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	3	6
		7	7	9
		5	9	5
		6	8	8
	Night	5	6	9
		8	8	7
		6	7	6
		6	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.

My SPSS stopped working and I am unable to complete my homework, help is appreciated!

Answer 1

a)

Anova: Two-Factor With Replication
SUMMARY	Low	Medium	High	Total
Morning
Count	6	6	6	18
Sum	33	38	43	114
Average	5.5	6.333333	7.166667	6.333333
Variance	1.1	4.666667	2.166667	2.823529
Night
Count	6	6	6	18
Sum	32	43	43	118
Average	5.333333	7.166667	7.166667	6.555556
Variance	3.066667	2.166667	1.366667	2.732026
Total
Count	12	12	12
Sum	65	81	86
Average	5.416667	6.75	7.166667
Variance	1.901515	3.295455	1.606061
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Sample	0.444444	1	0.444444	0.183486	0.671454	4.170877
Columns	20.05556	2	10.02778	4.139908	0.025839	3.31583
Interaction	1.722222	2	0.861111	0.355505	0.703732	3.31583
Within	72.66667	30	2.422222
Total	94.88889	35

Source of	SS	df	MS	F
Variation
Time of day	0.44	1	0.44	0.18
Intensity	20.06	2	10.03	4.14
Time of day *intensity	1.72	2	0.86	0.36
Error	72.67	30	2.42
Total	94.89	35

for the main effect of the time of day.          
          
p value=0.671>α,So          
          
-Retain the null hypothesis          
          
          
          
for the main effect of intensity.          
          
p value=0.026<α, So          
          
-Reject the null hypothesis.              
          
          
for the interaction effect.          
          
p value=0.704>α, So          
          
-Retain the null hypothesis.          

...............

q(3 , 30) = 3.4865

critical value = 1.53

.................

THANKS

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