PS296 Lecture Notes - Lecture 18: Null Hypothesis, Statistical Hypothesis Testing
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Ssbetween = 15. 000 df between groups (numerator) = k 1 = 2 df within groups (denominator) n k = 12. Now, look up the f value in the table e. 3 (alpha = . 05) or table e. 4 (alpha = . 01) For alpha = . 05, with df = 2, 12, f. 05 = 3. 89. For alpha = . 01, with df = 2, 12, f. 01 = 6. 93. Because our obtained value of f = 11. 25 exceeds the critical value of f. 05 = 3. 89 and f. 01 = 6. 93, we will reject the null hypothesis and conclude that at least one mean is different from another mean. Therefore, f(2, 12) = 11. 25, p < . 01. Our analysis of variance at alpha = . 05 testing the effect of room temperature (10, 20, and 30 degrees c) on learning performance yielded a significant result, f (2, 12) = 11. 25, p < . 01. At least one of the condition means (m10degrees = 1. 00, m20degrees = 4. 00,
