Chi-Square Test of Independence

There are web pages that will perform the chi-square test and . None of these web pages lets you set the degrees of freedom to the appropriate value for testing an intrinsic null hypothesis.

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The result is chi-square=7.26, 2 d.f., P=0.027. This indicates that you can reject the null hypothesis;the three genotypes have significantly different proportions of men with coronary artery disease.

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McDonald (1989) examined variation at the Mpi locus in the amphipod crustacean Platorchestia platensis collected from a single location on Long Island, New York. There were two alleles, Mpi90 and Mpi100 and the genotype frequencies in samples from multiple dates pooled together were 1203 Mpi90/90, 2919 Mpi90/100, and 1678 Mpi100/100. The estimate of the Mpi90 allele proportion from the data is 5325/11600=0.459. Using the Hardy-Weinberg formula and this estimated allele proportion, the expected genotype proportions are 0.211 Mpi90/90, 0.497 Mpi90/100, and 0.293 Mpi100/100. There are three categories (the three genotypes) and one parameterestimated from the data (the Mpi90allele proportion), so there is one degree of freedom. The result is chi-square=1.08, 1 d.f., P=0.299, which is not significant. You cannot reject the null hypothesis that the data fit the expected Hardy-Weinberg proportions.

How to test in Excel whether two categorical random variables are independent. Data is organized in a contingency table and tested using a chi-square test.

The chi-square test of goodness-of-fit is an alternative to the ; each of these tests has some advantages and some disadvantages, and the results of the two tests are usually very similar. You should read the section on near the bottom of this page, pick either chi-square or G–test, then stick with that choice for the rest of your life. Much of the information and examples on this page are the same as on the G–test page, so once you've decided which test is better for you, you only need to read one.

Definition of Mean Squared Error. Sample problem for finding the Mean Squared Error. Statistics explained simply. Step by step videos, articles.

The chi-square test of independence is an alternative to the , and they will give approximately the same results. Most of the information on this page is identical to that on the G–test page. You should read the section on , pick either chi-square or G–test, then stick with that choice for the rest of your life.

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As with most test statistics, the larger the difference between observed and expected, the larger the test statistic becomes. To give an example, let's say your null hypothesis is a 3:1 ratio of smooth wings to wrinkled wings in offspring from a bunch of Drosophila crosses. You observe 770 flies with smooth wings and 230 flies with wrinkled wings; the expected values are 750 smooth-winged and 250 wrinkled-winged flies. Entering these numbers into the equation, the chi-square value is 2.13. If you had observed 760 smooth-winged flies and 240 wrinkled-wing flies, which is closer to the null hypothesis, your chi-square value would have been smaller, at 0.53; if you'd observed 800 smooth-winged and 200 wrinkled-wing flies, which is further from the null hypothesis, your chi-square value would have been 13.33.