The null hypothesis H0 of interest in a two-way table is “There is no association between the row variable and the column variable.” In Example 9.3, this null hypothesis says that there is no association between political party preference and belief that vaccinations should be required. The alternative hypothesis Ha is that there is an association between these two variables. The alternative Ha does not specify any particular direction for the association. For two-way tables in general, the alternative includes many different possibilities. Because it includes all sorts of possible associations, we cannot describe Ha as either one-sided or two-sided.

In our example, the hypothesis H0 that there is no association between political party preference and opinions about requiring vaccinations is equivalent to the statement that the variables Required and Party are independent.

For other two-way tables, such as the lost wallet study of Example 9.2, we have samples from two populations, one for each condition. For each population there is a distribution of the categorical response variable Returned. For two-way tables like this, the columns correspond to independent samples from c distinct populations, there are c distributions for the row variable, one for each population. The null hypothesis then says that the c distributions of the row variable are identical. The alternative hypothesis is that the distributions are not all the same.

To test the null hypothesis in r×c tables, we compare the observed cell counts with expected cell counts calculated under the assumption that the null hypothesis is true. A numerical summary of the comparison will be our test statistic.

The reasoning of Example 9.7 leads to a simple formula for calculating expected cell counts. To compute the expected count of Democrats opposed to requiring vaccinations, we multiplied the proportion of adults opposed to requiring vaccinations (488/1696) by the number of Democrats (959). From Figure 9.1, we see that the numbers 488 and 959 are the row and column totals for the cell of interest and that 1696 is n, the total number of observations for the table. The expected cell count is, therefore, the product of the row and column totals divided by the table total.

The reasoning is similar for the wallet data in Example 9.2. Under the null hypothesis, the two populations defined by Condition have the same distribution. The common distribution is estimated by the marginal counts for Returned, 315 and 285. Expressed as percents, the distribution is 52.5% (315/600) for “No” and 47.5.5% (285/600) for “Yes.” The population “Money” has 300 samples, so the expected count is 52.5% of 300, or 157.5, for the cell defined by “Money” and “No.” Expected counts for the other three cells are calculated in the same way.

In Figure 9.3 (page 491), we used a mosaic plot to display the data for the vaccination and political party preference data. Looking at the two columns, we can see that the proportion in the lower region, corresponding to being opposed to required vaccinations, is smaller for the Democrats than for the Republicans. This illustrates graphically the difference in the conditional distributions for the two parties.

What would the mosaic plot look like if the null hypothesis were true? In this case, the two conditional distributions would be the same. Ideally, the observed counts would be equal to the expected counts. If we rerun the analysis with the expected counts in place of the observed counts, we obtain the mosaic plot in Figure 9.5. Notice that the proportions of each party responding no are now the same and equal to 28.77%, the marginal percent of adults who do not think vaccinations should be required.

To test H0 that there is no association between the row and column classifications, we use a statistic that compares the entire set of observed counts with the set of expected counts. To compute this statistic,

1. First, take the difference between each observed count and its corresponding expected count. Verify that the sum of the r×c differences is zero.
2. Then, square these values so that they are all either 0 or positive.
3. Because a large difference means less if it comes from a cell that is expected to have a large count, divide each squared difference by the expected count. This is a type of standardization.
4. Finally, sum over all cells.

The result is called the chi-square statistic X2. The chi-square statistic was proposed by the English statistician Karl Pearson (1857–1936) in 1900. It is the oldest inference procedure still used in its original form.

If the expected counts and the observed counts are very different, a large value of X2 will result. Large values of X2 provide evidence against the null hypothesis. To obtain a P-value for the test, we need the sampling distribution of X2 under the assumption that H0 (no association between the row and column variables) is true. The distribution is called the chi-square distribution, which we denote by χ2 (χ is the lowercase Greek letter chi).

