems.
  
--

- Work with a chi-square statistic and distribution.
  
--

- Use the `chisq.test` function in R to conduct hypothesis tests. 
---

# Video on Goodness of Fit

---

# Video on Two-Way Tables

---

# Motivating Example

Consider the following data:

<table class="table" style="margin-left: auto; margin-right: auto;">
 <thead>
  <tr>
   <th style="text-align:left;"> race </th>
   <th style="text-align:right;"> black </th>
   <th style="text-align:right;"> hispanic </th>
   <th style="text-align:right;"> white </th>
   <th style="text-align:right;"> other </th>
   <th style="text-align:right;"> total </th>
  </tr>
 </thead>
<tbody>
  <tr>
   <td style="text-align:left;"> Representation in juries </td>
   <td style="text-align:right;"> 26.00 </td>
   <td style="text-align:right;"> 25.00 </td>
   <td style="text-align:right;"> 205.00 </td>
   <td style="text-align:right;"> 19.00 </td>
   <td style="text-align:right;"> 275 </td>
  </tr>
  <tr>
   <td style="text-align:left;"> Registered voters </td>
   <td style="text-align:right;"> 0.07 </td>
   <td style="text-align:right;"> 0.12 </td>
   <td style="text-align:right;"> 0.72 </td>
   <td style="text-align:right;"> 0.09 </td>
   <td style="text-align:right;"> 1 </td>
  </tr>
</tbody>
</table>

We can compute the proportions for the data:

```
## # A tibble: 4 x 3
##   race         n      p
##   <fct>    <int>  <dbl>
## 1 black       26 0.0945
## 2 hispanic    25 0.0909
## 3 other       19 0.0691
## 4 white      205 0.745
```

- We would like to know if the jury is representative of the population.

- This problem illustrates "Given a sample of cases that can be classified into several groups, determine if the sample is representative of the general population."

---

# One-Way Tables

- If we were to take the bottom row of the table on the last slide as the assumed true proportions, then we would expect to get the following so-called **one-way table**:

<table class="table" style="margin-left: auto; margin-right: auto;">
 <thead>
  <tr>
   <th style="text-align:left;"> race </th>
   <th style="text-align:right;"> black </th>
   <th style="text-align:right;"> hispanic </th>
   <th style="text-align:right;"> white </th>
   <th style="text-align:right;"> other </th>
   <th style="text-align:right;"> total </th>
  </tr>
 </thead>
<tbody>
  <tr>
   <td style="text-align:left;"> Observed count </td>
   <td style="text-align:right;"> 26.00 </td>
   <td style="text-align:right;"> 25 </td>
   <td style="text-align:right;"> 205 </td>
   <td style="text-align:right;"> 19.00 </td>
   <td style="text-align:right;"> 275 </td>
  </tr>
  <tr>
   <td style="text-align:left;"> Expected count </td>
   <td style="text-align:right;"> 19.25 </td>
   <td style="text-align:right;"> 33 </td>
   <td style="text-align:right;"> 198 </td>
   <td style="text-align:right;"> 24.75 </td>
   <td style="text-align:right;"> 275 </td>
  </tr>
</tbody>
</table>

- From a one-way table we can produce a test statistic that follows a well-known distribution. Specifically, we compute

`$$\frac{(26-19.25)^2}{19.25} + \frac{(25-33)^2}{33} + \frac{(205-198)^2}{198} + \frac{(19-24.75)^2}{24.75}$$`

- This value is

```r
(26-19.25)^2/19.25 + (25-33)^2/33 + (205-198)^2/198 + (19-24.75)^2/24.75
```

```
## [1] 5.88961
```

---

# chi-Square Distributions

- In order to conduct inferences on data corresponding to one-way tables, we need to use a new distribution function called a **chi-square distribution**. Such distributions are characterized by a parameter called degrees of freedom. Below we plot a chi-square density function with 3 degrees of freedom:

```r
gf_dist("chisq",df=3)
```

![](lect_11_files/figure-html/unnamed-chunk-7-1.png)

---

# chi-Square for Different df's

![](lect_11_files/figure-html/unnamed-chunk-8-1.png)

---

# chi-Square Test Conditions

- In order to use a chi-square distribution to compute a p-value, we need to check two conditions:

- Independence. Each case that contributes a count to the table must be independent of all the other cases in the table. 
  
