In the vaccination example, the null hypothesis is that the proportion of children given thigh injections who have severe reactions is equal to the proportion of children given arm injections who have severe reactions.
The math of the chi-square test of independence is the same as for the chi-square test of goodness-of-fit , only the method of calculating the expected frequencies is different. For the goodness-of-fit test, you use a theoretical relationship to calculate the expected frequencies. For the test of independence, you use the observed frequencies to calculate the expected. Once you have each of the four expected numbers, you could compare them to the observed numbers using the chi-square test, just like you did for the chi-square test of goodness-of-fit.
To get the P value, you also need the number of degrees of freedom. Note that I'm just using the 3-to-6 year olds as an example; Jackson et al. While in principle, the chi-square test of independence is the same as the test of goodness-of-fit, in practice, the calculations for the chi-square test of independence use shortcuts that don't require calculating the expected frequencies.
MacDonald and Gardner use simulated data to test several post-hoc tests for a test of independence, and they found that pairwise comparisons with Bonferroni corrections of the P values work well. To illustrate this method, here is a study Klein et al. This is not quite significant by a tiny bit , but it's worthwhile to follow up to see if there's anything interesting. Because there are six comparisons, the Bonferroni-adjusted P value needed for significance is 0.
The P value for vitamin E vs. For this example, I tested all six possible pairwise comparisons. Klein et al. If they had decided ahead of time to just compare each of the three treatments vs. The important thing is to decide before looking at the results how many comparisons to do, then adjust the P value accordingly. If you don't decide ahead of time to limit yourself to particular pairwise comparisons, you need to adjust for the number of all possible pairs.
Another kind of post-hoc comparison involves testing each value of one nominal variable vs. The same principle applies: get the P value for each comparison, then apply the Bonferroni correction.
For example, Latta et al. They observed the following numbers lumping together the less common bird species as "Uncommon" :. The overall table yields a chi-square value of That tells us there's a difference in the species composition between the remnant and restored habitat, but it would be interesting to see which species are a significantly higher proportion of the total in each habitat.
Because there are 12 comparisons, applying the Bonferroni correction means that a P value has to be less than 0. When there are more than two rows and more than two columns, you may want to do all possible pairwise comparisons of rows and all possible pairwise comparisons of columns; in that case, simply use the total number of pairwise comparisons in your Bonferroni correction of the P value.
The chi-square test of independence, like other tests of independence, assumes that the individual observations are independent. Bambach et al. Gardemann et al. The data are:. The biological null hypothesis is that the apolipoprotein polymorphism doesn't affect the likelihood of getting coronary artery disease.
The statistical null hypothesis is that the proportions of men with coronary artery disease are the same for each of the three genotypes. This indicates that you can reject the null hypothesis; the three genotypes have significantly different proportions of men with coronary artery disease. You should usually display the data used in a test of independence with a bar graph, with the values of one variable on the X-axis and the proportions of the other variable on the Y-axis.
If the variable on the Y-axis only has two values, you only need to plot one of them. In the example below, there would be no point in plotting both the percentage of men with prostate cancer and the percentage without prostate cancer; once you know what percentage have cancer, you can figure out how many didn't have cancer. If the variable on the Y-axis has more than two values, you should plot all of them. Some people use pie charts for this, as illustrated by the data on bird landing sites from the Fisher's exact test page:.
But as much as I like pie, I think pie charts make it difficult to see small differences in the proportions, and difficult to show confidence intervals. In this situation, I prefer bar graphs:. There are several tests that use chi-square statistics. The one described here is formally known as Pearson's chi-square. Then, you compare the test statistic to a theoretical value from the Chi-square distribution.
The theoretical value depends on both the alpha value and the degrees of freedom for your data. Visit the pages for each test type for detailed examples. The Chi-Square Test. What is a Chi-square test? What are my choices? Types of Chi-square tests You use a Chi-square test for hypothesis tests about whether your data is as expected. How to perform a Chi-square test For both the Chi-square goodness of fit test and the Chi-square test of independence , you perform the same analysis steps, listed below.
Define your null and alternative hypotheses before collecting your data. This can limit the flexibility that researchers have in terms of the processes that they use. We use cookies to track how our visitors are browsing and engaging with our website in order to understand and improve the user experience.
Review our Privacy Policy to learn more. What is the Chi-Square Test? Market researchers use the Chi-Square test when they find themselves in one of the following situations: They need to estimate how closely an observed distribution matches an expected distribution.
A contingency table also known as a cross-tabulation , crosstab , or two-way table is an arrangement in which data is classified according to two categorical variables. The categories for one variable appear in the rows, and the categories for the other variable appear in columns.
Each variable must have two or more categories. Each cell reflects the total count of cases for a specific pair of categories. There are several tests that go by the name "chi-square test" in addition to the Chi-Square Test of Independence. Look for context clues in the data and research question to make sure what form of the chi-square test is being used.
The Chi-Square Test of Independence can only compare categorical variables. It cannot make comparisons between continuous variables or between categorical and continuous variables. Additionally, the Chi-Square Test of Independence only assesses associations between categorical variables, and can not provide any inferences about causation.
If your categorical variables represent "pre-test" and "post-test" observations, then the chi-square test of independence is not appropriate. This is because the assumption of the independence of observations is violated. In this situation, McNemar's Test is appropriate. The null hypothesis H 0 and alternative hypothesis H 1 of the Chi-Square Test of Independence can be expressed in two different but equivalent ways:.
H 0 : "[ Variable 1 ] is independent of [ Variable 2 ]" H 1 : "[ Variable 1 ] is not independent of [ Variable 2 ]". H 0 : "[ Variable 1 ] is not associated with [ Variable 2 ]" H 1 : "[ Variable 1 ] is associated with [ Variable 2 ]".
There are two different ways in which your data may be set up initially. The format of the data will determine how to proceed with running the Chi-Square Test of Independence. At minimum, your data should include two categorical variables represented in columns that will be used in the analysis. The categorical variables must include at least two groups. Your data may be formatted in either of the following ways:. An example of using the chi-square test for this type of data can be found in the Weighting Cases tutorial.
Recall that the Crosstabs procedure creates a contingency table or two-way table , which summarizes the distribution of two categorical variables. A Row s : One or more variables to use in the rows of the crosstab s. You must enter at least one Row variable. B Column s : One or more variables to use in the columns of the crosstab s. You must enter at least one Column variable. Also note that if you specify one row variable and two or more column variables, SPSS will print crosstabs for each pairing of the row variable with the column variables.
The same is true if you have one column variable and two or more row variables, or if you have multiple row and column variables. A chi-square test will be produced for each table. Additionally, if you include a layer variable, chi-square tests will be run for each pair of row and column variables within each level of the layer variable. C Layer: An optional "stratification" variable.
If you have turned on the chi-square test results and have specified a layer variable, SPSS will subset the data with respect to the categories of the layer variable, then run chi-square tests between the row and column variables.
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