11-2 Contingency Tables 595 Rationale for Expected Frequencies E To better understand expected frequencies, pretend that we know only the row and column totals in Table 11-6. Let’s assume that the row and column variables are independent and that 1 of the 1878 study subjects is randomly selected. First, note that 89 of the 1878 subjects had autism, and 387 of the 1878 subjects were unvaccinated, so we have the following: P1autism2 = 89>1878 and P1unvaccinated2 = 387>1878. If the row and column variables are independent, as we are assuming in the null hypothesis, we can use the multiplication rule for independent events (see Section 4-2) as follows: P1autismand unvaccinated2 = 89 1878 # 387 1878 = 0.00976584 With a probability of 0.00976584 for the first cell, we expect that among 1878 subjects, there are 1878# 0.00976584 = 18.340 subjects in the first cell. If we generalize these calculations, we get the following: Expected frequency E = 1grand total2 # 1row total2 1grand total2 # 1column total2 1grand total2 This expression can be simplified to E = 1row total2 1column total2 1grand total2 PART2 Test of Homogeneity, Fisher’s Exact Test, and McNemar’s Test for Matched Pairs Test of Homogeneity In Part 1 of this section, we focused on the test of independence between the row and column variables in a contingency table. In Part 1, the sample data are from one population, and individual sample results are categorized with the row and column variables. In a chi-square test of homogeneity, we have samples randomly selected from different populations, and we want to determine whether those populations have the same proportions of some characteristic being considered. (The word homogeneous means “having the same quality,” and in this context, we are testing to determine whether the proportions are the same.) Section 9-1 presented a procedure for testing a claim about two populations with categorical data having two possible outcomes, but a chi-square test of homogeneity allows us to use two or more populations with outcomes from several categories. DEFINITION A chi-square test of homogeneity is a test of the claim that different populations have the same proportions of some characteristics. Sampling from Different Populations In a typical test of independence, as described in Part 1 of this section, sample subjects are randomly selected from one population (such as people treated for stress fractures in a foot bone) and values of two different variables are observed (such as success>failure for people receiving different treatments). In a typical chi-square test of homogeneity, subjects are randomly selected from the different populations separately. Alternative to Clinical Trials Rheumatologist Jennifer Frankovich diagnosed a patient with lupus, but she noticed a specific combination of symptoms that had led to blood clots in the past. Her colleagues at the Stanford Packard Children’s Hospital recommended that she not treat with anti-clotting drugs, so she did research but could find no relevant studies. She then retrieved the data from all lupus patients treated in the hospital over the last five years and used basic statistics to find that her patient did have a higher risk of blood clots, so she then proceeded to treat with anti-clotting drugs. A randomized clinical trial with treatment and placebo groups would be better, but such trials are rarely conducted for such specific complications. ti f
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