Seasonal affective disorder (SAD) is a type of depression during seasons with le
ID: 3122848 • 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.
Night
5
6
9
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.)
State the decision for the main effect of the time of day.
Retain the null hypothesis.
Reject the null hypothesis.
State the decision for the main effect of intensity.
Retain the null hypothesis.
Reject the null hypothesis.
State the decision for the interaction effect.
Retain the null hypothesis.
Reject the null hypothesis.
(b) Compute Tukey's HSD to analyze the significant main effect.
The critical value is______ for each pairwise comparison.
Light Intensity Low Medium High Time ofDay Morning 5 5 7 6 6 8 4 4 6 7 7 9 5 9 5 6 8 8
Night
5
6
9
8 8 7 6 7 6 7 5 8 4 9 7 3 8 6 (a) Complete the Ftable 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.) Variation Time of day Intensity Time of day x Intensity Error TotalExplanation / Answer
See image:
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.) Source of Variation SS df MS F Time of day Intensity Time of day × Intensity Error Total
State the decision for the main effect of the time of day.
Calculated F=0.19, P=0.6634 which is > 0.05 level.
Retain the null hypothesis.
State the decision for the main effect of intensity.
Calculated F=4.06, P=0.0276 which is < 0.05 level.
Reject the null hypothesis.
State the decision for the interaction effect.
Calculated F=0.19, P=0.8253 which is > 0.05 level.
Retain the null hypothesis.
(b) Compute Tukey's HSD to analyze the significant main effect. The critical value is for each pairwise comparison.
Two factor ANOVA
Intensity
Means:
Low
Medium
High
Morning
5.5
6.5
7.2
6.4
Time of day
Night
5.5
7.2
7.2
6.6
5.5
6.8
7.2
6.5
6
replications per cell
ANOVA table
Source
SS
df
MS
F
p-value
Time of day
0.44
1
0.444
0.19
.6634
Intensity
18.67
2
9.333
4.06
.0276
Interaction
0.89
2
0.444
0.19
.8253
Error
69.00
30
2.300
Total
89.00
35
Post hoc analysis
Tukey simultaneous comparison t-values (d.f. = 30)
Low
Medium
High
5.5
6.8
7.2
Low
5.5
Medium
6.8
2.15
High
7.2
2.69
0.54
critical values for experimentwise error rate:
0.05
2.47
0.01
3.15
Tukey test shows that at 0.05 level of significance, low intensity is significantly different from high intensity. All other pairs are not significant.
Two factor ANOVA
Intensity
Means:
Low
Medium
High
Morning
5.5
6.5
7.2
6.4
Time of day
Night
5.5
7.2
7.2
6.6
5.5
6.8
7.2
6.5
6
replications per cell
ANOVA table
Source
SS
df
MS
F
p-value
Time of day
0.44
1
0.444
0.19
.6634
Intensity
18.67
2
9.333
4.06
.0276
Interaction
0.89
2
0.444
0.19
.8253
Error
69.00
30
2.300
Total
89.00
35
Post hoc analysis
Tukey simultaneous comparison t-values (d.f. = 30)
Low
Medium
High
5.5
6.8
7.2
Low
5.5
Medium
6.8
2.15
High
7.2
2.69
0.54
critical values for experimentwise error rate:
0.05
2.47
0.01
3.15
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.