Why does the test for homogeneity follow the same procedures a...

Question

Why does the test for homogeneity follow the same procedures as the test for independence?

Accepted Answer

Welcome back, everyone. In this problem, which of the following best explains why the chi square test for independence and the chi square test for homogeneity yield the same test statistic? A says it's because both tests use the mean and standard deviation of the data. B Both tests require equal sample sizes in each group. Both tests assume the data are measured on an interval scale. And the D that both tests compare observed and expected frequencies in a contingency table using the same formula. Now, for us to find the best explanation, let's think through what we do for both I square tests for independence and for homogeneity. Now if we think about it in both tests we are organizing categorical data into a contingency table or we have a cross tabulation of the counts, OK? So first, both tests we can write that here, both tests organize the categorical data. Into a contingency table. OK. Next, if we think about the name for both, both are, both use the name chi square test, and that's because they both use the test statistic formula. OK? Both use the test statistic formula. They use the same exact formula. To calculate the chi squared value which measures the overall discrepancy between the observed cones and the expected cones. In other words, they both use the formula, OK. That says we're finding the sum of the differences between the observed and expected values squared divided by the expected values. And now if we think about it, both also use the same formula for the degrees of freedom. OK? Now there's both use the formula that says the degrees of freedom are equal to the number of rows minus 1 multiplied by the number of columns minus 1. Then after that, they use the observed unexpected frequencies from a contingency table and calculate the expected cones as the row total multiplied by the column total divided by the grand total, OK. So they use the expected cones as the road total. Multiplied by the column total. OK. Divided by the grand total. So really if we consider both tests, their mathematical procedures are identical. So with that in mind, which of these would best explain why they yield the same test statistic? Well, if we look at A, where it says both tests use the mean and standard deviation of the data, neither tests use the mean and standard deviation. So that would be incorrect. In B, it says both tests require equal sample sizes in each group, but again neither test needs sample sizes. C says both tests assume the data are measured on an interval scale, but that's also incorrect because our case. Whereas tests are used for categorical, not interval data, so D must be the correct answer. Both tests compare observed and expected frequencies in a contingency table using the same formula, and we know that that is true. Thanks a lot for watching everyone. I hope this video helped.

Why does the test for homogeneity follow the same procedures as the test for independence?

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