13: ANOVA-B Grading Key

Assigned 2/17; Due 2/24; Returned and discussed 3/3; [Grading sample points]

Comment: The grading key will be posted, but will not be distributed in order to save natural resources. I will mark and annotate students' papers. This was a long HW assignment. The class did exceedingly well as a whole.

(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 acceptable]
[1] (C) Pr(all 20 correct) = .99²⁰ = .8179; Pr(at least one wrong) = 1 - .1821 = .1821 [family-wise error rate too big]

Note: Probabilities are restricted to the range 0 to 1; list between two and four significant figures when reporting 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 is greater than 1, report it as 1.00 (I'm surprised there wasn't more chatter on the electronic BBS on this one)

Interpretation: Aspirin is marginally better than the aspirin/acetaminophen combination and is significantly better than acetaminophen alone. [An interpretation was not asked for explicitly; in the future, always interpret your findings.

(13.8) Influence of smoking on birth weight

(A) Rebuttal to a charge of unequal variance

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) Non-parametric test of H₀: population medians are equal

[9] Presentation of K-W output from SPSS

Test results may be reported as c²₃ = 9.25, p = .026 [modified APA style; df in parentheses seems archaic.]

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

(C) Post hoc comparisons - I went ahead with a straight LSD post hoc comparisons. This can be justified in terms of prior planned comparisons (I would want to compare all groups ahead of time.) You could also use a Bonferroni method of post hoc comparison, or any other way as long as your method was stated. Before doing the post hoc comparison I would remind myself what I was comparing by listing group means:

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:

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

With this background, I computed (with SPSS) the following LSD p values for each comparison:

1 v 2 (p = .56)
1 v 3 (p = .020)*
1 v 4 (p = .0064)*
2 v 3 (p = .084) �
2 v 4 (p = .036)*
3 v 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 (not shown), only one of the comparison is significant (1 v 4; p = .0064 � 6 = .038) and all the rest are insignificant
This may seem paradoxical, but recall that the p value is not the probability the null hypothesis is correct; it is the probability of the observed data assuming the null hypothesis is true; it is a probability of confidence in your evidence [to be discussed further in class]
Education is more valuable than training.