Chi-Square Test

CHI-SQUARE TEST

Studies often collect data on categorical variables that can be summarized as a series of counts. These counts are commonly arranged in a tabular format known as a contingency table. For example, a study designed to determine whether or not there is an association between cigarette smoking and asthma might collect data that could be assembled into a 2−2 table. In this case, the two columns could be defined by whether the subject smoked or not, while the rows could represent whether or not the subject experienced symptoms of asthma. The cells of the table would contain the number of observations or patients as defined by these two variables.

The chi-square test statistic can be used to evaluate whether there is an association between the rows and columns in a contingency table. More specifically, this statistic can be used to determine whether there is any difference between the study groups in the proportions of the risk factor of interest. Returning to our example, the chi-square statistic could be used to test whether the proportion of individuals who smoke differs by asthmatic status.

The chi-square test statistic is designed to test the null hypothesis that there is no association between the rows and columns of a contingency table. This statistic is calculated by first obtaining for each cell in the table, the expected number of

Table 1

events that will occur if the null hypothesis is true. When the observed number of events deviates significantly from the expected counts, then it is unlikely that the null hypothesis is true, and it is likely that there is a row-column association. Conversely, a small chi-square value indicates that the observed values are similar to the expected values leading us to conclude that the null hypothesis is plausible. The general formula used to calculate the chi-square (X²) test statistic is as follows:

where O = observed count in category; E = expected count in the category under the null hypothesis; df = degrees of freedom; and c, r represent the number of columns and rows in the contingency table.

The value of the chi-square statistic cannot be negative and can assume values from zero to infinity. The p-value for this test statistic is based on the chi-square probability distribution and is generally extracted from published tables or estimated using computer software programs. The p-value represents the probability that the chi-square test statistic is as extreme as or more extreme than observed if the null hypothesis were true. As with the t and F distributions, there is a different chi-square distribution for each possible value of degrees of freedom. Chi-square distributions with a small number of degrees of freedom are highly skewed; however, this skewness is attenuated as the number of degrees of freedom increases. In general, the degrees of freedom for tests of hypothesis that involve an r×c contingency table is

Table 2

equal to (r7minus;1)×(c−1); thus for any 2×2 table, the degrees of freedom is equal to one. A chi-square distribution with one degree of freedom is equal to the square root of the normal distribution, and, consequently, either the chi-square or standard normal table can be used to determine the corresponding p-value.

The chi-square test is most widely used to conduct tests of hypothesis that involve data that can be presented in a 2×2 table. Indeed, this tabular format is a feature of the case-control study design that is commonly used in public health research. Within this contingency table, we could denote the observed counts as shown in Table 1. Under the null hypothesis of no association between the two variables, the expected number in each cell under the null hypothesis is calculated from the observed values using the formula outlined in Table 2.

The use of the chi-square test can be illustrated by using hypothetical data from a study investigating the association between smoking and asthma among adults observed in a community health clinic. The results obtained from classifying 150 individuals are shown in Table 3. As Table 3 shows, among asthmatics the proportion of smokers was 40 percent (20/50), while the corresponding proportion among asymptomatic individuals was 22 percent (22/100). By applying the formula presented in Table 2, for the observed cell counts of 20, 30, 22, and 78 (Table 3) the corresponding expected counts are 14, 36, 28, and 72. The observed and expected counts can then be used to calculate the chi-square test statistic as outlined in Equation 1. The resulting value of the chi-square

Table 3

test statistic is approximately 5.36, and the associated p-value for this chi-square distribution that has one degree of freedom is 0.02. Therefore, if there was truly no association between smoking and asthma, there is a 2 out of 100 probability of observing a difference in proportions that is at least as large as 18 percent (40%–22%) by chance alone. We would therefore conclude that the observed difference in the proportions is unlikely to be explained by chance alone, and consider this result statistically significant.

Because the construction of the chi-square test makes use of discrete data to estimate a continuous distribution, some authors will apply a continuity correction when calculating this statistic. Specifically,

where ^O_i−E_i is the absolute value of the difference between O_i and E_i and the term 0.5 in the numerator is often referred to as Yates correction factor. This correction factor serves to reduce the chi-square value, and, therefore, increases the resulting p-value. It has been suggested that this correction yields an overly conservative test that may fail to reject a false null hypothesis. However, as long as the sample size is large, the effect of the correction factor is negligible.

When there is a small number of counts in the table, the use of the chi-square test statistic may not be appropriate. Specifically, it has been recommended that this test not be used if any cell in the table has an expected count of less than one, or if 20 percent of the cells have an expected count that is greater than five. Under this scenario, the Fisher's exact test is recommended for conducting tests of hypothesis.

PAUL J. VILLENEUVE

(SEE ALSO: Normal Distributions; Probability Model; Sampling; Statistics for Public Health; T-Test)

BIBLIOGRAPHY

Cohran, W. G. (1954). "Some Methods for Strengthening the Common X² Test." Biometrics 10:417–451.

Grizzle, J. E. (1967). "Continuity Correction in the X2 Test for 2×2 Tables." The American Statistician 21:28–32.

Pagano, M., and Gauvreau, K. (2000). Principles of Biostatistics, 2nd edition. Pacific Grove, CA: Duxbury Press.

Rosner, B. (2000). Fundamentals of Biostatistics, 5th edition. Pacific Grove, CA: Duxbury Press.