Question

The data in the table below were collected from a repeated‑measures study of muscle growth related to exercise. There were 8 people in the study altogether! They each got the four exercise conditions. Fill in the table below

	Interval training	Aerobic exercise	Weight training	No exercise	SSwithin for the study
SSwithin for each group	36	18	26	15	95

Source	SS	DF	MS	F
Between groups
Within groups	95		-----	----
Between persons	16		-----	----
Error				----
Total	112	31	-----	----

1. What are the degrees of freedom that accompany that F value?

2. What is the critical value of F at alpha = .05? (From the table in the back of your book.)

3. Is your finding significant or not?

4. What is the value of η²_p?

5. What does the value of η²_p you calculated mean?

Answer 1

Source	SS	DF	MS	F
Between groups	17	3	5.67	1.51
Within groups	95	28	-----	----
Between persons	16	7	-----	----
Error	79	21	3.76	----
Total	112	31	-----	----

DF Between groups = Number of groups -1 = 4 - 1 = 3

DF Within groups = DF Total - DF Between groups = 31 - 3 = 28

DF Between persons = Number of persons - 1 = 8-1 = 7

DF Error =  DF Within groups - DF Between persons = 28 - 7 = 21

SS Error =  SS Within groups - SS Between persons = 95 - 16 = 79

SS Between groups = SS Total - SS Within groups = 112 - 95 = 17

MS = SS / DF

F = MS Between groups / MS Error = 5.67 / 3.76 = 1.51

1.

DF Between groups = 3

DF Error = 21

2.

Critical value of F at alpha = .05 and df = 3, 21 is 3.07

3.

Since the observed F (1.51) is less than the critical value, we fail to reject the null hypothesis H0 and conclude that the findings are not significant.

4.

η²_p = SS Between groups / SS Total = 17 / 112 = 0.1518

e.

It means that the 15.18% of variation in muscle growth is explained by exercise type.