Inference for Two Way Tables

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# Inference for Two Way Tables
## IS381 - Statistics and Probability with R
### Jason Bryer, Ph.D.
### April 13, 2026

---
# One Minute Paper Results

.pull-left[
**What was the most important thing you learned during this class?**
<img src="11-Two_way_tables_files/figure-html/unnamed-chunk-2-1.png" alt="" style="display: block; margin: auto;" />
]
.pull-right[
**What important question remains unanswered for you?**
<img src="11-Two_way_tables_files/figure-html/unnamed-chunk-3-1.png" alt="" style="display: block; margin: auto;" />
]

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# Weldon's dice

* Walter Frank Raphael Weldon (1860 - 1906), was an English evolutionary biologist and a founder of biometry. He was the joint founding editor of Biometrika, with Francis Galton and Karl Pearson.

* In 1894, he rolled 12 dice 26,306 times, and recorded the number of 5s or 6s (which he considered to be a success).

* It was observed that 5s or 6s occurred more often than expected, and Pearson hypothesized that this was probably due to the construction of the dice. Most inexpensive dice have hollowed-out pips, and since opposite sides add to 7, the face with 6 pips is lighter than its opposing face, which has only 1 pip.

---
# Labby's dice

In 2009, Zacariah Labby (U of Chicago), repeated Weldon’s experiment using a homemade dice-throwing, pip counting machine. http://www.youtube.com/watch?v=95EErdouO2w

* The rolling-imaging process took about 20 seconds per roll.
	* Each day there were ∼150 images to process manually.
	* At this rate Weldon’s experiment was repeated in a little more than six full days.

.center[![](images/labbyPipCounts.png)]
---
# Summarizing Labby's results

The table below shows the observed and expected counts from Labby's experiment.

| Outcome   | Observed	  | Expected    |
|-----------|-------------|-------------|
| 1	        | 53,222      | 52,612      | 
| 2	        | 52,118      | 52,612      | 
| 3	        | 52,465      | 52,612      | 
| 4	        | 52,338      | 52,612      | 
| 5	        | 52,244      | 52,612      | 
| 6	        | 53,285      | 52,612      | 
| Total     | 315,672     | 315,672     |

---
# Setting the hypotheses

Do these data provide convincing evidence of an inconsistency between the observed and expected counts?

* `$H_0$`: There is no inconsistency between the observed and the expected counts. The observed counts follow the same distribution as the expected counts.

* `$H_A$`: There is an inconsistency between the observed and the expected counts. The observed counts **do not** follow the same distribution as the expected counts. There is a bias in which side comes up on the roll of a die.

---
# Evaluating the hypotheses

* To evaluate these hypotheses, we quantify how different the observed counts are from the expected counts.

* Large deviations from what would be expected based on sampling variation (chance) alone provide strong evidence for the alternative hypothesis.

* This is called a *goodness of fit* test since we're evaluating how well the observed data fit the expected distribution.

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# Anatomy of a test statistic

* The general form of a test statistic is:

`$$\frac{\text{point estimate} - \text{null value}}{\text{SE of point estimate}}$$`

* This construction is based on 
	1. identifying the difference between a point estimate and an expected value if the null hypothesis was true, and 
	2. standardizing that difference using the standard error of the point estimate.

* These two ideas will help in the construction of an appropriate test statistic for count data.

---
# Chi-Squared

When dealing with counts and investigating how far the observed counts are from the expected counts, we use a new test statistic called the chi-square ( `$\chi^2$` ) statistic.

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`$$\chi^2 = \sum_{i = 1}^{k} \frac{(O - E)^2}{E}$$`
where k = total number of cells

