A sample of N = 8 scores has sum X = 48 and sum X^2 = 351. Test the scores against a hypothesised mean of 9.00: what t value results?

Step-by-step solution

1. 1

   Write down the one-sample $t$ formula.

   $t = \frac{\bar{X} - \mu_0}{s/\sqrt{N}}$

   Everything needed is recoverable from $N$, $\sum X$ and $\sum X^2$ — the raw scores are never required.

2. 2

   Mean and sum of squares.

   $\bar{X} = \frac{48}{8} = 6.00, \qquad SS = \sum X^2 - \frac{\left(\sum X\right)^2}{N} = 351 - \frac{2304}{8} = 63$

3. 3

   Standard deviation and standard error. Using $N-1 = 7$ in the denominator:

   $s^2 = \frac{63}{7} = 9, \quad s = 3, \quad s_{\bar{X}} = \frac{s}{\sqrt{N}} = \frac{3}{\sqrt{8}} = \frac{3}{2.828427} = 1.060660$

4. 4

   Compute $t$, keeping the sign. The sample mean $6$ is below the hypothesised $9$, so $t$ must come out negative — a positive answer here signals a dropped sign:

   $t = \frac{6.00 - 9.00}{1.060660} = \frac{-3}{1.060660} = -2.8284 \approx -2.83$

5. 5

   Check against the plausible wrong answers. $-8.00$ is the mean difference divided by $s^2/\ldots$ style slips; $-2.65$ would need $s_{\bar X} = 1.132$; $3.02$ has the wrong sign and magnitude. Note also the exact value: $\dfrac{3}{3/\sqrt{8}} = \sqrt{8} = 2.8284$, so $t = -\sqrt{8}$ exactly.

6. 6

   Interpret. With $df = N - 1 = 7$, the two-tailed critical value at $\alpha = 0.05$ is $2.365$. Since $|-2.83| > 2.365$, the sample mean of $6.00$ differs significantly from $9.00$ at the $5\%$ level.