In: Statistics and Probability
If a researcher wanted to test whether attending class influences how students perform on an exam, he collected needed data and decided that the alpha level is 0.05, so the critical value is 3.841 (4 mark)
Pass |
Fail |
Total |
|
attended |
25 |
6 |
31 |
skipped |
8 |
15 |
23 |
Total |
33 |
21 |
54 |
The researcher calculated the expected frequency for each cell and came up with these numbers in parenthesis
Pass |
Fail |
Total |
|
attended |
25 (18.94) |
6 (12.05) |
31 |
skipped |
8 (14.05) |
15 (8.94) |
23 |
Total |
33 |
21 |
54 |
Calculate and write down the final conclusion of the researcher.
Chi-Square Independence test - Results |
(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H0: The two variables - Attending classes and performance are independent Ha: The two variables - Attending classes and performance are dependent This corresponds to a Chi-Square test of independence. (2) Degrees of Freedom The number of degrees of freedom is df = (2 - 1) * (2 - 1) = 1 (3) Critical value and Rejection Region Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df = (2 - 1) * (2 - 1) = 1, so the critical value is 3.8415. Then the rejection region for this test becomes R={χ2:χ2>3.8415}. (4)Test Statistics The Chi-Squared statistic is computed as follows: (5)P-value The corresponding p-value for the test is p=Pr(χ2>11.686)=0.0006 (6)The decision about the null hypothesis Since it is observed that χ2=11.686>χ2_crit=3.8415, it is then concluded that the null hypothesis is rejected. (7)Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables - Attending classes and performance are dependent, at the 0.05 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. |
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