--

- Sample size / distribution. Each particular scenario must have at least 5 expected cases.

---

# chi-Square Test for One-Way Table

Suppose we are to evaluate whether there is convincing evidence that a set of observed counts `$O_{1}$`, `$O_{2}$`, `$\ldots$`, `$O_{k}$` in `$k$` categories are unusually different from what we might expect under a null hypothesis. Denote the *expected counts* that are based on the null hypothesis by `$E_{1}$`, `$E_{2}$`, `$\ldots$`, `$E_{k}$`. If each expected count is at least 5 and the null hypothesis is true, then the test statistic

`$$X^{2} = \frac{(O_{1}-E_{1})^2}{E_{1}} + \frac{(O_{2}-E_{2})^2}{E_{2}} + \cdots + \frac{(O_{k}-E_{k})^2}{E_{k}}$$`
follows a chi-square distribution with `$k-1$` degrees of freedom. Note that this test statistic is always positive.

The p-value for this test statistic is found by looking at the upper tail of this chi-square distribution. We consider the upper tail because larger values of `$X^2$` would provide greater evidence against the null hypothesis.

---

# Example chi-Square Test

- We expect the statistic for the one-way table for the jury data to follow a chi-square distribution with 3 degrees of freedom since there are `$k=4$` categories.

- Then the probability of observing a value that is as or more extreme that the value obtained from the sample data is

```r
1 - pchisq(5.89,3)
```

```
## [1] 0.1170863
```

- Alternatively:

```r
pchisq(5.89,3,lower.tail = FALSE)
```

```
## [1] 0.1170863
```

- We just computed a p-value, but to what null hypothesis does this p-value provide an appropriate means of testing?

---

# Null Hypothesis for One-Way Tables

- In our example, we would want to test:

- `$H_{0}:$` The jurors are a random sample, that is, there is no racial bias in who serves on a jury, and the observed counts reflect natural sampling fluctuation. 
  
--

- `$H_{A}:$` The jurors are not randomly sampled, that is, there is racial bias in juror selection. 
  
--

- Using the data, we can conduct the chi-square test as follows:

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

```
## 
## 	Chi-squared test for given probabilities
## 
## data:  c(26, 25, 205, 19)
## X-squared = 5.8896, df = 3, p-value = 0.1171
```

---

# More Examples

- Let's see some more examples.

- Let's look at exercise 6.33 from the textbook on page 239. The table for the data will look as follows:

<table class="table" style="margin-left: auto; margin-right: auto;">
 <thead>
  <tr>
   <th style="text-align:left;"> textbook </th>
   <th style="text-align:right;"> purchased </th>
   <th style="text-align:right;"> printed </th>
   <th style="text-align:right;"> online </th>
   <th style="text-align:right;"> total </th>
  </tr>
 </thead>
<tbody>
  <tr>
   <td style="text-align:left;"> Method </td>
   <td style="text-align:right;"> 71.0 </td>
   <td style="text-align:right;"> 30.00 </td>
   <td style="text-align:right;"> 25.00 </td>
   <td style="text-align:right;"> 126 </td>
  </tr>
  <tr>
   <td style="text-align:left;"> Expected percent </td>
   <td style="text-align:right;"> 0.6 </td>
   <td style="text-align:right;"> 0.25 </td>
   <td style="text-align:right;"> 0.15 </td>
   <td style="text-align:right;"> 1 </td>
  </tr>
</tbody>
</table>

- The expected counts will be

```
## purchased   printed    online 
##      75.6      31.5      18.9
```

- Thus, our `$X^2$` statistic is

```r
(X2 <- (71-75.6)^2/75.6  + (30-31.5)^2/31.5  + (25-18.9)^2/18.9)
```

```
## [1] 2.320106
```

---

# Example Continued

- The appropriate degrees of freedom is 2. Therefore, the tail area is

```r
pchisq(X2,2,lower.tail = FALSE)
```

```
## [1] 0.3134696
```

- If our hypothesis is: `$H_{0}:$` the distribution of the format of the book used follows the expected distribution, vs. `$H_{A}:$` the distribution of the format of the book used does not follow the expected distribution, then we will fail to reject the null hypothesis at the `$\alpha = 0.05$` level of significance.

