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Tenured Nontenured
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Do these data provide good statistical evidence that tenured and nontenured faculty differ in their attitudes toward the proposed salary increase
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Inference for Categorical Data 289
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solution: I Let p1 = the proportion of tenured faculty who favor the plan and let p2 = the proportion of nontenured faculty who favor the plan H0: p1 = p2 HA: p1 p2 II We will use a chi-square test for homogeneity of proportions The samples of tenured and nontenured instructors are given as random We determine that the expected values are given by the matrix 1278 1222 Because all expected values are 1022 978 greater than 5, the conditions for the test are present III X 2 = 178, df = (2 1)(2 1) = 1 015 < P-value < 020 (from Table C; on the TI-83/84: P = 2 cdf(178,1000,1) = 0182) IV The P-value is not small enough to reject the null hypothesis These data do not provide strong statistical evidence that tenured and nontenured faculty differ in their attitudes toward the proposed salary plan X 2 = z2 The example just completed could have been done as a two-proportion z-test where p1 and p2 are defined the same way as in the example (that is, the proportions of tenured and nontenured staff that favor the new plan) Then 15 p1 = 25 and 8 p2 = 20 Computation of the z-test statistics for the two-proportion z-test yields z = 1333 Now, z2 = 178 Because the chi-square test and the two-proportion z-test are testing the same thing, it should come as little surprise that z2 equals the obtained value of X 2 in the example For a 2 2 table, the X 2 test statistic and the value of z2 obtained for the same data in a two-proportion z-test are the same Note that the z-test is somewhat more flexible in this case in that it allows for the possibility of a one-sided test, whereas the chi-square test is strictly two sided (that is, it is simply a rejection of H0) However, this advantage only holds for a 2 2 table since there is no way to use a simple z-test for situations with more than two rows and two columns Digression: You aren t required to know these, but you might be interested in the following (unexpected) facts about a 2 distribution with k degrees of freedom:
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The mean of the 2 distribution=k The median of the 2 distribution (k > 2) k 2 3 The mode of the 2 distribution=k 2 The variance of the 2 distribution=2k
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290 U Step 4 Review the Knowledge You Need to Score High
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1 A study yields a chi-square statistic value of 20 (X 2 = 20) What is the P-value of the test if a the study was a goodness-of-fit test with n = 12 b the study was a test of independence between two categorical variables, the row variable with 3 values and the column variable with 4 values Answer: a n = 12 df = 12 1 = 11 0025 < P < 005 (Using the TI-83/84: 2cdf (20,1000,11) = 0045) b r = 3, c = 4 df = (3 1)(4 1) = 6 00025 < P < 0005 (Using the TI-83/84: 2 cdf(20,1000,6) = 00028) 2 4 The following data were collected while conducting a chi-square test for independence: Preference
BRAND A BRAND B BRAND C
Male Female
16 18 (X)
22 30
15 28
2 What null and alternative hypotheses are being tested Answer: H0: Gender and Preference are independent (or: H0: Gender and Preference are not related) HA: Gender and Preference are not independent (HA: Gender and Preference are related) 3 What is the expected value of the cell marked with the X Answer: Identifying the marginals on the table we have 16 18 (X) 34 22 30 52 15 28 43 53 76 129
Since there are 34 values in the column with the X, we expect to find 34 of each row 129 total in the cells of the first column Hence, the expected value for the cell containing 34 X is (76) = 2003 129 4 How many degrees of freedom are involved in the test Answer: df = (2 1)(3 1) = 2
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