In: Statistics and Probability
11. The Scheffe test
Sleep apnea is a sleep disorder characterized by pauses in breathing during sleep. Children with sleep apnea have behavior problems, including hyperactivity, inattention, and aggression, as well as impaired learning and diminished academic performance. The removal of tonsils and adenoids that are enlarged, causing the obstruction of the airways, is one of the most common treatments for pediatric sleep apnea.
A clinical psychologist studies the effects of tonsillectomy and adenoidectomy on aggressive behavior. Her quasi-experiment includes three groups of 9 children. The first group of children does not have sleep apnea, the second group has untreated sleep apnea, and the third group has sleep apnea treated by tonsillectomies and adenoidectomies. Aggression was measured using parent reports on the Conners Rating Scale. The sample means and sums of squared deviations of the scores for each of the three groups are presented in the table that follows.
Group |
Sample Mean |
Sum of Squares |
---|---|---|
No sleep apnea | 0.26 | 0.2888 |
Untreated sleep apnea | 0.62 | 0.3872 |
Treated sleep apnea | 0.33 | 0.2312 |
After collecting the data, the clinical psychologist analyzes the data using an ANOVA. The results of her analysis are presented in the ANOVA table that follows.
ANOVA Table
Source of Variation |
Sum of Squares |
Degrees of Freedom |
Mean Square |
F |
---|---|---|---|---|
Between Treatments | 0.6561 | 2 | 0.3281 | 8.66 |
Within Treatments | 0.9072 | 24 | 0.0378 | |
Total | 1.5633 | 26 |
The critical value of F when α = 0.01 is 5.614, meaning the critical region consists of all F-ratios greater than 5.614. The F-ratio is greater than this critical value, so you know that at least one difference exists among the treatments. Since more than two groups are involved, the psychologist is interested in determining which groups are different. The Scheffe test will be used to evaluate the pairs. Call the no sleep apnea group A, the untreated sleep apnea group B, and the treated sleep apnea group C.
Start with the calculations you will need to evaluate the difference between the no sleep apnea group (A) and the untreated sleep apnea group (B). The SSbetween ABbetween AB is . The FA versus BA versus B is .
(Hint: Recall that you can use the sample size for each treatment (n = 9) and the treatment mean to compute each treatment total (T).)
At α = 0.01, the psychologist conclude that the population means for children without sleep apnea and children with untreated sleep apnea differ.
Next evaluate the difference between the no sleep apnea group (A) and the treated sleep apnea group (C). The SSbetween ACbetween AC is . The FA versus CA versus C is .
At α = 0.01, the psychologist conclude that the population means for children without sleep apnea and children with treated sleep apnea differ.
Next calculate the values necessary to evaluate the difference between the untreated sleep apnea group (B) and the treated sleep apnea group (C). The SSbetween BCbetween BC is . The FB versus CB versus C is .
At α = 0.01, the psychologist conclude that the population means for children with untreated sleep apnea and children with treated sleep apnea differ.
group | sample size | sample mean |
No sleep Apnea (A) | 9 | 0.26 |
Untreated Sleep Apnea (B) | 9 | 0.62 |
Treated Sleep Apnea ( C ) | 9 | 0.33 |
Df between= | 2 | |
MSwithin= | 0.0378 | |
Critical F = | 5.614 |
grand mean AB, x̅̅ = Σni*x̅i/Σni =
(9*0.26+9*0.62)/(9+9)=
0.44
SS(between AB) = Σn( x̅ - x̅̅)² =
9*(0.26-0.44)²+9*(0.62-0.44)²=
0.5832
MS(between AB)=SS(between AB)/df between=
0.5832/2= 0.2916
F (A vs B) = 0.2916/0.0378=
7.7143
since, F stat > critical F value, Reject the null
hypothesis
The SSbetween AB is 0.5832
The F (A versus B) is 7.7143
the psychologist can conclude that
the population means for children without sleep apnea and children
with untreated sleep apnea differ
============================================================================================
grand mean AC, x̅̅ = Σni*x̅i/Σni =
(9*0.26+9*0.33)/(9+9)=
0.295
SS(between AC) = Σn( x̅ - x̅̅)² =
9*(0.26-0.295)²+9*(0.33-0.295)²=
0.0221
MS(between AC)=SS(between AC)/df between=
0.0221/2= 0.0110
F (A vs C) = 0.011/0.0378=
0.2917
Since, F stat < critical F value , Do not reject null
hypothesis
The SSbetween AC is 0.0221
The F (A versus C) is 0.2917
the psychologist cannot conclude
that the population means for children without sleep apnea and
children with treated sleep apnea differ
=================================================================================================
grand mean BC, x̅̅ = Σni*x̅i/Σni =
(9*0.62+9*0.33)/(9+9)=
0.475
SS(between BC) = Σn( x̅ - x̅̅)² =
9*(0.62-0.475)²+9*(0.33-0.475)²=
0.3785
MS(between BC)=SS(between BC)/df between=
0.3785/2= 0.1892
F (B vs C) = 0.1892/0.0378=
5.0060
Since, F stat < critical F value , Do not reject null
hypothesis
The SSbetween BC is 0.3785
The F (B versus C) is 5.0060
the psychologist cannot conclude
that the population means for children with untreated sleep apnea
and children with treated sleep apnea differ