A sample of N = 8 scores has sum X = 48 and sum X^2 = 351. For a hypothesised mean of 2.00, what is the effect size index d?

Step-by-step solution

1. 1

   Know which formula is being asked for. Cohen's $d$ for a one-sample design divides the mean difference by the sample standard deviation:

   $d = \frac{\bar{X} - \mu_0}{s}$

   The $t$ statistic divides by the standard error $s/\sqrt{N}$ instead. Confusing the two is the single most common error here, and with $N=8$ it inflates the answer by a factor of $\sqrt{8} \approx 2.83$.

2. 2

   Find the sample mean.

   $\bar{X} = \frac{\sum X}{N} = \frac{48}{8} = 6.00$

3. 3

   Compute the sum of squares with the computational formula.

   $SS = \sum X^2 - \frac{\left(\sum X\right)^2}{N} = 351 - \frac{48^2}{8} = 351 - \frac{2304}{8} = 351 - 288 = 63$

4. 4

   Get the sample standard deviation. Divide by $N-1 = 7$ for a sample:

   $s^2 = \frac{SS}{N-1} = \frac{63}{7} = 9, \qquad s = \sqrt{9} = 3$

   The numbers are engineered to be clean, which is a good sign the $SS$ step is right.

5. 5

   Divide the mean difference by $s$.

   $d = \frac{6.00 - 2.00}{3} = \frac{4}{3} = 1.3333 \approx 1.33$

6. 6

   Interpret and cross-check. By Cohen's conventions $d = 0.2$ is small, $0.5$ medium and $0.8$ large, so $1.33$ is a large effect. Cross-check against the distractor $3.77$: that is $t = \dfrac{4}{3/\sqrt{8}} = \dfrac{4}{1.06066} = 3.7712$ — the $t$ statistic, not the effect size. And $1.25$ would come from dividing by the population-style $\sqrt{63/8} = 2.806$, i.e. using $N$ instead of $N-1$.