variance/confidence interval of development patterns I was wondering if there is a technique to calculate the variance of developments.
eg. I have paired data eg: 100 110, 200 225, 78 92, 33 20...
This means the developments are: 1.1, 1.125, 1.18, 0.6...
I was wondering if there is a technique to calculate a confidence interval around this. And I was also wondering if there is a way to weight this as well (eg. instead of assuming each development is equal, weigh the 200-225 development twice as much as the 100-110 one.
 A: In the situation you describe it seem there is a good chance that
the development scores will be approximately normally distributed.
If you have software available, you can use a standard normality
test (such as Anderson-Darling or Shapiro-Wilk) to see if data are
consistent with sampling from a normal distribution. 
For normal data a standard kind of confidence interval for the
population variance is based on the chi-squared distribution.
Let $S^2$ be the sample variance of $n$ normal observations.
Then $Q = (n-1)S^2/\sigma^2 \sim \mathsf{Chisq}(n-1).$ So you
can use printed tables of the chi-squared distribution (or
software) to find values $L$ and $U$ such that
$$P\left(L < Q=\frac{(n-1)S^2}{\sigma^2} < U\right) = 0.95.$$
Then, after some straightforward algebra, a 95% CI for $\sigma^2$
is if the form $$\left(\frac{(n-1)S^2}{U},\, \frac{(n-1)S^2}{L}\right).$$
To find a 95% CI for $\sigma,$ take square roots of both endpoints of the
CI for $\sigma^2.$
For example, if $n = 25$ and $S^2 = 0.85$, then a 95% CI is $(0.518, 1.645)$,
where the computation in R statistical software is shown below. [Notice that
$S^2 = 0.85$ lies within the interval, but (because the chi-squared distribution is not symmetrical) not at its center.]
Num = 24*0.85
Num/qchisq(c(.975, .025), 24)
## 0.518239 1.645009

Other methods, including boostrap CIs are available if it is clear that
data are not normal. If you need help with a sample of reasonable size,
you can list it in your Question in the format $(1.1, 1.125, 1.18, 0.6, \dots),$
[that is, horizontally, with commas between values], leave me a Comment, and I will take a look.
