The study I am reviewing reports mean height for 20 subjects as 1.70 metres with a standard deviation of 0.0. Does this mean all 20 are exactly 1.70 metres? Or is this a reporting error?

**Answer**

According to this biology SE thread, the standard deviation of male adult height is about $0.07$ metres, and of females is about $0.06$ metres.

Rounding these to one decimal place would give $0.1$ metres. The fact that the standard deviation is reported as $0.0$ metres indicates a standard deviation below $0.05$ metres … but a standard deviation of, say, $0.048$ metres would still be consistent with the reported figure as it would round to $0.0$, yet would indicate a variation in heights in the sample only slightly less than the variability we observe everyday in the general population.

Is the figure well-reported? Well, it would be far more useful if the standard deviation had been reported to two decimal places, as the mean was. It may also be a simple numerical or rounding error; for example $0.07$ could have been *truncated* to $0.0$ rather than *rounded*. But could it be possible the figure refers to the standard error instead? I often see figures written in a way that makes it ambiguous whether a standard deviation or standard error is being quoted — for example, “the sample mean is $1.62 (\pm 0.06)$”.

Just how plausible is it for the correct standard deviation to round to $0.0$ to one decimal place? The following R code simulates one million samples of size twenty taken from a population of standard deviation $0.06$ (as has been reported elsewhere for female height), finds the standard deviation for each sample, plots a histogram of the results, and calculates the proportion of samples in which the observed standard deviation was below $0.05$:

```
set.seed(123) #so uses same random numbers each time code is run
x <- replicate(1e6, sd(rnorm(20, sd=0.06)))
hist(x)
sum(x < 0.05)/1e6
[1] 0.170691
```

Hence a standard deviation that rounds to $0.0$ is not implausible, occurring about seventeen percent of the time if heights are normally distributed with true standard deviation $0.06$.

Subject to these assumptions we can also calculate, rather than simulate, that probability as approximately seventeen percent, as follows:

$$P(S^2 < 0.05^2) = P\left(\frac{19 S^2}{0.06^2} < \frac{19 \times 0.05^2}{0.06^2}\right) = P\left(\frac{19 S^2}{0.06^2} < 13.194\right) = 0.1715$$

where we have used the fact that ${(n-1) S^2}/{\sigma^2} = {19 S^2}/{0.06^2}$ follows the chi-squared distribution with $n-1 = 19$ degrees of freedom. You can calculate the probability in R using `pchisq(q = 19*0.05^2/0.06^2, df = 19)`

; if you replace $0.06$ by $0.07$ in line with published figures for male standard deviations, the probability is reduced to about four percent. As @whuber points out in the comments below, this kind of small “rounds to zero” SD is more likely to occur if the group sampled from was more homogeneous than the general population. If the population standard deviation is about $0.06$ metres, then the probability of obtaining such a small sample standard deviation would also have declined if the sample size had been larger.

```
curve(pchisq(q = 19*0.05^2/x^2, df = 19), from=0.005, to=0.1,
xlab="Population SD", ylab="Probability sample SD < 0.05 if n = 20")
```

```
curve(pchisq(q = (x-1)*0.05^2/0.06^2, df = x-1), from=2, to=50, ylim=c(0,0.6),
xlab="Sample size", ylab="Probability sample SD < 0.05 if population SD = 0.06")
```

**Attribution***Source : Link , Question Author : Lee , Answer Author : Silverfish*