TheCaseoftheCarelessZookeeper.pdf

# Since the students should have some background

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questions 17 to 21, which will outline a chi-square test of independence. Since the students should have some background knowledge on chi-square tests of independence, they should be able to write the hypotheses correctly. The null hypothesis is 𝐻?: 𝐴???𝑎? ???? 𝑎?? ?????? ??𝑎??? 𝑎?? ??????????? and the alternative hypothesis is 𝐻𝑎: 𝐴???𝑎? ???? 𝑎?? ?????? ??𝑎??? 𝑎?? ????????? . With the hypotheses stated, the students need to determine if the chi-square test is appropriate. The two major assumptions the students

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_____________________________________________________________________________________________ ST atistics E ducation W eb: Online Journal of K-12 Statistics Lesson Plans 5 Contact Author for permission to use materials from this STEW lesson in a publication need to check revolve around expected cell counts. The expected cell counts stem from the idea of conditional probabilities and the definition of independence from a probability perspective. In general, to find the expected cell count for any cell, the students need to multiply the row total and column total for a given cell and then divide that by the overall total. So the expected cell counts for the table of example data would be: Expected Cell Counts Injury Status Row Totals Type of Animal Damaged Not Damaged Carnivore (f) 7 ∗ 8 23 = 2.435 (g) 7 ∗ 15 23 = 4.565 (b) 7 Herbivore (h) 16 ∗ 8 23 = 5.565 (i) 16 ∗ 15 23 = 10.435 (c) 16 Column Totals (d) 8 (e) 15 (a) 23 All of these expected cell counts need to be greater than 1 and at least 80% of the cell counts must be greater than 5. Clearly, in this example, 50% of the expected cell counts are less than 5 so the assumptions are not met. Because a basic statistics c ourse does not address Fisher’s Exact Test, have the students continue with the activity noting that the assumptions aren’t met and the results may not be appropriate. The students will need these expected cell counts to calculate the test statistic by hand. This is rather simple to do because the chi-square test statistic is calculated using the equation, 2 2 (observed expected) expected . Applying this equation to the sample data, the test statistic is, 2 2 2 2 2 3 2.435 4 4.565 5 5.565 11 10.435 .289. 2.435 4.565 5.565 10.435 On the TI-84 PLUS calculator, students can use the test statistic value to find the corresponding p-value. Select 2 nd DISTR 2 cdf( ENTER. Within the parentheses, the students need to enter the lower bound, upper bound, degrees of freedom. The lower bound will always be the test statistic due to the shape of the chi-square distribution. For the upper bound, students can enter any very large number such as 10000000. The degrees of freedom are found using the formula ( 1)( 1), df r c where r is the number of rows in the table and c is the number of columns. Note that in a two-by-two table, the degrees of freedom always equal 1. For this
_____________________________________________________________________________________________ ST atistics E ducation W eb: Online Journal of K-12 Statistics Lesson Plans 6 Contact Author for permission to use materials from this STEW lesson in a publication example, the p-value is .591. Since this value is far greater than .05, students should fail to reject the null hypothesis.

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