In: Statistics and Probability
Use the following contingency table to complete.
A | B | C | Total | |
1 | 10 | 25 | 55 | 90 |
2 | 40 | 50 | 60 | 150 |
Total | 50 | 75 | 115 | 240 |
a. Compute the expected frequencies for each cell.
b. Set up the null and alternative hypotheses to test.
c. Compute X2STAT Is it significant at α=0.01?
d. Compute P-Value
e.Is X2STAT significant at α=0.01?
(a) The second table contains the expected values.
Chi-Square Independence test - Results |
(b) Null and Alternative
Hypotheses The following null and alternative hypotheses need to be tested: H0: The two variables are independent Ha: The two variables are dependent This corresponds to a Chi-Square test of independence. Degrees of Freedom The number of degrees of freedom is df = (2 - 1) * (3 - 1) = 2 Critical value and Rejection Region Based on the information provided, the significance level is α=0.01, the number of degrees of freedom is df = (2 - 1) * (3 - 1) = 2, so the critical value is 9.2103. Then the rejection region for this test becomes R={χ2:χ2>9.2103}. (c)Test Statistics The Chi-Squared statistic is computed as follows: (d)P-value The corresponding p-value for the test is p=Pr(χ2>12.3208)=0.0021 (e)The decision about the null hypothesis Since it is observed that χ2=12.3208>χ2_crit=9.2103, it is then concluded that the null hypothesis is rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.01 significance level. Conditions: a. The sampling method is simple random sampling. b. The data in the cells should be counts/frequencies c. The levels (or categories) of the variables are mutually exclusive. |