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 lightly 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 lightbox, 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	5	8
		4	4	6
		7	6	9
		5	9	5
		6	8	8
	Night	5	6	9
		7	8	7
		6	7	6
		7	5	7
		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 experiment wise alpha equal to 0.05.)

Source of Variation	SS	df	MS	F
Time of day		1
Intensity		2
Time of day × Intensity		2
Error	64.83	30	2.16
Total	86.97	35

(b) Compute Tukey's HSD to analyze the significant main effect.

The critical value is ______ for each pairwise comparison.

(c) 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	37	43	113
Average	5.5	6.166667	7.166667	6.277778
Variance	1.1	3.766667	2.166667	2.565359
Night
Count	6	6	6	18
Sum	32	43	42	117
Average	5.333333	7.166667	7	6.5
Variance	2.666667	2.166667	1.2	2.5
Total
Count	12	12	12
Sum	65	80	85
Average	5.416667	6.666667	7.083333
Variance	1.719697	2.969697	1.537879

Source of	SS	df	MS	F
Variation
Time of day	0.44	1	0.44	0.20
Intensity	18.06	2	9.03	4.15
Time of day *intensity	2.72	2	1.36	0.62
Error	65.33	30	2.18
Total	86.56	35

for the main effect of the time of day.          
          
p value=0.65>α,So          
          
-Retain the null hypothesis          
          
          
          
for the main effect of intensity.          
          
p value=0.03<α, So          
          
-Reject the null hypothesis.              
          
          
for the interaction effect.          
          
p value=0.54>α, So          
          
-Retain the null hypothesis.          

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

Level of significance	0.05
no. of treatments,k	3
DF error =N-k=	30
MSE =	2.1800
q-statistic value(α,k,N-k)	3.486

			confidence interval
population mean difference	critical value	lower limit	upper limit	result
µ1-µ2	-1.25	1.49	-2.74	0.24	means are not different
µ1-µ3	-1.666	1.49	-3.15	-0.18	means are different
µ2-µ3	-0.42	1.49	-1.90	1.07	means are not different

.......

THANSK

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