A sample of N = 8 scores has sum X = 48 and sum X^2 = 351. Test the scores against a hypothesised mean of 2.00: find the t value, decide whether to reject the null hypothesis, and compute the effect size.

Step-by-step solution

1. 1

   Get the sample mean straight from the sum.

   $\bar X=\frac{\sum X}{N}=\frac{48}{8}=6$

   Raw data are never needed for a $t$ test — $N$, $\sum X$ and $\sum X^{2}$ carry all the information the test uses.

2. 2

   Use the computational formula for the sum of squares. The definition $\sum(X-\bar X)^2$ would require the individual scores, but the algebraically identical form does not:

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

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

   Dividing by $N-1$ rather than $N$ is what makes $s^2$ an unbiased estimate of the population variance, and it is also what sets the degrees of freedom at $7$.

3. 3

   Convert the sample standard deviation into a standard error. The test statistic compares the mean to $\mu_0$ in units of how much a sample mean would bounce around:

   $SE=\frac{s}{\sqrt N}=\frac{3}{\sqrt 8}=\frac{3}{2.8284}=1.0607$

4. 4

   Compute the $t$ statistic.

   $t=\frac{\bar X-\mu_0}{SE}=\frac{6-2}{1.0607}=3.771\qquad(df=N-1=7)$

   If your multiple-choice list does not contain a value near $3.77$, the correct response is "none of the alternatives" — do not round your way onto a neighbouring option.

5. 5

   Compare with the critical values on 7 degrees of freedom. Two-tailed critical values are

   $t_{.05}=2.365,\qquad t_{.01}=3.499,\qquad t_{.001}=5.408$

   Since $3.771$ exceeds $2.365$ and $3.499$ but not $5.408$, the null hypothesis is rejected at the .05 and .01 levels, but not at .001. The exact two-tailed $p$ value is $0.0070$, which sits between $.01$ and $.001$ exactly as that comparison predicts.

6. 6

   Report the effect size, which significance alone does not give. For a one-sample test, Cohen's $d$ measures the distance in standard-deviation units:

   $d=\frac{\bar X-\mu_0}{s}=\frac{6-2}{3}=1.33$

   By the usual benchmarks ($0.2$ small, $0.5$ medium, $0.8$ large) this is a large effect. Note that $d$ uses $s$, not $SE$ — dividing by $1.0607$ instead of $3$ would just reproduce the $t$ value.