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Lesson 12 - 3

Lesson 12 - 3. Tests for Independence and the Homogeneity of Proportions. Objectives. Perform a test for independence Perform a test for homogeneity of proportions. Vocabulary.

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Lesson 12 - 3

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  1. Lesson 12 - 3 Tests for Independence and the Homogeneity of Proportions

  2. Objectives • Perform a test for independence • Perform a test for homogeneity of proportions

  3. Vocabulary • Chi-Squared Test for Independence – used to determine if there is an association between a row variable and a column variable in a contingency table constructed from sample data • Expected Frequencies – row total * column total / table total • Chi-Squared Test for Homogeneity of Proportions – used to test if different populations have the same proportions of individuals with a particular characteristic

  4. Terms (row total)(column total) Expected Frequency = ---------------------------------- table total

  5. Requirements Goodness-of-fit test: • All expected counts are greater than or equal to 1 (all Ei ≥ 1) • No more than 20% of expected counts are less than 5

  6. Σ (Oi – Ei)2 Test Statistic: χ20 = ------------- Ei Goodness-of-Fit Test P-Value is thearea highlighted P-value = P(χ2 0) χ2α Critical Region where Oi is observed count for ith category and Ei is the expected countfor the ith category (Right-Tailed)

  7. Chi-Square Tests on TI • Access MATRX menu (2nd X-1) • Highlight EDIT menu and select 1: [A] • Enter the number of rows and columns of the matrix • Enter the cell entries for the observed data and press 2nd QUIT • Repeat steps for expected values in matrix B • Press STAT, highlight TESTS and select C: χ²-Test • With cursor after the Observed: enter matrix [A] by accessing the MATRX menu, highlighting NAMES, and selecting 1: [A] • With cursor after the Expected: enter matrix [B] • Highlight Calculate and press ENTER

  8. Example

  9. Summary and Homework • Summary • Often, in contingency tables, we wish to test specific relationships, or lack of, between the two variables • The test for independence analyzes whether the row and column variables are independent • The test for homogeneity analyzes whether the observed proportions are the same across the different categories • Homework • pg 662 - 667: 1, 4, 5, 11, 12, 16

  10. Even Homework Answers • 4: a) the chi-square test statistic is 13.049b) chi-square critical value (df = (3-1)*(2-1)=2, α=0.05) = 5.991 so we would rejectc) p-value = 0.001467 • 12: a) FTR H0, not enough evidence to support the claim that abortion opinion and gender are independent ; the chi-square test statistic is 0.036318; chi-square critical value (df = (2-1)*(2-1)=1, α=0.1) = 2.706 so we would reject b) p=value = 0.84886 Pro-Life M49.62% F48.98% M50.38% F51.02% Pro-Choice Male Female

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