The standard error of a mean is SE = s/√n. Error bars of ±2 SE give roughly a 95% confidence interval; when such bars for two groups do not overlap, the difference is likely real.
In reading graphs and data you compared bars by their heights. But a mean from a few measurements is only an estimate. Grow 10 more bean plants and the mean height will come out a little different. Error bars show how much we trust a mean. To read them, you need two ideas that sound alike: the standard deviation and the standard error. You met the first in Standard Deviation.
Standard deviation (SD, $s$) describes the data: how spread out the individual measurements are.
Standard error of the mean (SE) describes the mean: how far the sample mean is likely to be from the true mean of the whole population.
🔑 Key idea
$$SE = \frac{s}{\sqrt{n}}$$ A bar of the mean ± 2 SE is roughly a 95% confidence interval. If the ±2 SE bars of two groups do not overlap, the difference between the groups is likely real.
💡 Why it works
Why divide by $\sqrt{n}$? A single measurement can land far from the true mean. But when you average many, the highs and lows partly cancel. The spread of the means you would get from repeated samples shrinks as the sample grows, in proportion to $1/\sqrt{n}$. So four times as many measurements cut the SE in half. The SD does not shrink, because the plants are no less varied just because you measured more of them.
Drag the sample size. The red bar (±1 SD) stays about the same width; the green bar (±2 SE) narrows.
Worked examples
✏️ Worked example: Standard error from SD and n
Sixteen bean plants have a mean height of 24.0 cm with a standard deviation of 4.0 cm. Find the SE and the ±2 SE interval.
$SE = s/\sqrt{n} = 4.0/\sqrt{16} = 4.0/4 = 1.0$ cm.
2 SE = 2.0 cm.
Interval: 24.0 − 2.0 to 24.0 + 2.0, that is 22.0 cm to 26.0 cm.
Answer: SE = 1.0 cm; the interval is about 22.0 to 26.0 cm. We are about 95% confident the true mean lies in it.
✏️ Worked example: Do the bars overlap?
Plants given fertilizer A: mean 30.0 cm, SE 1.2 cm. Plants given fertilizer B: mean 25.0 cm, SE 1.0 cm. Is the difference likely real?
A: 30.0 ± 2(1.2) gives 27.6 to 32.4 cm.
B: 25.0 ± 2(1.0) gives 23.0 to 27.0 cm.
The top of B (27.0) is below the bottom of A (27.6): the bars do not overlap.
Answer: The difference is likely real: the plants with fertilizer A grew taller.
The second example as a graph: the ±2 SE bars do not overlap.
⚠️ Common mistake
Two mix-ups. First, reading overlapping bars as proof that there is no difference. Overlap only means these data do not show a clear difference; a larger sample, with smaller SE, might. Second, swapping SD and SE. Check which one a graph uses (it should say in the caption), and remember the test: more data shrinks the SE, but not the SD.
The "no overlap" rule of thumb is a quick visual check, not a formal test. Its formal partners are the null hypothesis and tests like chi-square, which come next.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.