Three groups of seven participants gave sum X = 36, 17 and 13 with sum X^2 = 200, 59 and 35 (Exercise, Comedy, Control). Find Cohen's d comparing the Comedy group with the Control group.

Step-by-step solution

1. 1

   Note which groups are involved. Cohen's $d$ compares two groups, so the Exercise column plays no part here — its numbers are distractors for this particular question (they matter only for the overall ANOVA).

2. 2

   Compute the two means.

   $M_{\text{Comedy}}=\frac{17}{7}=2.4286,\qquad M_{\text{Control}}=\frac{13}{7}=1.8571$

   so the mean difference is $2.4286-1.8571=0.5714$ (exactly $\tfrac47$).

3. 3

   Compute each group's sum of squares. Using the computational formula $SS=\sum X^{2}-\dfrac{(\sum X)^{2}}{n}$:

   $SS_{\text{Comedy}}=59-\frac{289}{7}=59-41.2857=17.7143$

   $SS_{\text{Control}}=35-\frac{169}{7}=35-24.1429=10.8571$

4. 4

   Pool the two variances. With equal group sizes, add the sums of squares and divide by the combined degrees of freedom $df=(7-1)+(7-1)=12$:

   $s_{p}^{2}=\frac{17.7143+10.8571}{12}=\frac{28.5714}{12}=2.3810$

   $s_{p}=\sqrt{2.3810}=1.5430$

   Note the pooled standard deviation, not the standard error, goes in the denominator — $d$ measures a difference in raw score units, so it must not shrink as $n$ grows.

5. 5

   Form Cohen's d.

   $d=\frac{M_{1}-M_{2}}{s_{p}}=\frac{0.5714}{1.5430}=0.370$

   $\boxed{d\approx 0.37}$

6. 6

   Interpret against the conventional benchmarks. Cohen's rough guides are $0.20$ small, $0.50$ medium and $0.80$ large, so $0.37$ is between small and medium: the comedy group scored higher than the control group by about a third of a standard deviation. For comparison, the overall ANOVA on all three groups gives $SS_{\text{between}}=43.14$ out of $SS_{\text{total}}=86.57$, i.e. $\eta^{2}=0.498$ — a large overall effect driven mainly by the Exercise group, which is why the Comedy-versus-Control contrast alone is much smaller.