Assume that Professor Wiseman has developed a new memory training technique. He wants to do an experiment to test the effectiveness of his new technique by comparing his technique to a rote...

1. Assume that Professor Wiseman has developed a new memory training technique. He wants to do an experiment to test the effectiveness of his new technique by comparing his technique to a rote memorization group and a control group. Therefore, there are 3 groups: 1) Professor Wiseman's group; 2) rote memorization group; 3) control group.

1) If Professor Wiseman wanted to use a between subjects design to do his experiment, how could he design the study? What type of ANOVA would he use to analyze his data? Explain why.

2) If Professor Wiseman wanted to use a repeated measures (within subjects) design to do his experiment, how could he design the study? What type of ANOVA would he use to analyze his data? Explain why.

llltkl | Student

For a between-subjects study of the effect (if any) of Prof. Wiseman’s technique, a one way randomized ANOVA has to be designed as follows:

Imagine that subjects of three groups, viz. control, rote memorization and Prof. Wiseman’s technique, are asked to study a list of 10 words using one of the techniques. Obviously, subjects of the control group are not advised to follow any technique. A total of 27 participants are assigned to each condition, nine to each. Following table lists the number of words correctly recalled by each participant:

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CONTROL        ROTE MEMORIZATION      PROF. WISEMAN’S TECHNIQUE

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2                             5                                     6

3                             4                                     5

3                             4                                     5

5                            3                                      9

7                            5                                      10

6                            7                                      8

2                             3                                     5

4                             9                                     7

3                             5                                     7

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Mean=3.86       mean=5                     mean=6.89         [Grand mean=5.26]

To ascertain whether the observed difference in results is due to error variance or due to a definitive effect of conditions, ANOVA summary table has to be completed from the definitional formulae as follows:

Source          df (formula)         SS          MS         F(obt.)              [Fcv .05, .01]

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Between       2 (k-1)             41.4074      20.704     6.076                (3.40, 5.61)

Within         24 (N-k)             81.7778     3.407

Total           26 (N-1)            123.1852

--------------------------------------------------------------------------   --------------------

`eta^2` =41.4074/123.1852=0.336

So, approximately 34% of the variance is attributable to the effect of memorization condition.

Post hoc test: HSD (.01) `=4.55(sqrt(3.407/9)) ` =2.8

So, a mean difference of 2.8 or more is significant at the 0.01 level.

Decision: Therefore, Prof. Wiseman’s memorizing technique is significantly effective when compared with the control group, but not so much when compared with the rote memorization group (p<.01).