Target Accuracy and Precision

[Chipperzs]

I’m a novice when it comes to applying statistics to practical applications. Sorry if this has been covered in another post somewhere. I compete in competitions where accuracy and precision is compared between two competitors.

My question is...
How many shots at a target are required to determine the accuracy?
What attributes, factors or terms should we be talking about when comparing one competitor’s results (precision) to another?

Some times competitors take 3 shots and brag about how accurate they are. But I know that it’s the repeatability of that group size that really matters. Sometimes 5 shots are taken and some times 10 or 20 are taken.

I know that precision is dependent on the number of shots taken. A 3 shot group that measures .3” is a lot less impressive than a 20 shot .5” group.

#1) Is there a better number of shots that I should use? Should I include a confidence number (or some other attribute) when comparing group sizes with a larger or smaller number of shots used?

#2) What are the attributes of the number of shots taken vs groups size that we should be talking about? I typically shoot 5 shot groups. I feel that 3 is too little and 10 is too much when running tests.

 

[Chipperzs]

I found some info on measuring the Coefficient of Variation. Let me know if this works for my application...

Take measurements of the shots variation from the center of the group. Use that to estimate the standard deviation of distance from center then divide the number of shots by that standard deviation.

 

[Chipperzs]

Would this be a better calculation...

CV = (Standard Deviation of the shots distance to group center) / (the average shot distance to the group center)

 

[Miner]

There is no cut and dried answer to your question as posed. You can use confidence intervals to quantify the uncertainty in both the accuracy (difference between mean and target) and precision (standard deviation). As you increase your number of shots, the uncertainty will decrease. If there is an acceptable level of uncertainty, you can calculate the number of shots necessary to reach that. In order to compare accuracy between two shooters, use a 2-sample t-test. To compare precisions, use a test for equal variances.

 

[katxt]

Are you looking for some some way practical way to compare groups from different numbers of shots on the range, without having to resort to calculations?

 

[Chipperzs]

I fully expect to have to do some sort of calculation. I’m trying to work with one of the phone app programmers to have this additional information included with the group size (MOA) that they provide.

As @Miner stated earlier, a confidence level or some other attribute (maybe histogram shape?) should be included with the analysis so that a fair comparison of a 3 shot group can be made to a ten (or more) shot group.

 

[katxt]

A phone app - that's very ingenious. Do you take a photo and the app works out the measurements and does the calculations?

 

[Chipperzs]

Yes, there are several on the market now. The one I use makes you select a couple points to calibrate distances. Then choose the point of aim, then select locations for impacts. Example is attached.

 

[katxt]

A very useful screenshot. Things are becoming clearer.
Can you explain why it is that a 3 shot group that measures .3” is a lot less impressive than a 20 shot .5” group?
One approach to this is to calculate and draw a 95% circle for each case which would, in theory, contain 95% of all future shots. The shooter with the smallest 95% circle is the most precise.

 

[Chipperzs]

Group sizes never decrease, they can only get bigger or stay the same. It’s not hard to “get lucky” and shoot 3 right next to each other that measure .3” maximum diameter. But it’s rather difficult to do it 20 times in a row and have it measure less than .5”

I think it’s one of the flaws when comparing group sizes. I think there should be some leniency given and one shot every [8?] dropped from the aggregate. Maybe this could be an option in the software.

I was thinking of something similar to your 95% circle. Is there a way to use that to predict a maximum and minimum group size as you add more shots to the calculation? Predicted groups would look like a donut (maybe a transparent donut overlaid on the image). The 3 shot group would equal the inside diameter and the outside diameter equal to the expected maximum group size for [20 shots?]. With each additional shot the donut (group uncertainty size) would get thinner and thinner as the uncertainty decreases. Then a [95%?] circle could be used to compare a 3 shot group to a 5, 10, etc.

 

[katxt]

These things are possible, at least approximately, so long as the number of shots is large. Unfortunately, most of the theory breaks down at sample sizes like 3 and 5.

Here is another possibility - assume that one particular consistent shooter fires a group of many thousands of shots and that the distribution of these shots is normal and random in both x and y directions (bivariate normal) with sd = 1. Almost all of the shots will be in a circle of about 6 units. We now select many of groups of 3 at random and find the spread of each group by simply measuring the distance between the two most distant points. The average of all these spreads is a measure of a typical 3 shot spread. This average spread is is impossible to compute by formulas, but it is very easy to estimate it by simulation. Then the same for a large number of 5 shot groups taken at random from the bivariate distribution. They also have a typical average spread. And for say, 10 and 20 shot groups. Here's what we get from an Excel simulation of 20000 iterations at each group size (d is the average diameter) -

Now, if we want to see what the 20 group equivalent of 3 shots in 0.3" we go 0.3x4.46/2.43 = 0.55 which is not as tight as 20 in 0.5" which confirms your thought that 20 at 0.5" is better than 3 at 0.3"

 

[Chipperzs]

This is awesome. And shortly after writing the post I came to the same conclusion as you, it’s difficult to calculate with formulas and could be easier with a simulation.

Did you create a spreadsheet that runs this simulation? Could you help me create one that does? I was obsessing over this the other night and I drafted an outline for the simulation. Another addition to the simulation Would be to run it for different types of target group densities. I would like to create sample data with an average distribution (platykurtic), a dense center distribution (leptokurtic), and normal distribution (kurtosis). This would generate 3 different conversation tables. Which would also be useful for characterizing the type of precision one shooter is achieving achieving vs another shooters.

 

[katxt]

I will attach the file I used. It is based on a spreadsheet you can find here -
https://sie.scholasticahq.com/article/4672-build-your-own-monte-carlo-spreadsheet (look in the data sets) Forget the p values.
It is set up for 5 shots. Copy down the formulas and adjust them to cover the right range. H2 and down are array formulas. If you're not used to them, they must be entered using ctrl-shift-enter. Once the top one is in the rest can be copied down.
I'm not sure how you would go about generating different distributions. or even how you would know which shooters were producing what. Normal is easy. Excel will produce random beta variables and they can be tuned to give distributions of different shapes.
Cheers, kat

 

[ChuckS]

I watched this video.

Then did the following
So I have a target with X and Y coordinates. I find the minimum circle indicated by thick green line. I have the center to that circle X,Y. I get the distribution of the bullets from the X,Y position in Radius terms. What I want to do is predict the probabilty of them falling into the 90%, 95% and 99.5% groups. I beleive i did it correctly, but i am not a statistics guy. in this case the sample size was 53. If i get the T Value (52,.9) it is equal to 1.675. I then multiply 1.675 to the STDev to get 0.228. I take the mean of 0.2345 and add it to 0.228 to get .462 radius from the center as my 90% confidence. I used the 2 tail table
which is posted here
http://www.talkstats.com/threads/confidence-interval-problem.74752/

 

[katxt]

Unfortunately Radius is not normally distributed, so your analysis doesn't work. X and y are normal so r^2 which is x^2+y^2 will be chi square with 2 df.