Contrast

A contrast is used to test the hypothesis that the mean of the three smile conditions is different from the mean of the neutral condition. The results of the analysis using SAS JMP are shown below. The coefficients for each of the three smile conditions is 0.333; the coefficient for the neutral condition is -1. Note that the sum of the coefficients is 0.


How this was done

The value obtained by applying the coeffients to the sample mean is 0.9461. JMP labels this as the "Estimate" since it is an estimate of what you would get if you applied the coefficents to the population means.

What can be deduced from an 'estimate' of 0.9461?
The smile means are lower (on average) than the neutral mean.
The smile means are statistically signficantly lower than the neutral mean.
The smile means are higher (on average) than the neutral mean.
The smile means are statistically signficantly higher than the neutral mean.
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The question is whether the estimate is significantly different from 0. This is tested by dividing the estimate by its standard error. This produces a t Ratio of 2.9354 on (4)(34-1) =132 degrees of freedom. An alternate method based on sums of squares yields an F of 8.61647 which is, of course, t squared).

Can the null hypothesis that the average of the three smiling conditions is equal to the neutral condition be rejected?
Yes
No
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Once it is established that the average of the three smile means is significantly different from the neutral mean, it is logical to compare each experimental mean with the neutral control. Computing a separate t test to test the difference between each smile mean and the neutral contol would lead to three chances for a type I error and thus an inflated experiment-wise error rate. To control the error rate, Dunnett's test should be used. Below is shown the results of the Dunnett's test as computed by SAS.

Dunnett's Test

                      General Linear Models Procedure
                 Dunnett's T tests for variable: LENIENCY

        NOTE: This tests controls the type I experimentwise error for
              comparisons of all treatments against a control.
           Alpha= 0.05  Confidence= 0.95  df= 132  MSE= 2.648897
                    Critical Value of Dunnett's T= 2.376
                   Minimum Significant Difference= 0.938
     Comparisons significant at the 0.05 level are indicated by '***'.
 

                      General Linear Models Procedure
                               Simultaneous            Simultaneous
                                   Lower    Difference     Upper
               COND             Confidence    Between   Confidence
            Comparison             Limit       Means       Limit
      false     - neutral         0.3120      1.2500      2.1880   ***
      felt      - neutral        -0.1439      0.7941      1.7321
      miserable - neutral        -0.1439      0.7941      1.7321

How this was done


Which comparison of conditions is significant at the 0.05 level?
false - neutral
felt - neutral
miserable - neutral
none of the above
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So far, the analysis has shown that (a) the mean of the three smile conditions is significantly higher than the mean of the neutral condition, (b) the mean of the false smile condtion is significantly higher than the mean of the neutral condition, and (c) neither the difference between the felt-smile condition and the neutral condition nor the difference between the miserable-smile condition and the neutral condition is significant. The next question is whether the three smile conditions differ significantly from each other. This can be tested by computing a t for each of the three pairwise comparisons among the three smile conditions. To control the type I error rate, the Bonferroni correction can be applied. This correction uses the 0.05/3 = 0.01667 significance level for each comparison.

The analysis from SAS is shown below. Note that none of the comparisons approached significance even without the correction. Although it may seem strange that the comparison of felt and miserable results in a sum of squares of exactly 0, this can happen when the dependent variable is discrete rather than continuous. In this case the means for false and miserable were exactly the same.



Dependent Variable: LENIENCY

Contrast                DF    Contrast SS   Mean Square   F Value     Pr > F

false versus felt        1      3.5330882     3.5330882      1.33     0.2502

false versus miserab     1      3.5330882     3.5330882      1.33     0.2502

felt versus miserabl     1      0.0000000     0.0000000      0.00     1.0000

How this was done