Question

A researcher examined whether the time of day someone exercises affects memory retention in college courses.  Participants were assigned to one of three exercise groups: morning, afternoon, evening, and their performance on a memorization task was measured.  This data is below:

Morning	Afternoon	Evening
6	4	7
7	5	6
8	6	8
5	4	6
6	5	5
Mean = 6.40	Mean = 4.80	Mean = 6.40
s = 1.14	s =.84	s = 1.14

Assuming the researcher wants to know whether the performance was different in these groups (alpha = .05).  What is the Mean Squares Within (SS_W)?  Round to two decimal places.

Answer 1

given data are:-

[ i have considered the morning group as group 1, evening group as group 2 and night group as group 3 ]

hypothesis:-

mean group 1 = mean group 2 = mean group 3

mean group 1 ≠ mean group 2 ≠ mean group 3

necessary calculation:-

grand mean :-

sum of squares between the group be:-

= 8.5333

sum of squares within group be:-

= 13.2192

number of groups (k) = 3

total number of subjects (N) = 5*3 =15

the ANOVA table be :-

source	df	sum of squares (SS)	mean of squares (MS)	F statistic	F critical
between groups	(k-1) = (3-1) = 2	8.5333	8.5333 / 2 =4.2667	= 3.8732	= 3.89 [ from f table ]
within groups	(N-k) =(15-3) = 12	13.2192	13.2192 / 12 =1.1016
total	14	21.7525

decision:-

F statistic = 3.8732 < F critical = 3.89

so, we fail to reject the null hypothesis.

we conclude that,

there is not sufficient evidence to claim that , the performance was different in these groups at  0.05 level of significance.

the mean squares within(MSW) = 1.11 [ rounded off to 2 decimal places ]

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