Row TotalNeitherSignificance Level: let’s use 0.01Topic (16) – Further Tests Comparing Two Populations 16-1 Topic (16) – FURTHER TESTS COMPARING TWO POPULATIONS A) Testing Two Population Variances, versus 21σ22σ In the t-test of two means using independent samples we had to address the issue of whether the two populations had the same variance or not. Here we learn the test for determining whether the variances of two populations are the same using two independently collected samples. An underlying assumption to the following test is that the two populations being tested are Normally distributed. We just saw a test for determining whether the sample support the argument that the population has a Normal distribution and can be used to test that before running this test. To test hypotheses about population variances we look at the ratio of the two sample variances: 2min2max*ssF = where > 2maxs2mins . Hence, the larger sample variance is always put in the numerator.Topic (16) – Further Tests Comparing Two Populations 16-2 This test statistic, F*, has a sampling distribution known as the F-distribution with two sets of degrees of freedom, the numerator and the denominator degrees of freedom. The numerator df are nmax – 1 (sample size for minus 1) and the denominator df are n2maxsmin – 1 (sample size for 2mins minus 1). Note that nmax need not be larger than nmin! The F-distribution is positively skewed with a long right tail and whose shape depends on the two df values. It is a probability distribution for random variables whose values are > 0 (like the Chi-Square distribution). Like the chi-square distribution, we use a table of cutoff values to determine whether to reject the null hypothesis, H0: . Here, we are using a slight variant of the F – test, known as Hartley’s F-max test. Hartley’s test can be used for more than two variances, in fact (we’ll see an example later). 2221σσ= Hartley’s F-Max Test Of Equality Of Two Population Variances Based On Two Independent Samples: Null hypothesis: H0: 2221σσ=Alternative Hypothesis: HA: 2221σσ≠Topic (16) – Further Tests Comparing Two Populations 16-3 Test Statistic: 2min2max*ssF= where > 2maxs2mins The numerator df are the df for and the denominator 2maxsdf are the df for 2mins . Decision Rule: a) reject H0 if F* > tabulated value for α, and den df. Denominator df α = 0.05 α = 0.012 39.0 199 3 15.4 47.5 4 9.6 23.2 5 7.15 14.9 6 5.82 11.1 7 4.99 8.89 8 4.43 7.5 9 4.03 6.54 10 3.72 5.85 12 3.28 4.91 15 2.86 4.07 20 2.46 3.32 30 2.07 2.63 60 1.67 1.96 • 1 1 EXAMPLE A wildlife biologist is interested in comparing the variability in weights for two populations of deer: those raised in the wild and those raised in a zoo. She randomly selected eight deer from each population and weighed them (lbs) at the age of 1 year. The data are:Topic (16) – Further Tests Comparing Two Populations 16-4 W 114.7 W 128.9 W 111.5 W 116.4 W 134.5 W 126.7 W 120.6 W 129.6 Z 103.1 Z 90.7 Z 129.5 Z 75.8 Z 182.5 Z 76.8 Z 87.3 Z 77.3 In JMP, several tests that variances are equal 010203040Std DevWZPopulation Level Count Std Dev MeanAbsDif to MeanMeanAbsDif to MedianW 8 8.23424 7.06250 7.06250Z 8 36.85286 26.61875 23.57500 Test F Ratio DFNum DFDen Prob > F O'Brien[.5] 1.9667 1 14 0.1826 Brown-Forsythe 2.2853 1 14 0.1528 Levene 5.4718 1 14 0.0347 Bartlett 11.1616 1 . 0.0008Topic (16) – Further Tests Comparing Two Populations 16-5 These are valid tests, especially for testing equality of variances in several populations, but let’s stick with the F-max test. Test statistic: 05.2023.885.36222min2max*===ssF Choose 05.0=α. From the table, with the denominator df = 8-1 = 7, we have a cutoff value of 4.99. Decision: Since F* = 20.05 > cutoff = 4.99, we reject the null hypothesis and conclude that the two populations of deer, those raised in the wild and those raised in zoos, differ in the variability of their weights at 1 year of age.Topic (16) – Further Tests Comparing Two Populations 16-6 B) Testing Homogeneity: Comparing Multinomial Proportions for Two (Or More) Populations Recall that earlier in the semester we learned how to compare the population proportion of successes for two populations. Here we extend that to comparing two populations where the variable of interest has several categories, not just success and failure. This is a single expansion of the hypothesis test we learned earlier in topic 14. First, let’s discuss how to display the type of information we are interested in testing: We are interested in displaying and summarizing the RELATIONSHIP between two categorical variables. 1) Tabular Summary Recall in topic 3 that we displayed a single categorical variable in a one-way frequency table as counts in each category. Bivariate categorical data are displayed in TWO-WAY FREQUENCY or CONTINGENCY TABLES. The table is called an R×C CONTINGENCY TABLE when one variable has R categories and the other has C categories.Topic (16) – Further Tests Comparing Two Populations 16-7 The rows are the categories for one variable (X) and the columns are the categories for the other variable (Y). Each cell in the table contains the frequency of sampled units having that (x,y) combination of categories. The variables do not have to be designated as the response and the explanatory variable. EXAMPLE: College students and marijuana use The original dataset would have looked like:Topic (16) – Further Tests Comparing Two Populations 16-8 IDnum Student Parent 11002 never neither 11003 never one 11004 regular neither 11005 occasional both Observed Cell Counts Grand Total Column Row Marginal Marginal Totals TotalsTopic (16) – Further Tests Comparing Two Populations 16-9 Hard to interpret these raw counts. Hence, all of the counts are usually converted to various kinds of relative frequencies as well. “Frequency” = the observed cell count “Percent” = percent of
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