- We can confirm our result using R:

```r
chisq.test(c(71,30,25),p=c(0.6,0.25,0.15))
```

```
## 
## 	Chi-squared test for given probabilities
## 
## data:  c(71, 30, 25)
## X-squared = 2.3201, df = 2, p-value = 0.3135
```

---

# Two-Way Tables

- A one-way table describes counts for each outcome in a single categorical variable.

- A two-way table describes counts for combinations of two categorical variables where at least one of the two has more than 2 levels.

- When we consider a two-way table, we often would like to know, are these variables related in any way? That is, are they dependent versus independent?

---

# Null Hypothesis for Two-Way Tables

- For a two-way table problem, the typical hypotheses are of the form

- `$H_{0}:$` The two variables are independent. 
  
--

- `$H_{A}:$` The two variables are dependent.

- Let's look at an example.

---

# Offshore Drilling Example

- Consider the data with first few rows shown below:

```
## # A tibble: 6 x 2
##   position college_grad
##   <fct>    <fct>       
## 1 support  yes         
## 2 support  yes         
## 3 support  yes         
## 4 support  yes         
## 5 support  yes         
## 6 support  yes
```

- Let's look at the corresponding two-way table:

```r
addmargins(table(my_offshore_drilling$position,my_offshore_drilling$college_grad))
```

```
##              
##                no yes Sum
##   do_not_know 131 104 235
##   oppose      126 180 306
##   support     132 154 286
##   Sum         389 438 827
```

---

# Hypothesis Test for Offshore Drilling

- We would like to test the hypothesis:

- `$H_{0}:$` College graduate status and support for offshore drilling are independent. 
  
--

- `$H_{A}:$` College graduate status and support for offshore drilling are not independent. 
  
--

- This is easily done with

```r
chisq.test(my_offshore_drilling$position,my_offshore_drilling$college_grad)
```

```
## 
## 	Pearson's Chi-squared test
## 
## data:  my_offshore_drilling$position and my_offshore_drilling$college_grad
## X-squared = 11.461, df = 2, p-value = 0.003246
```

- Thus, we will reject the null hypothesis at the `$\alpha=0.05$` significance level.

---

# Tests By Hand

- Let's see how to conduct the previous test by hand.

- The two things we need to know are the value of the `$X^2$` statistic and the appropriate number of degrees of freedom to use.

- When applying the chi-square test to a two-way table, we use

`$$df = (R-1) \times (C-1)$$`

where `$R$` is the number of rows in the table and `$C$` is the number of columns.

- Thus, in our example, `$df = (3-1)\times (2-1) = 2$`.

---

# Computing `$X^2$`

- As before

`$$X^2 = \sum \frac{(O-E)^2}{E}$$`

- The question is, how do we compute the expected counts ( `$E_{ij}$` ) for a two-way table? The answer is

`$$\text{Expected Count}_{\text{row }i \text{ col }j} = E_{ij} = \frac{\text{row }i \text{ total} \times \text{column }j \text{ total}}{\text{table total}}$$`

- Let's work this out on the board for our example.

---

# Hypothesis Testing Summary

- To date, we have covered the following tests:

- Single proportion and difference of proportions using a "z-test."
  
--

- Single mean, paired mean, difference of means using a "t-test."
  
--

- Comparing many means with ANOVA.
  
--

- One-way and two-way table tests for categorical variables with chi-square. 
  
--

- Simple linear regression via ordinary least squares for a pair of numeric variable. 
 
--

- We also know how to construct confidence intervals for parameter estimates for proportions, difference of proportions, mean, difference of means, and intercept and slope parameters for a linear model.

---

# Next Topic

- Our next topic discusses more regarding regression. This video will get you started: