Can you show me how to do this in detail using Excel? Seasonal affective disorde
ID: 3239869 • Letter: C
Question
Can you show me how to do this in detail using Excel?
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.)
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.
Summarize the results for this test using APA format.
Day Morning 5 5 7 6 6 8 4 4 6 7 7 9 5 8 5 6 8 8 Night 5 6 9 8 8 7 6 7 6 7 4 8 4 9 7 3 8 6
Explanation / Answer
Answer:
State the decision for the main effect of the time of day.
Retain the null hypothesis..
State the decision for the main effect of intensity.
Reject the null hypothesis.
State the decision for the interaction effect.
Retain the null hypothesis.
(b) Compute Tukey's HSD to analyze the significant main effect.
The critical value is for each pairwise comparison.
Critical value =2.47
Summarize the results for this test using APA format.
There was a significant Intensity effect, F(2, 30) = 3.76, p= .035.
There was no significant Time effect, F(1, 30) = 0.19, p= .666.
There was no interaction effect, F(2, 30) = 0.19, p= .828.
Post hoc test indicating low intensity is significant from high intensity, p<.05.
Two factor ANOVA
Intensity
Means:
Low
Medium
High
Morning
5.5
6.3
7.2
6.3
Time of day
Night
5.5
7.0
7.2
6.6
5.5
6.7
7.2
6.4
6
replications per cell
ANOVA table
Source
SS
df
MS
F
p-value
Time of day
0.44
1
0.444
0.19
.6656
Intensity
17.56
2
8.778
3.76
.0349
Interaction
0.89
2
0.444
0.19
.8276
Error
70.00
30
2.333
Total
88.89
35
Post hoc analysis
p-values for pairwise t-tests for Factor 2
Low
Medium
High
5.5
6.7
7.2
Low
5.5
Medium
6.7
.0712
High
7.2
.0121
.4290
Tukey simultaneous comparison t-values (d.f. = 30)
Low
Medium
High
5.5
6.7
7.2
Low
5.5
Medium
6.7
1.87
High
7.2
2.67
0.80
critical values for experimentwise error rate:
0.05
2.47
0.01
3.15
Two factor ANOVA
Intensity
Means:
Low
Medium
High
Morning
5.5
6.3
7.2
6.3
Time of day
Night
5.5
7.0
7.2
6.6
5.5
6.7
7.2
6.4
6
replications per cell
ANOVA table
Source
SS
df
MS
F
p-value
Time of day
0.44
1
0.444
0.19
.6656
Intensity
17.56
2
8.778
3.76
.0349
Interaction
0.89
2
0.444
0.19
.8276
Error
70.00
30
2.333
Total
88.89
35
Post hoc analysis
p-values for pairwise t-tests for Factor 2
Low
Medium
High
5.5
6.7
7.2
Low
5.5
Medium
6.7
.0712
High
7.2
.0121
.4290
Tukey simultaneous comparison t-values (d.f. = 30)
Low
Medium
High
5.5
6.7
7.2
Low
5.5
Medium
6.7
1.87
High
7.2
2.67
0.80
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.