Like the t distributions, the χ2 distributions form a family described by a single parameter, the degrees of freedom. We use χ2 (df) to indicate a particular member of this family. Figure 9.6 displays the density curves of the χ2(2) and χ2(4) distributions. As you can see in the figure, χ2 distributions take only positive values and are skewed to the right. Table F in the back of the book gives upper critical values for the χ2 distributions.

Now that we have our test statistic and its sampling distribution under H0, we can describe the test.

The chi-square test always uses the upper tail of the χ2 distribution because any deviation from the null hypothesis makes the statistic larger. The approximation of the distribution of X2 by χ2 becomes more accurate as the cell counts increase. Moreover, it is more accurate for tables larger than 2×2 tables.

Example 9.8 Chi-square significance test.

The results of the chi-square significance test for the vaccination example appear in the computer outputs in Figure 9.7, labeled Pearson (JMP) or Pearson Chi-square (Minitab and SPSS). Because all the expected cell counts are large (5 or more), the χ2 distribution provides an accurate P-value. We see that X2=24.71, df=1, and P<0.0001. Note that Minitab and SPSS report the P-value as 0.000 or .000. These are rounded numbers and potentially misleading. The P-value is small, but it is not zero. For this reason, we prefer to report P<0.0001.

As a check, we verify that the degrees of freedom are correct for a 2×2 table:

df=(r-1)(c-1)=(2−1)(2−1)=1

The chi-square test confirms that the data provide evidence against the null hypothesis that there is no relationship between political party preference and vaccination opinion. Under H0, the chance of obtaining a value of X2 greater than or equal to the calculated value of 24.71 is small, less than 0.0001—fewer than 1 time in 10,000.

The outputs in Figure 9.7 also report results for testing the hypothesis of no association using alternatives to the chi-square significance test. Fisher’s exact test is preferred by many, particularly when the counts are small and the chi-square approximation is not very accurate. Its results are provided in each of the software outputs.

The significance test result does not provide insight into the nature of the relationship between the variables. It is up to us to see that the data show Republicans are more likely to believe that vaccinations should not be required. You should always accompany a chi-square test by percents such as those in Example 9.5 and Figure 9.2 and by a description of the nature of the relationship.

Observational studies such as the one in Example 9.3 cannot tell us whether or not an explanatory variable is a cause of a pattern in a response variable. For the vaccine and party preference setting, a causal association does not seem plausible. Often, association can be explained by confounding with other variables. Similarly, the results for the wallet data in Exercise 9.2 tell us that wallets with money are treated differently, but they do not explain the underlying reason for this behavior.

The calculations required to analyze a two-way table are straightforward but tedious. In practice, we recommend using software, but it is possible to do the work with a calculator or a spreadsheet, and some insight can be gained by examining the details. Here is an outline of the steps required.

The next few examples illustrate these steps.

Example 9.9 Health habits of college students.

Physical activity generally declines when students leave high school and enroll in college. This suggests that college is an ideal setting to promote physical activity. One study examined the level of physical activity and other health-related behaviors in a sample of 1184 college students.² Let’s look at the data for physical activity and consumption of fruits. The study categorized physical activity as low, moderate, or vigorous and fruit consumption as low, medium, or high. Here is the two-way table that summarizes the data:

	Physical activity
Fruit consumption	Low	Moderate	Vigorous	Total
Low	69	206	294	569
Medium	25	126	170	321
High	14	111	169	294
Total	108	443	633	1184

This table includes the marginal totals obtained by summing across rows and columns. For example, the first-row total is 69+206+294=569. The grand total, the number of students in the study, can be computed by summing the row totals (569+321+294=1184) or the column totals (108+443+633=1184). It is easy to make an error in these calculations, so if doing this by hand, it is a good idea to do both as a check on your arithmetic.

First, we summarize the observed relation between physical activity and fruit consumption. We expect a positive association, but there is no clear distinction between an explanatory variable and a response variable in this setting. If we have such a distinction, then the clearest way to describe the relationship is to compare the conditional distributions of the response variable for each value of the explanatory variable. Otherwise, we can compute the conditional distribution each way and then decide which gives a better description of the data.

