# One Way Anova

**Topics:**Statistical significance, Null hypothesis, Statistical hypothesis testing

**Pages:**3 (727 words)

**Published:**February 25, 2013

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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