For the sample with scores 1, 2, 3, find the 95 percent confidence interval about the sample mean.

Step-by-step solution

1. 1

   Compute the sums first. $n = 3$, $\sum X = 6$, $\sum X^2 = 1 + 4 + 9 = 14$. From these:

   $\bar{X} = \frac{6}{3} = 2, \qquad SS = 14 - \frac{6^2}{3} = 14 - 12 = 2$

2. 2

   Get $s$ and the standard error. Divide $SS$ by $n-1 = 2$:

   $s^2 = \frac{2}{2} = 1, \quad s = 1, \quad s_{\bar{X}} = \frac{s}{\sqrt{n}} = \frac{1}{\sqrt{3}} = 0.57735$

   The $\sqrt{n}$ division is essential — leaving it out is what produces the widest distractor below.

3. 3

   Use $t$, not $z$, and get the right df. With $n = 3$ the degrees of freedom are $df = 2$, and the two-tailed $95\%$ critical value is

   $t_{0.025,\,2} = 4.3027$

   This is far larger than $z = 1.96$; with only two degrees of freedom the $t$ distribution has very heavy tails.

4. 4

   Build the interval.

   $\bar{X} \pm t\,s_{\bar{X}} = 2 \pm 4.3027 \times 0.57735 = 2 \pm 2.4849$

   $LL = -0.4841, \qquad UL = 4.4841$

   So the interval is roughly $-0.48$ to $4.48$.

5. 5

   Compare with the offered options. The choices $-2.30$ to $6.30$, $-0.43$ to $3.43$ and $1.20$ to $2.80$ all miss. In particular $-2.30$ to $6.30$ is $2 \pm 4.3027 \times 1$ — the same $t$ value applied to $s$ instead of $s_{\bar{X}}$, i.e. the missing $\sqrt{3}$. Since none matches $(-0.48,\ 4.48)$, the correct response is "none of the other alternatives are correct".

6. 6

   Interpret the width. An interval stretching from below zero to almost $4.5$ around a mean of $2$ is the honest cost of $n = 3$: three observations simply cannot locate a mean precisely. Verification: $4.3027 \times 0.57735 = 2.48490$, and $2 - 2.48490 = -0.48490$, $2 + 2.48490 = 4.48490$ ✓.