1. PSY 307 – Statistics for the Behavioral Sciences Chapter 16 – One-Way ANOVA (Cont.)

2. Sleep Deprivation Example  Research Problem: On average, are subjects’ aggression scores in a controlled social situation affected by sleep deprivation periods of 0, 24, or 48 hours?  State the hypotheses: H 0 :  0 =  24 =  48 H 1 : H 0 is false

3. Format for Data Entry 1 0.00.0 2 0.04.0 3 0.02.0 4 24.03.0 5 24.06.0 6 24.06.0 7 48.06.0 8 48.08.0 9 48.010.0

4. Example (Cont.)  Decision Rule: Reject H 0 at.05 level of significance if F ≥ 5.14 (from F table), given df between = 2 and df within = 6.  Calculations: Use formulas to calculate F(2,6) = 7.36  Decision: Reject H 0 at the.05 level of significance because 7.36 exceeds 5.14

5. SPSS Results (Example)

6. Example (Cont.)  Interpretation of Results: Hours of sleep deprivation do affect the subjects’ mean aggression scores in a controlled social situation.  Calculate Effect Size: Divide SS between by SS total to get  2  Graph the data to see where the significant group differences are.

7. Graph of Data (Example)

8. Example (Cont.)  Perform post-hoc t-tests or planned comparisons to find differences between pairs of means.  Hypotheses for Post-Hoc test: H 0 :  0 –  48 = 0 H 1 :  0 –  48 ≠ 0  Decision Rule: Reject H 0 at the.05 level of significance if t ≥ 2.78 (from t-table).

9. SPSS Output for t-Test This is the relevant result.

10. Example (Cont.)  Calculations: Use formula for independent samples t- test to calculation t(6) = -3.67.  Decision: Reject H 0 at the.05 level of significance because -3.67 exceeds -2.78.  Interpretation: Sleep deprivation for 48 hrs results in greater aggression.

11. Multiple Comparisons  The regular t-test assumes that a single comparison is being made.  Use of multiple comparisons increases the likelihood of a Type I error.  To prevent this, the cumulative probability of all tests is used when determining the critical value for multiple t-tests.