Summary statistics from a sample were sum x^2 = 60, sum x = 10, and a sample size of 5. What is the standard error of the mean?

Step-by-step solution

1. 1

   Set up the two-stage calculation. The standard error of the mean is $s_{\bar{x}} = s/\sqrt{n}$, where $s$ is the sample standard deviation. So the work is: sum of squares, then variance with $n-1$, then square root, then divide by $\sqrt{n}$.

2. 2

   Compute the sum of squares.

   $SS = \sum x^2 - \frac{\left(\sum x\right)^2}{n} = 60 - \frac{10^2}{5} = 60 - 20 = 40$

3. 3

   Variance and standard deviation. Divide by $n-1 = 4$:

   $s^2 = \frac{40}{4} = 10, \qquad s = \sqrt{10} = 3.16228$

4. 4

   Divide by $\sqrt{n}$.

   $s_{\bar{x}} = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2} = 1.41421 \approx 1.414$

   The answer is exactly $\sqrt{2}$, because $\sqrt{10}/\sqrt{5} = \sqrt{10/5}$.

5. 5

   Check whether any offered option matches. The listed choices are $1.79$, $0.89$, $0.57$, $0.80$ and $0.40$. None equals $1.414$, so the correct response is "none of the other alternatives are correct". For completeness, the population-style version $\sqrt{40/5}/\sqrt{5} = 2.8284/2.2361 = 1.2649$ does not match either, so the mismatch is not caused by the $n$ versus $n-1$ choice.

6. 6

   Confirm with explicit data. The five values $0, 0, 0, 4, 6$ have $\sum x = 10$ and $\sum x^2 = 52$ — not a match, so instead take $-2, 0, 2, 4, 6$: $\sum x = 10$ ✓ and $\sum x^2 = 4+0+4+16+36 = 60$ ✓. Their mean is $2$, deviations $-4,-2,0,2,4$, $SS = 40$ ✓, $s = \sqrt{10}$, and $s_{\bar{x}} = \sqrt{2} = 1.414$ ✓.