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	5	8
		4	4	6
		7	7	9
		5	9	5
		6	7	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 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 or Reject the null hypothesis.    

State the decision for the main effect of intensity.

Retain the null hypothesis or Reject the null hypothesis.    

State the decision for the interaction effect.

Retain the null hypothesis or 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

data

	Low	Medium	High
Morning	5	5	7
	6	5	8
	4	4	6
	7	7	9
	5	9	5
	6	7	8
Night	5	6	9
	7	8	7
	6	7	6
	7	5	7
	4	9	7
	3	8	6

Using Excel

data -> data analysis -> Anova: Two-Factor With Replication

Anova: Two-Factor With Replication
SUMMARY	Low	Medium	High	Total
Morning
Count	6	6	6	18
Sum	33	37	43	113
Average	5.5	6.166666667	7.166666667	6.277777778
Variance	1.1	3.366666667	2.166666667	2.447712418
Night
Count	6	6	6	18
Sum	32	43	42	117
Average	5.333333333	7.166666667	7	6.5
Variance	2.666666667	2.166666667	1.2	2.5
Total
Count	12	12	12
Sum	65	80	85
Average	5.416666667	6.666666667	7.083333333
Variance	1.71969697	2.787878788	1.537878788
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Sample	0.4444	1	0.4444	0.2105	0.6497	4.17087679
Columns	18.0556	2	9.0278	4.2763	0.0232	3.3158295
Interaction	2.7222	2	1.3611	0.6447	0.5319	3.3158295
Within	63.3333	30	2.1111
Total	84.5556	35

if p-value < alpha, we reject the null hypothesis

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

p-value for time of day is 0.6497 > 0.05

hence Retain the null hypothesis

State the decision for the main effect of intensity.

p-value of intensity = 0.0232 < 0.05

hence Reject the null hypothesis.    

State the decision for the interaction effect.

p-value = 0.5319 > 0.05

hence Reject the null hypothesis.    

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