# How to work with an asymmetrical distribution

I'm looking at estimated and actual time taken for a range of projects, as these vary in length quite a lot I've normalised them so they are just estimate/actual, so an estimate of 16 days work that took 12 days shows as 0.75 and estimate of 5 days work that took 7 shows as 1.40.

The total estimate/actual for all the projects is 1.09 so on average the estimates are ok and many are within 10% (45/127), but individually there is a very wide range. I can see the data is asymmetrical, of course nothing can be completed in less than zero time, but more generally there are about as many projects in the 1.0 -> 2.0 range as there are in the 0.5 -> 1.0 range. I believe this may follow for any 1.0 -> x vs 1/x -> 1.0.

0.23 0.36 0.48 0.50 0.50 0.50 0.58 0.58 0.60 0.63 0.63 0.65 0.69 0.69 0.70 0.71 0.72 0.73 0.75 0.75 0.75 0.80 0.81 0.83 0.83 0.83 0.83 0.83 0.86 0.86 0.88 0.88 0.88 0.89 0.90 0.91 0.91 0.91 0.92 0.94 0.95 0.95 0.96 0.96 0.97 0.97 0.97 0.98 0.98 0.99 0.99 0.99 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.02 1.02 1.03 1.04 1.05 1.06 1.06 1.08 1.09 1.09 1.10 1.11 1.11 1.14 1.14 1.15 1.15 1.17 1.17 1.19 1.20 1.20 1.20 1.21 1.21 1.21 1.22 1.24 1.24 1.25 1.26 1.28 1.28 1.29 1.32 1.33 1.34 1.36 1.39 1.40 1.41 1.42 1.46 1.58 1.60 1.61 1.63 1.75 1.81 1.82 1.94 2.12 2.15 2.44 2.56 2.67 2.73

If this were a standard distribution I would work out the sigma value and put a value on the accuracy of the estimates. I could then compare the sigma values between years to see if estimating accuracy is improving.

Is there an equivalent or a transformation I can use to apply the same to this asymmetrical distribution?

• There are many possibilities including a Box-Cox transformation. Sometimes a log or square root transformation can work. – Michael R. Chernick May 7 '17 at 15:13
• The "sigma-algebra" tag appears irrelevant here (the tag does not refer to the "standard deviation"("sigma") that appears to be what the OP is interested in). – Alecos Papadopoulos May 7 '17 at 18:04

Your data exhibits some strange behavior. One choice is to fit a skewed distribution to the data. First, we can examine the Cullen-Frey plot. Since the blue dot appears to be on the dashed line, we might try fitting a Gamma distribution to the data. Using the $\texttt{fitdist}$ function in R, we find MLE for the shape and rate parameters to be: $\alpha = 7.8$ and $\beta=7.1$. Here is a histogram of your data, where the red curve is the fitted Gamma distribution. Your data seems to have a peak at 1, which is going to be hard to capture with a parametric distribution.

Since your primary interest is looking at the estimation over time, you might try the following. Create data by doing $x_i = est_i - act_i$. Your sample size is large enough so that the 1 sample t-procedures should apply. Search for Matched Pairs t-procedures on the web for details. You could construct a confidence interval for the difference by doing:

$$\bar X \pm t^*\frac{S_X}{\sqrt{n}}$$

Where $\bar{X}$ and $S_X$ are the sample mean and standard deviation of the data and where $t^*$ is your critical value which controls the confidence and should come from a t-distribution with $n-1$ degrees of freedom. For your problem, (with $n=127$) a 95% confidence interval would require $t^* = 1.979$.

Do this for various years and see how it changes. This may be a good start.

• There's a clear issue with a lognormal model -- see this more detailed histogram of the logs for example. It might or might not be a particular problem depending on what the model is being used for. – Glen_b -Reinstate Monica May 8 '17 at 1:50