Seasonal affective disorder (SAD) is a type of depression during seasons with le

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 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. Summarize the results for this test using APA format.

Explanation / Answer

There are two factors-Light intensity and Time of the day. The Light intensity factor has 3 levels and Time of day factor has 2 levels. Thus, this is a complete 3*2 factorial design. The design will contain 3*2=6 factors. The complete model is as follows:

E(y)=beta0+beta1x1+beta2x2+beta3x3+beta4x4+beta5x5

where, x1={1 if low intensity, 0 if not; x2={1 if medium intensity, 0 if not ;

x3={1 if morning, 0 if not

For ANOVA model, enter data in Minitab by entering Light intensity and Time of day in column C1 C2, and mood score in C3-Stat-ANOVA-Balanced ANOVA-enter C3 in Response- C1 C2 C1*C2 in Model-click OK. The session window will display the factorial ANOVA output. Based on the screen shot, the ANOVA table is as follows:

Per rejection rule based on P value, reject null hypothesis if P value is less than alpha=0.05. For main factor Light intensity, the P value is less than 0.05, therefore, reject null hypothesis and conclude that light intensity has significant effect on mood score. Equivalently, the P value for main factor Time of day is not less than 0.05, therefore, fail to reject null hypothesis and conclude that time of day has no significant effect on mood score. The P value corresponding to interaction is higher than 0.05, therefore fail to reject null hypothesis, and conclude that time of day and light intensity do not interact.

b.

The experiment wise error rate is alpha=0.05, there are v=30 degrees of freedom (df associated with MSE), s=sqrt MSE=1.49, nt=12 observations per treatment, number of sample means, p=3, the critical value of Studentized range q0.05(3,30)=3.49. Substitute the values in following formula to obtain the distance omega.

Omega=qalpha(p.v){s/sqrt nt}

=3.49{1.49/sqrt 12}

=8.11

Rank the sample means:

Treatment1: Low: 5.4167, Treatment 2: Medium:6.8333, Treatment 3:High:7.1667

Place a bar over those pairs of treatment means that differ by by less than omega=8.11. All the treatment means differ by less than omega. Any pair of treatment means not connected by bar is implies a difference in population means. There is no significant main effect.

Source SS DF MS F P Light intensity 20.722 2 10.361 4.65 0.017 Time of day 0.694 1 0.694 0.31 0.581 Light intensity*Time of day 0.722 2 0.361 0.16 0.851 Error 66.833 30 2.228 Total 88.972 35

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