Explanation Given info: The contingency table shows the number of U.S adults (in millions) ages 25 over by employment status and educational attainment. Calculation: The chi-square independence test will use when all the expected frequencies are greater than 5. The expected frequency is the joint frequencies in the contingency table represent the observed frequencies in the interior of a contingency table. Thus, the formula to find the expected frequency is, E = ( row total ) ( column total ) ( grand total ) Where, row total represents the sum of the frequencies along each row, the column total represents the sum of the frequencies along each column and the grand total represents the sum of all frequencies. Expected frequency for the cell E 11 is, E 11 = 132.3 × 23.5 215.1 = 3109.05 215.1 = 14.454 Expected frequency for the cell E 12 is, E 12 = 132.3 × 62 215.1 = 8202.6 215.1 = 38.134 Expected frequency for the cell E 13 is, E 13 = 132.3 × 36 215.1 = 4762.8 215.1 = 22.142 Expected frequency for the cell E 14 is, E 14 = 132.3 × 93.6 215.1 = 12 , 383.28 215.1 = 57.570 Expected frequency for the cell E 21 is, E 21 = 6 × 23.5 215.1 = 381 215.1 = 1.771 Expected frequency for the cell E 22 is, E 22 = 6 × 62 215.1 = 372 215

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ISBN: 9781323839577

Author: Larson

Publisher: Pearson Custom Publishing

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Chapter 10.2, Problem 34E

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To explain: The reason for the chi-square independence test is not using for the given data.

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The reason for the chi-square independence test is not using for the given data.

Solution Summary: The author explains why the chi-square independence test is not using for the given data. The expected frequency is the joint frequencies in the contingency table.

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