Seasonal affective disorder (SAD) is a type of depression during seasons with le
ID: 3243793 • Letter: S
Question
Seasonal affective disorder (SAD) is a type of depression during seasons with less daylight (e.g., winter months). One therapy for SAD is phototherapy, which is increased exposure to light used to improve mood. A researcher tests this therapy by exposing a sample of SAD patients to different intensities of light (low, medium, high) in a light box, either in the morning or at night (these are the times thought to be most effective for light therapy). All participants rated their mood following this therapy on a scale from 1 (poor mood) to 9 (improved mood). The hypothetical results are given in the following table.
(a) Complete the F-table and make a decision to retain or reject the null hypothesis for each hypothesis test. (Round your answers to two decimal places. Assume experimentwise alpha equal to 0.05.)
(b) Compute Tukey's HSD to analyze the significant main effect.
The critical value is ________ for each pairwise comparison.
Summarize the results for this test using APA format.
Day Morning 5 5 7 6 6 8 4 4 6 7 7 8 5 9 5 6 8 8 Night 5 6 9 8 8 7 6 7 6 7 5 8 4 9 7 2 8 6
Explanation / Answer
a)Step 1: Click the “Data” tab and then click “Data Analysis
Step 2: Click “ANOVA two factor with replication” and then click “OK.”
The two way ANOVA window will open.
Step 3: Type an Input Range into the Input Range box. We include all of our data, including headers and group names.
Step 4: Type a number in the “Rows per sample” box. It means how many individuals are in each group.
Step 5: Select an Output Range.
Step 6: Select alpha level=0.05.
Step 7: Click “OK” to run the two way ANOVA. The data will be returned where we specified in Step 5.
Step 8: We read the results. To figure out if we are going to reject the null hypothesis or not, we’ll basically be looking at two factors:
1. 1. If the F-value ( f )is larger than the f critical value ( f crit )
2. If the p-value is smaller than your chosen alpha level.
Source of
Variation
SS
df
MS
F
F
critical
Decision
Time of day
0.44
1
0.44
0.18
4.17
Accept
Intensity
19.39
2
9.69
4.04
3.32
Reject
Time of
day ×
Intensity
1.06
2
0.53
0.22
3.32
Accept
Error
72
30
2.4
Total
92.89
35
This means :
1. The hypothesis that effect of time of day is statistically insignificant is accepted.
2. The hypothesis that effect of intensity is statistically insignificant is rejected.
3. The hypothesis that interaction effect of time of day and intensity is statistically insignificant is accepted.
b) The formula for HSD :
HSDi,j =(Mi-Mj) / (MSW/n),
Mi – Mj is the difference between the pair of means.
MSw is the Mean Square Within, and n is the number in the group or treatment.
Here, since the hypothesis that effect of intensity is statistically insignificant is rejected, we analyse the significant main effect.
MSW=2.4,n=12,M1=5.42,M2=6.83,M3=7.08.
HSD(low,medium) = 3.17
HSD(medium,high) = 0.56
HSD(low,high) = 3.73
Critical value = 3.49, for no. of treatments = 3, df of error term = 30.
For pair (low,high), the observed HSD > critical value, hence the means are significantly different and for other pairs, the means are not significantly different.
Source of
Variation
SS
df
MS
F
F
critical
Decision
Time of day
0.44
1
0.44
0.18
4.17
Accept
Intensity
19.39
2
9.69
4.04
3.32
Reject
Time of
day ×
Intensity
1.06
2
0.53
0.22
3.32
Accept
Error
72
30
2.4
Total
92.89
35
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