In: Math
Researchers in mental health fields are often interested in evaluating the effectiveness of using food images to enhance positive mood. Adapting a typical design from such studies, suppose we have participants rate their mood change on a standard self-report affect scale after viewing images of "comfort" foods, fruits/vegetables (F/V), and random non-food images (used as a control group). The results are given in the table at right for this hypothetical study.
Images | ||
---|---|---|
Control | F/V |
Comfort Foods |
0 | 0 | 3 |
1 | 1 | 2 |
1 | 2 | 3 |
3 | 3 | 5 |
0 | 1 | 0 |
(a) Complete the F-table. (Round your answers to two decimal places.)
Source of Variation |
SS | df | MS | Fobt |
---|---|---|---|---|
Between groups |
||||
Between persons |
||||
Within groups (error) |
||||
Total |
(b) Compute a Bonferroni procedure and interpret the results.
(Assume experimentwise alpha equal to 0.05. Select all that
apply.)
There were no significant differences between any of the groups.Participants rated a significantly larger mood change after viewing images of fruits/vegetables compared with the mood change after viewing control images.Participants rated a significantly larger mood change after viewing images of comfort foods compared with the mood change after viewing images of fruits/vegetables.Participants rated a significantly larger mood change after viewing images of comfort foods compared with the mood change after viewing control images.
A.
Data Summary from Excel:
ANOVA: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Control | 5 | 5 | 1 | 1.5 | ||
F/V | 5 | 7 | 1.4 | 1.3 | ||
Comfort Foods | 5 | 13 | 2.6 | 3.3 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | Fobt | P-value | F crit |
Between Groups | 6.933333333 | 2 | 3.466666667 | 1.704918033 | 0.222997177 | 3.885293835 |
Within Groups | 24.4 | 12 | 2.033333333 | |||
Total | 31.33333333 | 14 |
B.
The Bonferroni Correction = *100/(s(s-1)*0.5)
where, s is the no. of groups compared
So,
Bonferroni Correction = 0.05*100/(3(2-1)*0.5)
= 1.66%
Since p-value = 0.223 > 0.0167 i.e. we can't reject H0.
Interpretation: There were no significant differences between any of the groups.
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