Chapter 13

• 13.1 For the one-way between-groups ANOVA, we calculate one kind of variability that occurs within groups: within-groups variability. But for the one-way within-groups ANOVA, the variability that occur within groups is divided into two types: subjects variability and within-groups variability. Subjects variability assesses how much each person’s overall mean differs from the others’, assessed by comparing each person’s mean score to the grand mean. We can remove this from the within-groups variability because we have the same people in each group, and they would be the same across levels of the independent variable on many variables that may affect the dependent variable. This makes the within-groups variability smaller.
• 13.2
  • a. df_between = N_groups − 1 = 3 − 1 = 2
  • b. df_subjects = n − 1 = 3 − 1 = 2
  • c. df_within = (df_between)(df_subjects) = (2)(2) = 4
  • d. df_total = df_between + df_subjects + df_within = 2 + 2 + 4 = 8; or we can calculate it as df_total = N_total − 1 = 9 − 1 = 8
• 13.3
  • a. SS_total = Σ(X − GM)² = 24.886
    

    GROUP	RATING (X)	X − GM	(X − GM)2
    1	7	0.111	0.012
    1	9	2.111	4.456
    1	8	1.111	1.234
    2	5	−1.889	3.568
    2	8	1.111	1.234
    2	9	2.111	4.456
    3	6	−0.889	0.790
    3	4	−2.889	8.346
    3	6	−0.889	0.790
    GM = 6.889                  Σ(X − GM)2 = 24.886

  • b. SS_between = Σ(M − GM)² = 11.556
    

    GROUP	RATING (X)	GROUP MEAN (M)	M − GM	(M − GM)2
    1	7	8	1.111	1.234
    1	9	8	1.111	1.234
    1	8	8	1.111	1.234
    2	5	7.333	0.444	0.197
    2	8	7.333	0.444	0.197
    2	9	7.333	0.444	0.197
    3	6	5.333	−1.556	2.421
    3	4	5.333	−1.556	2.421
    3	6	5.333	−1.556	2.421
    GM = 6.889                  Σ(M − GM)2 = 11.556

  • c. SS_subject = Σ(M_participant − GM)² = 4.221
    

    PARTICIPANT	GROUP	RATING (X)	PARTICIPANT MEAN (MPARTICIPANT)	MPARTICIPANT − GM	(MPARTICIPANT − GM)2
    1	1	7	6	−0.889	0.790
    2	1	9	7	0.111	0.012
    3	1	8	7.667	0.778	0.605
    1	2	5	6	−0.889	0.790
    2	2	8	7	0.111	0.012
    3	2	9	7.667	0.778	0.605
    1	3	6	6	−0.889	0.790
    2	3	4	7	0.111	0.012
    3	3	6	7.667	0.778	0.605
    GM = 6.889                        Σ(Mparticipant − GM)2 = 4.221

  • d. SS_within = SS_total − SS_between − SS_subjects = 24.886 − 11.556 − 4.221 = 9.109
  • 13.4
    

    SOURCE	SS	df	MS	F
    Between-groups	11.556	2	5.778	2.54
    Subjects	4.221	2	2.111	0.93
    Within-groups	9.109	4	2.277
    Total	24.886	8

• 13.5
  • a. Null hypothesis: People rate the driving experience of these three cars the same, on average—H₀: μ₁ = μ₂ = μ₃. Research hypothesis: People do not rate the driving experience of these three cars the same, on average.
  • b. Order effects are addressed by counterbalancing. We could create a list of random orders of the three cars to be driven. Then, as a new customer arrives, we would assign him or her to the next random order on the list. With a large enough sample size (much larger than the three participants we used in this example), we could feel confident that this assumption would be met with this approach.
  • c. The critical value for the F statistic for a p level of 0.05 and 2 and 4 degrees of freedom is 6.95. The between-groups F statistic of 2.54 does not exceed this critical value. We cannot reject the null hypothesis, so we cannot conclude that there are differences among mean ratings of cars.
• 13.6 In both cases, the numerator in the ratio is SS_between, but the denominators differ in the two cases. For the between-groups ANOVA, the denominator of the R² calculation is SS_total. For the within-groups ANOVA, the denominator of the R² calculation is SS_total − SS_subjects, which takes into account the fact that we are subtracting out the variability caused by differences among participants (subjects sum of square) from the measure of error.
• 13.7 There are no differences in the way that the Tukey HSD is calculated. The formula for the calculation of the Tukey HSD is exactly the same for both the between-groups ANOVA and the within-groups ANOVA.
• 13.8
  • a. First, we calculate s_M:
    
    Next, we calculate HSD for each pair of means. For time 1 versus time 2:
    
    For time 1 versus time 3:
    
    For time 2 versus time 3:
    
  • b. We have an independent variable with three levels and df_within = 10, so the q cutoff value at a p level of 0.05 is 3.88. Because we are performing a two-tailed test, the cutoff values are 3.88 and − 3.88.
  • c. We reject the null hypothesis for all three of the mean comparisons because all of the HSD calculations exceed the critical value of − 3.88. This tells us that all three of the group means are statistically significantly different from one another.
• 13.9
• 13.10
  • a.
    This is a large effect size.
  • b. Because the F statistic did not exceed the critical value, we failed to reject the null hypothesis. As a result, Tukey HSD tests are not necessary.