Testing for independence in two-way tables

We all buy used products – cars, computers, textbooks, and so on – and we sometimes assume the sellers of those products will be forthright about any underlying problems with what they’re selling. This is not something we should take for granted. Researchers recruited 219 participants in a study where they would sell a used iPod¹ that was known to have frozen twice in the past. The participants were incentivized to get as much money as they could for the iPod since they would receive a 5% cut of the sale on top of $10 for participating. The researchers wanted to understand what types of questions would elicit the seller to disclose the freezing issue.

Unbeknownst to the participants who were the sellers in the study, the buyers were collaborating with the researchers to evaluate the influence of different questions on the likelihood of getting the sellers to disclose the past issues with the iPod. The scripted buyers started with “Okay, I guess I’m supposed to go first. So you’ve had the iPod for 2 years ...” and ended with one of three questions:

• General: What can you tell me about it?

• Positive Assumption: It doesn’t have any problems, does it?

• Negative Assumption: What problems does it have?

The question is the treatment given to the sellers, and the response is whether the question prompted them to disclose the freezing issue with the iPod. The results are shown in Figure 6.14, and the data suggest that asking the, What problems does it have?, was the most effective at getting the seller to disclose the past freezing issues. However, you should also be asking yourself: could we see these results due to chance alone, or is this in fact evidence that some questions are more effective for getting at the truth?

The hypothesis test for the iPod experiment is really about assessing whether there is statistically significant evidence that the success each question had on getting the participant to disclose the problem with the iPod. In other words, the goal is to check whether the buyer’s question was independent of whether the seller disclosed a problem.

Expected counts in two-way tables

with one-way tables, we will need to compute estimated counts for each cell in a two-way table.

We can compute the expected number of sellers who we would expect to disclose or hide the freezing issue for all groups, if the questions had no impact on what they disclosed, using the same strategy employed in Example 6.38 and Guided Practice 6.39. These expected counts were used to construct Figure 6.15, which is the same as Figure 6.14, except now the expected counts have been added in parentheses.

The examples and exercises above provided some help in computing expected counts. In general, expected counts for a two-way table may be computed using the row totals, column totals, and the table total. For instance, if there was no difference between the groups, then about 27.85% of each column should be in the first row: $$\begin{aligned} 0.2785{}\times (\text{column 1 total}) &= 20.33{} \\ 0.2785{}\times (\text{column 2 total}) &= 20.33{} \\ 0.2785{}\times (\text{column 3 total}) &= 20.33{} \end{aligned}$$ Looking back to how 0.2785 was computed – as the fraction of sellers who disclosed the freezing issue ($158{}/219{}$) – these three expected counts could have been computed as $$\begin{aligned} \left(\frac{\text{row 1 total}}{\text{table total}}\right) \text{(column 1 total)} &= 20.33{} \\ \left(\frac{\text{row 1 total}}{\text{table total}}\right) \text{(column 2 total)} &= 20.33{} \\ \left(\frac{\text{row 1 total}}{\text{table total}}\right) \text{(column 3 total)} &= 20.33{} \end{aligned}$$ This leads us to a general formula for computing expected counts in a two-way table when we would like to test whether there is strong evidence of an association between the column variable and row variable.

The chi-square test for two-way tables

The chi-square test statistic for a two-way table is found the same way it is found for a one-way table. For each table count, compute $$\begin{aligned} &\text{General formula} && \frac{(\text{observed count } - \text{expected count})^2} {\text{expected count}} \\ &\text{Row 1, Col 1} && \frac{(2- 20.33)^2}{20.33} = 16.53 \\ &\text{Row 1, Col 2} && \frac{(23- 20.33)^2}{20.33} = 0.35 \\ & \vdots && \vdots \\ &\text{Row 2, Col 3} && \frac{(37- 52.67)^2}{52.67} = 4.66 \end{aligned}$$ Adding the computed value for each cell gives the chi-square test statistic $X^2$: $$\begin{aligned} X^2 = 16.53 + 0.35 + \dots + 4.66 = 40.13 \end{aligned}$$ Just like before, this test statistic follows a chi-square distribution. However, the degrees of freedom are computed a little differently for a two-way table.³ For two way tables, the degrees of freedom is equal to $$\begin{aligned} df = \text{(number of rows minus 1)}\times \text{(number of columns minus 1)} \end{aligned}$$ In our example, the degrees of freedom parameter is $$\begin{aligned} df = (2-1)\times (3-1) = 2 \end{aligned}$$ If the null hypothesis is true (i.e. the questions had no impact on the sellers in the experiment), then the test statistic $X^2 = 40.13$ closely follows a chi-square distribution with 2 degrees of freedom. Using this information, we can compute the p-value for the test, which is depicted in Figure 6.16.

When analyzing 2-by-2 contingency tables, one guideline is to use the two-proportion methods introduced in Section 6.2.

Exercises