Summary statistics from a sample were sum X = 8, sum X^2 = 46, and a sample size of 4. What is the standard error of the mean?

Step-by-step solution

1. 1

   Separate the two divisions. The standard error is

   $s_{\bar{X}} = \frac{s}{\sqrt{n}}$

   and it needs two denominators: $n-1$ when turning $SS$ into $s^2$, and $\sqrt{n}$ when turning $s$ into the standard error. Mixing them up is what generates every distractor in this question.

2. 2

   Compute the sum of squares. With the computational formula:

   $SS = \sum X^2 - \frac{\left(\sum X\right)^2}{n} = 46 - \frac{8^2}{4} = 46 - 16 = 30$

3. 3

   Get the sample variance and standard deviation. Divide by $n - 1 = 3$:

   $s^2 = \frac{30}{3} = 10, \qquad s = \sqrt{10} = 3.16228$

4. 4

   Divide by the square root of $n$.

   $s_{\bar{X}} = \frac{3.16228}{\sqrt{4}} = \frac{3.16228}{2} = 1.58114 \approx 1.581$

5. 5

   Trace the distractors back. $3.162$ is $s$ itself, with the $\sqrt{n}$ step skipped. $1.826$ is $\sqrt{10/3} = \sqrt{3.333}$, from dividing the variance by $n-1$ twice. $1.369$ is $\sqrt{30/16}$, and $0.791$ is $1.581/2$, one extra halving. Only $1.581$ follows both divisions exactly once.

6. 6

   Sanity check with explicit data. The four values $-1, 0, 3, 6$ have $\sum X = 8$ ✓ and $\sum X^2 = 1+0+9+36 = 46$ ✓. Their mean is $2$, the deviations are $-3, -2, 1, 4$, and $SS = 9+4+1+16 = 30$ ✓ — matching the computational formula. So $s = \sqrt{30/3} = 3.162$ and $s_{\bar{X}} = 3.162/2 = 1.581$ ✓.