0
$\begingroup$
            Loss$         Total Weight LB     $/GT
       2009 $74,563.07     127,938.00         0.58 
       2010 $962,970.22    226,424.00         4.25 
       2011 $1,773,942.47 295,228.00          6.01 
       2012 $350,768.36    319,619.00         1.10 
       2013 $1,056,278.57 395,904.00          2.67 
       2014 $763,223.54    350,201.00         2.18 
       2015 $34,895.84     290,466.00         0.12 
       Total 5,016,642.07 2,005,780.00        2.50

The above data is made up, but its along the lines of what I would like to do. I have the average loss there at 2.5$/LB,using the total data, I would like to run a monte carlo simulation on it to justify more rigorously the 2.5 avg. I was thinking maybe for each year in the Loss/GT, I should do =Normalvalue(2.5,1.9555), where 1.955 is the standard deviation of all the values. After that, run it a few times taking the average. I am not sure if this is the right way to ago about it, but I need to figure out what the actual average should roughly be with confidence if possible. On top of the above idea, I should make the loss column values = LB* Loss/GT since they would change with each iteration.

Anyone know how to do what I want?

$\endgroup$

closed as unclear what you're asking by leonbloy, Milo Brandt, Leucippus, Fabian, Yagna Patel Jan 14 '16 at 18:50

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

Wanting to "justify more rigorously" the average does not give me a very good view of what you want to accomplish, or why you would want to use a Monte Carlo procedure.

If you have only 8 observation as in your made up data, then there is not much that can be done. No statistical or Monte Carlo procedure will increase what you know from these 8 observations. These are computational procedure that manipulate the dataset you already have to clarify its meaning. (Some people have the idea that Monte Carlo procedures produce more data about the problem at hand, but the values they produce are for computational purposes only, and add no information about the original situation.)

If the 'losses' are nearly normal, then you could get a 95% confidence interval $(0.78, 4.06)$ for 'loss' using a standard t procedure. So you would be reasonably sure that the mean of the population of losses lies between 0.78 and 4.06. If you don't know about t confidence intervals you could find the formula and an explanation for it in a basic statistics text or online.

In R, the procedure looks like this:

 loss = c(.58, 4.25, 6.01, 1.10, 2.67, 2.18, .12, 2.50)
 t.test(loss)

         One Sample t-test

 data:  loss 
 ...
 95 percent confidence interval:
 0.7910042 4.0614958 
 sample estimates:
 mean of x 
   2.42625 

If you have more data than the 8 observations you show and you think the distribution is far from normal, then you could use a 'nonparametric bootstrap', a Monte Carlo procedure which would give you a confidence interval.

Another posting on this site shows one bootstrap method that might be applicable.

If you can tell me more about your data and what your objective is, please edit your Question appropriately. I (or someone else) might be able to give you more targeted advice.

$\endgroup$

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