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 6 8 4 4 6 6 7 9 5 9 5 6 8 8 Night 5 6 9 8 8 6 6 7 6 7 5 8 3 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

Hypothesis test fot time of the day

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: ?1 = ?2

Alternative hypothesis: At-least one of the ? is not equal.

Formulate an analysis plan. For this analysis, the significance level is 0.05.

Analyze sample data.

F statistics is given by:-

F = 0.1089

Fcritical = 4.17

The P-value = 0.7436

Interpret results. Since the P-value (0.7436) is greater than the significance level (0.05), we have to accept the null hypothesis.

Conclusion:-

Reject H0, There is insufficient evidence for significant differences between the two different times of the day.

Hypothesis test for Intensity

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: ?1 = ?2 = ?3

Alternative hypothesis: At-least one of the ? is not equal.

Formulate an analysis plan. For this analysis, the significance level is 0.05.

Analyze sample data.

F statistics is given by:-

F = 4.515

Fcritical = 3.315

The P-value = 0.019

Interpret results. Since the P-value (0.019) is less than the significance level (0.05), we have to reject the null hypothesis.

Conclusion:-

Reject H0, There is sufficient evidence for significant differences between the different light intensities.

b)

Tukey's HSD to analyze the significant main effect.

The critical value for each pairwise comparison is Qcritical?=0.05 = 3.4696

Hypothesis test fot time of the day

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: ?1 = ?2

Alternative hypothesis: At-least one of the ? is not equal.

Formulate an analysis plan. For this analysis, the significance level is 0.05.

Analyze sample data.

F statistics is given by:-

F = 0.1089

Fcritical = 4.17

The P-value = 0.7436

Interpret results. Since the P-value (0.7436) is greater than the significance level (0.05), we have to accept the null hypothesis.

Conclusion:-

Reject H0, There is insufficient evidence for significant differences between the two different times of the day.

Hypothesis test for Intensity

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: ?1 = ?2 = ?3

Alternative hypothesis: At-least one of the ? is not equal.

Formulate an analysis plan. For this analysis, the significance level is 0.05.

Analyze sample data.

F statistics is given by:-

F = 4.515

Fcritical = 3.315

The P-value = 0.019

Interpret results. Since the P-value (0.019) is less than the significance level (0.05), we have to reject the null hypothesis.

Conclusion:-

Reject H0, There is sufficient evidence for significant differences between the different light intensities.

b)

Tukey's HSD to analyze the significant main effect.

The critical value for each pairwise comparison is Qcritical?=0.05 = 3.4696