Consider the following 3 times 3 contingency table: You use these data to test t

Question

Consider the following 3 times 3 contingency table: You use these data to test the null hypothesis that the genotypes at the A-locus are independent of levels of fitness. Here, fitness is defined as the ability to live and produce lots of offspring. Using these data, the chi-squared statistic is The degrees of freedom are If the corresponding critical value is 9.487729, you would the null hypothesis. For the chi-squared statistic, please carry your answer to two decimal places and follow normal rounding procedures (5 and below, round down; > 5 round up). For the degrees of freedom, provide a numeric values that is an integer (e.g. 1, 2, 3, 4, 5, 6, etc.).

Explanation / Answer

144  (Grand Total

The chi-square statistic is 72.3542. The p-value is < 0.00001. The result is significant at p < .05.

Degrees of freedom is number of categories minus 1 = 3-1=2

Since, chi-square calculated value is greater than thechi-square critical value, then we reject your null hypothesis.

Category 1 Category 2 Category 3 Row Totals Group 1 39  (18.03)  [24.37] 8  (15.65)  [3.74] 2  (15.31)  [11.57] 49 Group 2 11  (14.35)  [0.78] 18  (12.46)  [2.47] 10  (12.19)  [0.39] 39 Group 3 3  (20.61)  [15.05] 20  (17.89)  [0.25] 33  (17.50)  [13.73] 56 Column Totals 53 46 45

144  (Grand Total

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