View Full Version : Find aggs

James M.

12-03-2014, 12:33 PM

Since the "Cabin Fever" season has begun for many shooters, please allow me to offer a little quiz. There has always been some problems with measuring groups with accuracy, but I contend that there are serious problems with our misuse of statistical rules when averaging groups. Using the statistical rules of significant digits, please average (or find the agg.) for the following groups. O.278, 0.197, 0.212, 0.235, and 0.159. What is that fictional shooters agg? According to the statistical rules mentioned above, what should it be? Good luck...James Mock PS- if shooter B had the following: 0.279, 0.197, 0.213, 0.235, and 0.160, what would his agg be? Who won the match? Who would win if the statistical rules were faithfully followed?

James M.

12-03-2014, 04:23 PM

No math people? Did you get 0.2162 for the first shooter and 0.2168 for shooter B. If so, you got what the NBRSA or IBS would have gotten. However, using the rules of statistical averages where measured 3 significant digits cannot be averaged to 4 significant digits. In other words, both shooters shot a 0.216 agg. I know that this will probably never happen, but that is the way it would be done with proper averaging. Good shooting...James

Centerfire

12-03-2014, 04:34 PM

Just woke up and did the math.

1st answer .2162 2nd answer .2168

Hope my Victor 1228-2 did it correctly I would award to the .2162 total.

Centerfire

r44astro

12-03-2014, 04:36 PM

It is only a guess but if someone has to win you keep going until one number is smaller. I suppose it is possible to have true mathematical tie. Ask Steve Lee author of BugHole

bill

David Halblom

12-03-2014, 05:04 PM

No math people? Did you get 0.2162 for the first shooter and 0.2168 for shooter B. If so, you got what the NBRSA or IBS would have gotten. However, using the rules of statistical averages where measured 3 significant digits cannot be averaged to 4 significant digits. In other words, both shooters shot a 0.216 agg. I know that this will probably never happen, but that is the way it would be done with proper averaging. Good shooting...James

is to simply add up the decimal measurements and express the agg as a numerical sum. The first shooter would have still won, but it would be expressed as a truer "number". Personally, I don't see ANY way of getting to a mathematically "true" way of expressing the agg, but someone has to win and this has proven to be pretty damn long lived way of doing it. Maybe not right, but it works. Even as we get smaller and smaller, groups that is.:cool:

P.Ericson

12-03-2014, 05:11 PM

I have studied math at university, and I don't really get what the big problem is. The shooter with the .2162" agg had a 1.081" total spread, the other one had a 1.084". Shooter ones "score" is 1.081/5, shooter two came second with a 1.084/5...

For me, it's like saying that 3 divided by 2 isn't 1,5.:p

David Halblom

12-03-2014, 05:41 PM

I have studied math at university, and I don't really get what the big problem is. The shooter with the .2162" agg had a 1.081" total spread, the other one had a 1.084". Shooter ones "score" is 1.081/5, shooter two came second with a 1.084/5...

For me, it's like saying that 3 divided by 2 isn't 1,5.:p

However, we run into some problems as we calculate the grand aggs and two guns...

we don't run into a problem in grands or multiple guns IF we express the aggregates as sums of the total. Just run a series of ten numbers, or 20 numbers or 40 numbers and you will see what I mean. Where we run into problems is if we average the averages, the individual agg averages and not the distinct targets, As we get smaller and smaller separations in competitors what just described becomes VERY significant. Using the 3 decimal measurement and 4 decimal expression there are some BIG matches where second place is losing by .000X averages. What method of averaging was used? Each way will yield a different number. For those who truly mathematics, much the same as where parentheses are placed in an equation.

r44astro

12-03-2014, 06:13 PM

coin toss anyone?

R.G. Robinett

12-03-2014, 06:25 PM

The preceding, all dismiss the reality that, with the capture and measurement tools we use, groups cannot be precisely measured: the default error being +/- the 0.009" change in measurement required by the protest rule! Further, different individuals (official scorer & referees), using the same tools, on the same day, may produce group measurements varying by that much, or, more.:p So, the REAL significant decimal place becomes the second! Therefore, the original example is definitely, "a statistical tie." ;) RG

afrench

12-03-2014, 06:34 PM

it looks like this has been addressed, at least for the group nationals, during the NBRSA BOD meeting at the nationals back in Sept.

from page 6:

Gulf Coast Items:

1. Electronic instruments round off after certain decimal places. There was an incident in the region where the two gun winner did not have the smallest aggregate due to rounding. Mr Hunter made a motion that for the Group Nationals only, if two people appear to be tied for multi-gun aggregates, the scorer will do an add-up with the lowest total to determine the winner. Second by Mr Neary and unanimously passed.

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