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 SAD patients 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	5	8
		4	4	6
		7	7	9
		5	9	5
		6	8	8
	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 experiment wise 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.

A. Retain the null hypothesis.

B. Reject the null hypothesis.    

State the decision for the main effect of intensity.

A. Retain the null hypothesis.

B. Reject the null hypothesis.    

State the decision for the interaction effect.

A. Retain the null hypothesis.

B. 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	38	43	114
Average	5.5	6.333333	7.166667	6.333333
Variance	1.1	3.866667	2.166667	2.588235
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	81	86
Average	5.5	6.75	7.166667
Variance	2.090909	2.931818	1.606061
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
time of the day	0.69	1	0.69	0.29	0.59	4.17
intensity	18.06	2	9.03	3.82	0.03	3.32
time of day*intensity	1.39	2	0.69	0.29	0.75	3.32
error	70.83	30	2.36
Total	90.97	35

State the decision for the main effect of the time of day.

A. Retain the null hypothesis. [p-value >0.05

State the decision for the main effect of intensity.

B. Reject the null hypothesis. [p-value <0.05]

State the decision for the interaction effect.

A. Retain the null hypothesis.    [p-value >0.05

b)

Level of significance	0.05
no. of treatments,k	3
DF error =N-k=	30
MSE	2.361
q-statistic value(α,k,N-k)	3.488

Tukey Kramer test      
      
critical value = q*√(MSE/n) = 1.5472

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

	low	medium	high
count, ni =	12	12	12
mean , x̅ i =	5.50	6.75	7.17

absolute mean difference		critical value			result
µ1-µ2	-1.250	1.5472			means are not different
µ1-µ3	-1.667	1.5472			means are different
µ2-µ3	-0.417	1.5472			means are not different