7. Comparing one- and two-way ANOVA Aa Aa Dr. Diane Gold and her colleagues stud

Question

7. Comparing one- and two-way ANOVA Aa Aa Dr. Diane Gold and her colleagues study how rotating shift work (switching back and forth between the night shift and the day shift, for example) contributes to disrupted sleep cycles, accidents, and nodding off at work. Suppose that a graduate student is studying a group of police officers to see whether those with a particular circadian type are sleepier during their shift, particularly when they are assigned to different shift rotations. Circadian types include larks, who are most active in the early morning; hummingbirds, who are active in the middle of the day; and owls, who are active late into the night. The graduate student suspects that larks have the hardest time rotating and that there is an interaction between circadian type and shift rotation, such that hummingbirds have the easiest time with a day/evening rotation and owls have the easiest time with an evening/night rotation. The student administers a sleepiness test to 72 police officers (6 in each cell) each shift for a month and use an average score across the month for each person as the indication of each person's typical sleepiness. The means of the scores are shown in the following table, where a higher score indicates more sleepiness. Shift Rotation Day/Evening Evening/Night Night/Day Day/Night/Evening Lark! 11 = 0.50 ! Hummingbird |21 = 0.83 | | k1 = 0.42 |C2 = 0.67 a = 1.17 gm = 0.75 12 = 0.33 22 = 0.33 13 = 0.33 |X23 = 0.50| 14 = 0.50 4 = 0.50 Circadian Type Xs 1 = 1.00 Xs2 = 0.61 Xs3 = 0.61 Xs4 = 0.78 The graduate student performs a one-factor analysis of variance to test whether the average sleepiness scores are equal across the populations defined by the four shift rotations. That is, the graduate student wants to test if the average sleepiness scores are equal across the columns in the previous table.

Explanation / Answer

Answer:

Not rejected, does not have sufficient evidence to conclude.

source

SS

df

MS

F

C

7.0008

2

3.5004

5.21

S

1.8661

3

0.6220

0.93

CS

2.3214

6

0.3869

0.58

Error

40.3117

60

0.6719

Total

51.5000

71

source

SS

df

MS

F

Group

1.8661

3

0.6220

0.85

Error

49.6339

68

0.7299

Total

51.5

71

Critical F(3,68) at 0.01 level =4.083

Ho is not rejected.

Measure

Equal

small in one way

small in two way

MS

Yes

SS

Yes

Df error

Yes

MS error

Yes

FS

Yes

source

SS

df

MS

F

C

7.0008

2

3.5004

5.21

S

1.8661

3

0.6220

0.93

CS

2.3214

6

0.3869

0.58

Error

40.3117

60

0.6719

Total

51.5000

71

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