One Way Anova

Topics: Statistical significance, Null hypothesis, Statistical hypothesis testing Pages: 3 (727 words) Published: February 25, 2013
Example B: Raw Score Method
You may wish to test the effects of a number of experimental treatments (counselling approaches): group counselling, peer counselling and individual counselling on the self-concepts of students. In this case, the independent variable, counselling approach has three levels. Necessarily there should be three groups randomly selected from the school population which will be exposed to three different counselling approaches.

The dependent variable, self-concept, may be measured through a standardized self-concept instrument which yields interval scores for the subjects.
In this problem, application of the one-factor ANOVA will test the following hypothesis: There is no significant difference in self-concept among the three groups of students exposed to different counselling approaches. Step 1 Enter the data in a worksheet table. (See below.)

Step 2 Find the square of each raw score (X2).
Step 3 Compute the sum of N for each group, the total N, the sums of the raw scores and the sums of the squared scores. N1 = 6; N2 = 6; N3 = 6; Nt = 18
ƩX1 = 366; ƩX2 = 492: ƩX3 = 510: ƩXt = 1368
ƩX21 = 23, 866; ƩX22 = 40, 798; ƩX23 = 43, 652; ƩX2t = 108, 316 WORKSHEET TABLE for the One-Way ANOVA
Counselling Conditions
Group (1)Peer (2)Individual (3)
X1X21X2X22X3X23
786084775929786084
462116836889918281
411681979409979409
502500694761826724
694761796241857225
826724877569775929
SUMS366238664924079851043652ƩXt=1368
Means618285ƩX2t = 108, 316
N666Nt = 18
Step 4 Compute Sums of Squares.
a.SSt (SS for total variability)
=ƩXt2 – (ƩXt)2
N
SSt = 108,316 – 13682 = 4348
18
b.SSb (SS for between group variability)
= (ƩX1)2 + (ƩX2)2 + (ƩX3)2 – (ƩXt)2
N1 N2 N3 Nt
= 3662 + 4922 + 5102 - 13682
6 6 6 18
= 106,020 – 103,968 = 2052
c.SSw = (SS for within group...
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