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
		6	7	9
		5	9	5
		6	8	8
	Night	5	6	8
		8	8	7
		6	6	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

using excel data analysis tool for two factor anova, following o/p Is obtained : write data>menu>data>data analysis>anova :two factor with replication>enter required labels>ok, and following o/p Is obtained,

		Light Intensity				Anova: Two-Factor With Replication
		Low	Medium	High
Time of	Morning	5	5	7		SUMMARY	Low	Medium	High	Total
Day		6	6	8		Morning
		4	4	6		Count	6	6	6	18
		6	7	9		Sum	32	39	43	114
		5	9	5		Average	5.333333	6.5	7.166667	6.333333
		6	8	8		Variance	0.666667	3.5	2.166667	2.470588
	Night	5	6	8
		8	8	7		Night
		6	6	6		Count	6	6	6	18
		7	5	8		Sum	33	42	42	117
		4	9	7		Average	5.5	7	7	6.5
		3	8	6		Variance	3.5	2.4	0.8	2.5
						Total
						Count	12	12	12
						Sum	65	81	85
						Average	5.416667	6.75	7.083333
						Variance	1.901515	2.75	1.356061
						ANOVA
						Source of Variation	SS	df	MS	F	P-value	F crit
						Sample	0.25	1	0.25	0.11509	0.736786	4.170877
						Columns	18.66667	2	9.333333	4.296675	0.022863	3.31583
						Interaction	0.666667	2	0.333333	0.153453	0.85841	3.31583
						Within	65.16667	30	2.172222
						Total	84.75	35

a)

Source of	SS	df	MS	F
Variation
Time of day	0.25	1	0.25	0.12
Intensity	18.67	2	9.33	4.30
Time of day *intensity	0.67	2	0.33	0.15
Error	65.17	30	2.17
Total	84.75	35

for the main effect of the time of day.

p value=0.74>α,So

-Retain the null hypothesis

for the main effect of intensity.

p value=0.02<α, So

-Reject the null hypothesis.    

for the interaction effect.

p value=0.86>α, So

-Retain the null hypothesis.

b)

Tukey's HSD Test

	Low	Medium	High
count, ni =	12	12	12
mean , x̅ i =	5.417	6.75	7.08

Level of significance	0.05
no. of treatments,k	3
DF error =N-k=	30
MSE	2.172
q-statistic value(α,k,N-k)	3.488

critical value = q*√(MSE/2*(1/ni+1/nj))=1.4840

if absolute difference of means > critical value,means are significnantly different ,otherwise not                      
                      

population mean difference		critical value			result
µ1-µ2	1.333	1.4840			means are not different
µ1-µ3	1.667	1.4840			means are different
µ2-µ3	0.333	1.4840			means are not different