ANOVA Part B Grading Key 3 (Assigned 2/17; Due 2/24; Returned  3/3) 

(13.2) Multiple comparisons at a = .01
(A) Pr(all three correct) = .99³ = .9703; Pr(at least one wrong) = 1 - .9703 = .0297 [family-wise error rate acceptable]
(B) Pr(all 10 correct) = .99¹⁰ = .9044; Pr(at least one wrong) = 1 - .9044 = .0956 [family-wise error rate fair]
[1] (C) Pr(all 20 correct) = .99²⁰ = .8179; Pr(at least one wrong) = 1 - .1821 = .1821 [family-wise error rate too big]
{Comment: Use two to four significant figures when calculating probabilities.}

(13.6) Fever reduction trial.  Mean reduction in body temp (degrees F)
Group 1 (aspirin): 1.5
Group 2 (aspirin/acetaminophen combination): 0.36
Group 3 (acetaminophen): 0.08
n = 5 (each group)
[2] se_{mean dif} = sqrt(0.4338 * (1/5 + 1/5)) = 0.4166 for each pair-wise comparison
[3,4] Test 1: H₀: �₁ = �₂; t_stat = (1.5 - 0.36) / 0.4166 = 2.7364, df = 12; p_Bonf = .01804 � 3 @ .055
[5,6] Test 2: H₀: �₁ = �₃; t_stat = (1.5 - 0.08) / 0.4166 = 3.4085, df = 12; p_Bonf = .005188 � 3 @ .016
[7,8] Test 3: H₀: �₂ = �₃; t_stat = (0.36 - 0.08) / 0.4166 = 0.6721, df = 12; p_Bonf = .0.5143 � 3 = an answer that is more than 1, but since you cannot have a probability that exceeds "certainty," it should be reported as 1.0. {Many students accepted the numerical answer of 1.543 -- a Cargo Cult. The answer of p > 1.0 should have been self-revealed as meaningless. IMHO, there should have been some chatter on the electronic BBS about this problem.

Interpretation: Aspirin is a little better than the aspirin/acetaminophen combination, and is a lot better than acetaminophen alone.  

 (13.8) Influence of smoking on birth weight

(A) Rebuttals to unequal variance argument

• Sample standard deviations are about the same; largest s = 1.17, smallest s = 0.72
• Levene's test finds no significant difference: F_3,21 = 0.156; p = .92 [I have reported df as subscripts; APA recommends normal font in parentheses]
• ANOVA can withstand violations in the equal variance assumption 
• p value are not meant to be interpreted literally

(B) Kruskal-Wallis test of H₀: population medians are equal

[9] KW output should be shown. Test results may be reported as c²₃ = 9.25, p = .026 {modified APA style; df as subscript instead of in parentheses.}

[10] In using the K-W test, we reject the H₀ at a = .05; groups differ significantly. 

(C) Post hoc comparisons - I opted for an LSD post hoc. You might have opted for a different technique. The LSD method can be justified in terms of prior interested in all possible comparisons.  

Before doing the post hoc comparison, you should list group means (so you know what you are testing)

                          n    Mean    StdDev
nonsmokers   6   7.5833   1.0534
ex-smokers   5   7.2400   0.9127
half-pack    6   6.1833   1.1754
full pack    8   6.0125   0.7200
Total        25  6.6760   1.1352

and recall ANOVA results:

             SS   df      MS     F     p  
Between  11.508    3   3.836  4.15    .019
Within   19.417   21   0.925

LSD p values for each comparison are:

H₀: �₁ = �₂  (p = .56)
H₀: �₁ = �₃ (p = .020)*
H₀: �₁ = �₄ (p = .0064)*
H₀: �₂ = �₃ (p = .084) �
H₀: �₂ = �₄ (p = .036)*
H₀: �₃ = �₄ (p = .075) �

Comments:

• At the a = .05 level, only three comparisons are significant (identified with *)
• At a = .10, an addition 2 comparisons are significant (�)
• If we apply a Bonferroni adjustment, only one of the comparison is significant (H₀: �₁ = �₄; p_Bonf = .0064 � 6 = .038)
• How can we have these apparently paradoxical conclusions? Like any paradox, there is nothing paradoxical once we see what has happened. Two keys to seeing is to understand the p value as a nonobjective probability. It is not the probability the null hypothesis is correct. It is the probability of the observed data assuming very specific conditions, such as the assumption that H₀ is true (an absurd assumption). The probability changes whether we consider test-by-test error or family-wise error. 

1. You, the student, can appreciate this type and level of thought as a beautiful subject
2. I, the teacher, really want you to learn this level and way of thinking
3. You, the student, are capable of learning and using the subject with a measure of autonomy
4. The overall goal of this art is intellectual autonomy