# Calculate variance for effort estimation

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

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

$n$: 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.

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

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.

• 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? – JohnDoe Jan 14 '15 at 10:39
• Because in your question, you posted $\{13,13,13,8,13,20,13\}$. In this dataset $8$ has the third position. – Thijs Jan 14 '15 at 10:50
• Ok, thank you! I understand now. – JohnDoe Jan 14 '15 at 10:56