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 patients with SAD 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.

		Light Intensity
		Low	Medium	High
Time of Day	Morning	5	5	7
		6	6	8
		4	4	6
		7	7	9
		5	9	5
		6	8	7
	Night	5	6	9
		8	8	7
		6	7	6
		7	5	8
		4	9	7
		3	8	6

(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.

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:

Answer 1

Anova: Two-Factor With Replication
SUMMARY	Low	Medium	High	Total
Morning
Count	6	6	6	18
Sum	33	39	42	114
Average	5.5	6.5	7	6.333333
Variance	1.1	3.5	2	2.352941
Night
Count	6	6	6	18
Sum	33	43	43	119
Average	5.5	7.166667	7.166667	6.611111
Variance	3.5	2.166667	1.366667	2.722222
Total
Count	12	12	12
Sum	66	82	85
Average	5.5	6.833333	7.083333
Variance	2.090909	2.69697	1.537879
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Time of day	0.69	1	0.69	0.31	0.58	4.17
Intensity	17.39	2	8.69	3.83	0.03	3.32
Time of day x Intensity	0.72	2	0.36	0.16	0.85	3.32
Error	68.17	30	2.27
Total	86.97	35

For Time of day:  
p-value > α, Retain the null hypothesis  
  
For Intensity:  
p-value < α, Reject the null hypothesis.  
  
For Interaction:  
p-value > α, Retain the null hypothesis  
  
b)   
At α = 0.05, k = 3, N-K = 30, Q value = 3.49

Critical Range, CV = Q*√(MSW/n) = 3.49*√(2.2722/12) = 1.52