A clinical psychologist conducts an experiment to evaluate two methods for promo
ID: 3174043 • Letter: A
Question
A clinical psychologist conducts an experiment to evaluate two methods for promoting smoking cessation, nicotine patch and nicotine gum. Eighteen volunteers from a heavy smoking population are assigned six each to one of three groups. Conditions are similar for all groups, except for the treatment. Subjects in Group 1 get the nicotine patch. Subjects in Group 2 get the nicotine gum and subjects in Group 3 chew a placebo gum. All subjects are asked to do their best to stop smoking. The treatment for each group lasts for three months. On the last treatment day, the number of cigarettes smoked by each subject is recorded. The results are shown in the following table.
For convenience, the data and ANOVA summary table are shown below.
*With = 0.05, Fcrit = 3.68. Therefore, H0 is rejected.
Using this data,
(a) Test the planned comparisons that (1) nicotine patch and placebo gum have different effects and (2) that nicotine gum and placebo gum have different effects. Use = 0.051 tail. (In answering this part, subtract placebo gum scores from nicotine patch and nicotine gum scores.)
(1) Nicotine Patch and Placebo Gum
(2) Nicotine Gum and Placebo Gum
(b) Make all possible post hoc comparisons using the HSD test. Use = 0.05.
(1) Nicotine Patch and Placebo Gum
(2) Nicotine Gum and Placebo Gum
(3) Nicotine Patch and Nicotine Gum
Nicotine Patch Group 2
Nicotine Gum Group 3
Placebo Gum 10
8
6
7
10
9 11
6
9
7
8
5 15
12
14
16
15
12
Explanation / Answer
(a1) standerd error of difference==sqrt(2*mse/r)=sqrt(2*2.9/6)=0.9832
difference of mean of Nicotine Patch and Placebo Gum=8.5-14=-5.5
t-obtained=difference/standard error of difference=-5.5/0.9832=-5.59
critical t=t(alpha,error df)=t(0.05,15)=2.1314
(a2)standerd error of difference=sqrt(2*mse/r)=sqrt(2*2.9/6)=0.9832
mean difference of Nicotine Gum and Placebo Gum=8-14=-6
t=-6/0.9832=-6.1
critical t=t(alpha,error df)=t(0.05,15)=2.1314
(b)
where MSwg is the within-groups MS obtained in the original analysis and Np/s is the number of values of Xi per sample ("p/s"=per sample).
Np/s=3/(1/Na)+(1/Nb)+(1/Nc)=3/(1/6+1/6+1/6)=6
(1) Nicotine Patch and Placebo Gum
Q=(8.5-14)/sqrt(2.9/6)=-7.91
(2) Nicotine Gum and Placebo Gum
Q=(8-14)/sqrt(2.9/6)=-8.63
(3) Nicotine Patch and Nicotine Gum
Q=(8-8.5)//sqrt(2.9/6)=-0.72
Q=(M1 - M2)/sqrt[MSwg / Np/s]
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