0
$\begingroup$

I want to calculate the variance for effort estimation (Scrum), but unfortunately I get a wrong result.

$$ \text{Variance} = \displaystyle \frac{1}{n} \sum_{i=1}^n (X_i - ev)^2 $$

Please note:

$n$: number of involved estimators

$X_i$: position of an estimate within the estimation-scale

$ev$: final estimation value

Example:

Amount of single estimations: $\{13,13,13,8,13,20,13\}$

Final estimation value: $13$

Variance: $≈ 0,286$

My result is:

$$ \frac{1}{7} \left((13 - 13)^2 + (13 - 13)^2 + (13 - 13)^2 + (8 - 13)^2 + (13 - 13)^2 + (20 - 13)^2 + (13 - 13)^2\right) ≈ 10.57 $$

Any help would be greatly appreciated.

Paper: Identification of inaccurate effort estimates in agile software development (p. 69).

$\endgroup$
  • $\begingroup$ The average of the given set is $\frac{93}{7}=13+\frac{2}{7}$, not $13$ $\endgroup$ – Thijs Jan 14 '15 at 9:58
  • $\begingroup$ Thank you for your help. What do you mean? The example is taken from the paper and should be correct. $\endgroup$ – JohnDoe Jan 14 '15 at 10:02
0
$\begingroup$

First of all, the paper uses the word 'variance', but the quantity studied is not the variance.

Their calculation is the following: $$\frac{1}{7}((2-2)^2+(2-2)^2+(2-2)^2+(3-2)^2+(2-2)^2+(1-2)^2+(2-2)^2)=\frac{2}{7}\approx 0,286$$

Here they use the following definitions:

$X_i$: position of an estimate within the estimation-scale

$ev$: position of final estimate within the estimation-scale

In this example, $8$ has third position, $13$ has second position, $20$ has first position.

Given the estimates, the lower this quantity is, the better corresponds the final estimate with the given estimates.

$\endgroup$
  • $\begingroup$ Thank you for your help! However, the estimation scale is like this: {1,2,3,5,8,13,20,40}. So why has 8 the third position, 13 the second and 20 the first? $\endgroup$ – JohnDoe Jan 14 '15 at 10:39
  • $\begingroup$ Because in your question, you posted $\{13,13,13,8,13,20,13\}$. In this dataset $8$ has the third position. $\endgroup$ – Thijs Jan 14 '15 at 10:50
  • $\begingroup$ Ok, thank you! I understand now. $\endgroup$ – JohnDoe Jan 14 '15 at 10:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.