Question

seasonal affective disorder is a type of depression during seasons with less daylight 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 in a light box either in the morning or at night. All participants rated their mood following this therapy on a scale from 1 to 9. The hypothetical results are given in the following table.

Mo 6 Time of Day Night (a) Complete the F-table and make a decision to retain or reject the null hypothesis for each hypothesis test. (Ro alpha equal to 0.05.) MS 0.69 2 Time of day 1075x404 14 04 | Intensi 2.00 Time of day x Intensity 12.00 | 110.36 0.14EX 0.72 79.83X 10275 |x 13 レ ,-30 00 1 , 266 lx 30 00 266 35.00 Error 79 83 Total

Answer 1

(a)

$\begin{matrix} Source~of~variation & SS &df & MS & F&p-value\\ \hline Time~of~day& 0.69 &1.00 & 0.69 & 0.33&0.572 \\ Intensity &19.50 &2.00 &9.75 &4.58 & 0.018\\ Time~of~day\times Intensity& 0.72 &2.00 & 0.36 &0.17&0.845 \\ Error& 63.83& 30.00& 2.13 & \\ \hline Total& 84.75 &35.00 & & \end{matrix}\\ Since~p-values~corresponding~to~Time~of~day~and~Time~of~day\times~Intensity\\ >0.05~so~they~are~insignificant~whereas~Intensity~is~significantly~present\\ since~its~p-value<0.05.$

(b) Now we perform one way ANOVA for intensity.

Outputs are as follows:

One-way ANOVA: Rate versus Intensity

Source DF SS MS F P
Intensity 2 19.50 9.75 4.93 0.013
Error 33 65.25 1.98
Total 35 84.75

S = 1.406 R-Sq = 23.01% R-Sq(adj) = 18.34%

Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev ----+---------+---------+---------+-----
1 12 5.417 1.443 (-------*-------)
2 12 6.667 1.497 (--------*-------)
3 12 7.167 1.267 (--------*-------)

Since intensity is significant so we perform Tukey's test.

$Critical~value=q_{0.025,3,33}=3.47\\ Let~\bar{y}_1= 5.42=sample~mean~of~intensity~1\\ \bar{y}_2= 6.67=sample~mean~of~intensity~2\\ \bar{y}_3= 7.17=sample~mean~of~intensity~3\\ \begin{matrix} Absolute~Difference & q_{0.025,3,33}*\sqrt{\frac{MSE}{12}}&Remark \\ \hline |\bar{y}_1-\bar{y}_2|=1.25& 1.41& Reject~(value~of~column1>value~of~column2)\\ |\bar{y}_1-\bar{y}_3|=1.75& 1.41 &Reject~(value~of~column1>value~of~column2) \\ |\bar{y}_1-\bar{y}_2|=0.5& 1.41 &Accepted~(value~of~column1<value~of~column2) \end{matrix}\\ Hence~intensity~1~is~significantly~different~from~other~two~intensity~levels.$