The researchers perform an analysis of variance (ANOVA) at a 01 to test the hypo
ID: 3249045 • Letter: T
Question
The researchers perform an analysis of variance (ANOVA) at a 01 to test the hypothesis that the treatment means are equal. The results are presented in the following ANOVA table. ANOVA Table Source of Degrees of Sum of squares Freedom Mean square Variation 0.1863 Between Treatments 0.3725 5.65 0.9900 0.0330 Within Treatments 30 1.3625 Total 32 The ANOVA yielded a significant F statistic, so the The Studentized Range Statistic (q) null hypothesis is rejected. Since there are more d for Error Term 2 than two groups, the researchers are interested in determining which groups are different. The Tukey's 20 2.95 3.58 4.02 4.64 Honestly significant Difference (HSD) test will be 2.92 3.53 used to evaluate the pairs. First, use the table at the 3.96 4.55 right to determine the appropriate value of q at o- 30 2.89 3.49 3.89 4.45 .01. The q value for this problem is 2.86 3.44 3.82 4.37 2.83 3.40 60 Now, use the q value to calculate Tukey's HSD. 3.76 4.28 Tukey's HSD is Thus, the mean difference The top value is or 05 the bottom (bold) value is o 01. The number of between any two samples must be at least treatments is listed across. The df for the error term is in the left column, where to be significant. the error term' is another name for the within-treatments variance The researchers conclude that the population means for children without sleep apnea and children with untreated sleep apnea differExplanation / Answer
(first part) q=4.45
since there are 3 treatments groups and error df=30
so q=4.45 for alpha=0.01
(second part) Tukey HSD=0.2437
HSD(0.01)=q(0.01)*sqrt(MSwithin)/r)=4.45*sqrt(0.033/11)=0.2437
r=number of repetition of treatment groups=(total df +1)/(treatmeant df +1)=(32+1/(2+1)=11
(third part) mean difference between any sample must be atleast 0.2437 to be significant
for the remaing part, mean of different treatment groups is needed
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