ANOVA Table Coursework Example

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Category:
Statistics
Level:
High School
Type:
AP
Pages:
3
Words:
525
Given the aggression scores below for Outcome A of the sleep deprivation experiment, verify that, as suggested earlier, these mean differences shouldn’t be taken seriously by testing the null hypothesis at the .05 level of significance. Use the computation formulas for the various sums of squares and summarize results with an ANOVA table. HOURS OF SLEEP DEPRIVATION ZERO TWENTY-FOUR FORTY-EIGHT 342 584 766 564 Grand mean = 5 The F value was 0.75, while the p-value was 0.51. Since the p-value for the ANOVA was greater than 0.05 (p=0.51) it could be concluded that there was no significant depression in aggression scores across several levels of sleep deprivation. Hence, sleep deprivation should not be considered as a determinant of aggression scores with respect to study population that was included in the trial. *16.10 Another psychologist conducts a sleep deprivation experiment. For rea-sons beyond his control, unequal numbers of subjects occupy the differ-ent groups. (Therefore, when calculating in SS between and SS within, you must adjust the denominator term, n, to reflect the unequal numbers of subjects in the group totals.) (a) Summarize the results with an ANOVA table. You need not do a step-by-step hypothesis test procedure. ΣT2n HOURS OF SLEEPDEPRIVATION ZEROTWENTY-FOUR FORTY-EIGHT 147 3712 6610 29 1 ANOVA Summary  Source SS df MS F P Treatment[between groups] Error Ss/Bl (b) If appropriate, estimate the effect size with h2. (c) If appropriate, use Tukey’s HSD test (with –n 5 4 for the sample size, n) to identify pairs of means that contribute to the signifi cant F, given that –X0 5 2.60, –X24 5 5.33, and –X48 5 9.50. (d) If appropriate, estimate effect sizes with Cohen’s d. (e) Indicate how all of the above results would be reported in the literature, given sample standard deviations of s 0 5 2.07, s 24 5 1.53, and s 48 5 2.08. Turkey's HSD test results indicate that there was a significant differe...
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