]
.pull-right[

| Outcome   | Observed	  | Expected    | `$\frac{(O - E)^2}{E}$`
|-----------|-------------|-------------|-----------------------|
| 1	        | 53,222      | 52,612      | `$\frac{(53,222 - 52,612)^2}{52,612} = 7.07$` |
| 2	        | 52,118      | 52,612      | `$\frac{(52,118 - 52,612)^2}{52,612} = 4.64$` |
| 3	        | 52,465      | 52,612      | `$\frac{(52,465 - 52,612)^2}{52,612} = 0.41$` |
| 4	        | 52,338      | 52,612      | `$\frac{(52,338 - 52,612)^2}{52,612} = 1.43$` |
| 5	        | 52,244      | 52,612      | `$\frac{(52,244 - 52,612)^2}{52,612} = 2.57$` |
| 6	        | 53,285      | 52,612      | `$\frac{(53,285 - 52,612)^2}{52,612} = 8.61$` |
| Total     | 315,672     | 315,672     | 24.73 |

]

---
# Chi-Squared Distribution

Squaring the difference between the observed and the expected outcome does two things:

*  Any standardized difference that is squared will now be positive.
*  Differences that already looked unusual will become much larger after being squared.

In order to determine if the `$\chi^2$` statistic we calculated is considered unusually high or not we need to first describe its distribution.

* The chi-square distribution has just one parameter called **degrees of freedom (df)**, which influences the shape, center, and spread of the distribution.

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# Degrees of freedom for a goodness of fit test

When conducting a goodness of fit test to evaluate how well the observed data follow an expected distribution, the degrees of freedom are calculated as the number of cells (*k*) minus 1.

`$$df = k - 1$$`

For dice outcomes, `$k = 6$`, therefore `$df = 6 - 1 = 5$`

p-value = `$P(\chi^2_{df = 5} > 24.67)$` is less than 0.001

---
# Turns out...

.font140[
* The 1-6 axis is consistently shorter than the other two (2-5 and 3-4), thereby supporting the hypothesis that the faces with one and six pips are larger than the other faces.

* Pearson's claim that 5s and 6s appear more often due to the carved-out pips is not supported by these data.

* Dice used in casinos have flush faces, where the pips are filled in with a plastic of the same density as the surrounding material and are precisely balanced.
]

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# Recap: p-value for a chi-square test

* The p-value for a chi-square test is defined as the tail area **above** the calculated test statistic.

* This is because the test statistic is always positive, and a higher test statistic means a stronger deviation from the null hypothesis.

---
# Independence Between Groups

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Assume we have a population of 100,000 where groups A and B are independent with `$p_A = .55$` and `$p_B = .6$` and `$n_A = 99,000$` (99% of the population) and `$n_B = 1,000$` (1% of the population). We can sample from the population (that includes groups A and B) and from group B of sample sizes of 1,000 and 100, respectively. We can also calculate `$\hat{p}$` for group A independent of B.

``` r
propA <- .55    # Proportion for group A
propB <- .6     # Proportion for group B
pop.n <- 100000 # Population size
sampleA.n <- 1000
sampleB.n <- 100
```
]

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``` r
pop <- data.frame(
	group = c(rep('A', pop.n * 0.99),
			  rep('B', pop.n * 0.01) ),
	response = c(
		sample(c(1,0), 
			   size = pop.n * 0.99, 
			   prob = c(propA, 1 - propA), 
			   replace = TRUE),
		sample(c(1,0), 
			   size = pop.n * 0.01, 
			   prob = c(propB, 1 - propB), 
			   replace = TRUE) )
)
sampA <- pop[sample(nrow(pop), 
					size = sampleA.n),]
sampB <- pop[sample(which(pop$group == 'B'), 
					size = sampleB.n),]
```
] ]

---
# Independence Between Groups (cont.)

`$\hat{p}$` for the population sample

``` r
mean(sampA$response)
```

```
## [1] 0.551
```

`$\hat{p}$` for the population sample, excluding group B

``` r
mean(sampA[sampA$group == 'A',]$response)
```

```
## [1] 0.5488419
```

`$\hat{p}$` for group B sample

``` r
mean(sampB$response)
```

```
## [1] 0.62
```

---
# Independence Between Groups (cont.)

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# One Minute Paper

.pull-left[
1. What was the most important thing you learned during this class?
2. What important question remains unanswered for you?
]
.pull-right[
<img src="11-Two_way_tables_files/figure-html/unnamed-chunk-14-1.png" alt="" style="display: block; margin: auto;" />
]

https://forms.gle/bz8GvYWfKdMggRv38