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	8
	Night	5	6	9
		7	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.

Answer 1

using excel data analysis tool for two factor anova with replication,

Anova: Two-Factor With Replication
SUMMARY	Low	Medium	High	Total
Morning
Count	6	6	6	18
Sum	33	39	43	115
Average	5.5	6.5	7.166667	6.388889
Variance	1.1	3.5	2.166667	2.486928
Night
Count	6	6	6	18
Sum	32	43	43	118
Average	5.333333	7.166667	7.166667	6.555556
Variance	2.666667	2.166667	1.366667	2.614379
Total
Count	12	12	12
Sum	65	82	86
Average	5.416667	6.833333	7.166667
Variance	1.719697	2.69697	1.606061
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Sample	0.25	1	0.25	0.115681	0.736137	4.170877
Columns	20.72222	2	10.36111	4.794344	0.015605	3.31583
Interaction	1.166667	2	0.583333	0.269923	0.765273	3.31583
Within	64.83333	30	2.161111
Total	86.97222	35

a)

ANOVA
Source of Variation	SS	df	MS	F	P-value
time of the day	0.25	1	0.25	0.12	0.74
intensity	20.72	2	10.36	4.79	0.02
time *intensity	1.17	2	0.58	0.27	0.77
error	64.83	30	2.16
Total	86.97	35

p-value for  main effect of the time of day = 0.74 >α=0.05,

so, fail to reject null hypothesis

-----------------

p-value for  the main effect of intensity =0.02<α=0.05

s, reject the null hypothesis

-------------

p-value for the interaction effect = 0.77<α=0.05

so, fail to reject null hypothesis

------------------------------------------------------------------------------------------

b)

Level of significance	0.05
no. of treatments,k	3
DF error =N-k=	30
MSE	2.1611
q-statistic value(α,k,N-k)	3.4880

	Low	Medium	High
count, ni =	12	12	12
mean , x̅ i =	5.4167	6.83	7.17

critical value = q*√(MSE/2*(1/ni+1/nj)) = 1.4802

The critical value is_____1.4802____for each pairwise comparison.