The sum of the percents in each column should be 100, except for possible small roundoff errors. It is good practice to calculate each percent separately and then sum each column as a check. In this way, we can find arithmetic errors that would not be uncovered if, for example, we calculated the column percent for the “High” row by subtracting the sum of the percents for “Low” and “Medium” from 100.

Figure 9.8 compares the distributions of fruit consumption for each of the three physical activity levels. For each activity level, the highest percent is for students who consume low amounts of fruit. For low physical activity, there is a clear decrease in the percent when moving from low to medium to high fruit consumption. The patterns for moderate physical activity and vigorous physical activity are similar. Low fruit consumption is still dominant, but the percents for medium and high fruit consumption are about the same for the moderate and vigorous activity levels. The percent of low fruit consumption is highest for the low physical activity students compared with those who have moderate or vigorous physical activity. These plots suggest that there is an association between these two variables.

We observe a clear relationship between physical activity and fruit consumption in this study. The chi-square test assesses whether this observed association is statistically significant—that is, too strong to occur often just by chance. The test confirms only that there is some relationship. The percents we have compared describe the nature of the relationship.

The chi-square test does not in itself tell us what population our conclusion describes. The subjects in this study were college students from four midwestern universities. The researchers could argue that these findings apply to college students in general. This type of inference is important, but it is based on expert judgment and is beyond the scope of the statistical inference that we have been studying.

We can check our work by adding the expected counts to obtain the row and column totals, as in the table of Example 9.10 (page 497). These totals should be the same as those in the table of observed counts except for small roundoff errors.

A comparison of the proportions of “successes” in two populations leads to a 2×2 table. We can compare two population proportions either by using the chi-square test or by using the two-sample z test from Section 8.2. In fact, these tests always give exactly the same result because the X2 statistic is equal to the square of the z statistic, and χ2(1) critical values are equal to the squares of the corresponding N(0, 1) critical values. The advantage of the z test is that we can test either one-sided or two-sided alternatives. The chi-square test always tests the two-sided alternative. Of course, the chi-square test can compare more than two populations, whereas the z test compares only two.

Example 9.13 Meta-analysis for eating too much salt.

Evidence from a variety of sources suggests that diets high in salt are associated with risks to human health. To investigate the relationship between salt intake and stroke, information from 14 studies was combined in a meta-analysis.³ Subjects were classified based on the amount of salt in their normal diet. They were followed for several years and then classified according to whether or not they had developed cardiovascular disease (CVD). A total of 104,933 subjects were studied, and 5161 of them developed CVD. Here are the data from one of the studies:⁴

	Low salt	High salt
CVD	88	112
No CVD	1081	1134
Total	1169	1246

Let’s look at the relative risk for this study. We first find the proportion of subjects who developed CVD in each group. For the subjects with a low salt intake, the proportion who developed CVD is

881169=0.0753

or 75 per thousand; for the high-salt group, the proportion is

1121246=0.0899

or 90 per thousand. We can now compute the relative risk as the ratio of these two proportions. We choose to put the high-salt group in the numerator. The relative risk is

0.08990.0753=1.19

Relative risk greater than 1 means that the high-salt group developed more CVD than the low-salt group. For this study, the association is not statistically significant. The 95% confidence interval for the relative risk is (0.91, 1.56).

When the data from all 14 studies were combined, the relative risk was reported as 1.17, with a 95% confidence interval of (1.02, 1.32). Because this interval does not include the value 1, corresponding to equal proportions in the two groups, we conclude that the higher CVD rates are not the same for the two diets (P<0.05). The high-salt diet is associated with a 17% higher rate of CVD than the low-salt diet. Note that the relative risk for the individual study in this example was not statistically significant, even though it was higher than the overall estimate (1.19 versus 1.17). This illustrates the value of the meta-analysis where the conclusion is based on combining results from several studies.