# Error calculation

1. Aug 22, 2015

### desquee

(Sorry if this is the wrong place for this question, I wasn't sure where to put it)

I've been running an experiment with rabbits, and am trying to figure out the error of my feed measurement.
I fed them using a 1/3 cup measure, and recorded the number of scoops they got. I estimated that the error for each individual scoop is 10%. If I then want to determine the error of the total amount of feed (850 scoops), would that also be 10%?

Intuitively that seems too high, since it's incredible unlikely that every single scoop was high by 10% (or low by 10%). In fact, I can say for sure that some scoops were high (over 1/3 cup) and some were low (under 1/3 cup), so that its impossible for the total feed to be off by 10%.

Is there a way of more accurately calculating the error?

2. Aug 22, 2015

### mathman

A good approximation to the error (assuming independent errors) is $\frac{10}{\sqrt{850}}$ %.

3. Aug 22, 2015

### desquee

Thanks mathman. That comes to 0.34%. I want to better understand how that equation is derived, do you by any chance have a link to an explanation?

4. Aug 23, 2015

### Hornbein

5. Aug 23, 2015

### desquee

OK, so reading that, it looks like the equation given by mathman is for the std of the mean of the sample. But what I'm trying to figure out is the std of the sum of the sample, and I don't see that the link explains how to do that.

6. Aug 23, 2015

### Hornbein

The std of the sum of the sample is equal to
the std of the mean of the sample times the sample size.

sigma_sum = n * sigma_samplemean

Since you are using percentages, you divide n * sigma_samplemean by n to get the percentage.

n/n = 1.

Voila! sigma_sum% = sigma_samplemean%

It is worth mentioning that your actual error is in all likelyhood greater than 0.34%. I'd bet that you have systematic error that is greater than this unsystematic error.

Last edited: Aug 23, 2015
7. Aug 23, 2015

### mathman

The variance of the sum of independent random variables is the sum of the variances.

Proof:
Let $(X_k, k=1,n)$ be a set of n independent random variables, each mean $=m_k$. Let $Y_k=X_k-m_k$. The variance of the sum is $E((\sum_{k=1}^{n}Y_k)^2)=\sum_{k=1}^{n}E(Y_k^2)+\sum_{k=1}^{n}\sum_{j\ne k}E(Y_jY_k)$. However, due to independence, the $E(Y_kY_j)=0 \ for \ j\ne k$. Therefore the variance of the sum is the sum of the variances.

Last edited: Aug 23, 2015
8. Aug 23, 2015

### desquee

Hornbein: What sort of systemic errors should I be looking for? The feeding method was: fill scoop from feed bucket, pour feed from scoop into feeding container in cage, count the number of scoops I added. And the number I'm interested in is the total amount of feed I gave to the rabbits (so feed that gets knocked out of the cage by the rabbits still counts).

I'm just now learning how to work with errors, so any advice on what to be looking for would be helpful.

9. Aug 23, 2015

### Hornbein

Oh, you likely tend to add a little over or a little under a third cup. I bet that that tendency is greater than 1%.

10. Aug 23, 2015

### desquee

My original assumption is that the error for each scoop was 10%, from which the 0.34% for 850 scoops was calculated. Was that 1% a typo and you meant to say that you bet the tendency is greater than 10%?

11. Aug 23, 2015

### desquee

You were right Hornbein. I just tried measured the greatest likely error of a scoop. I measured the weight of exactly 1/3 cup of feed (or as exact as I could make it given that the feed was solid). Then I took a scoop as overflowing as it could be without feed falling out, removed the excess, and weighed that. Then I took a scoop as under-full as it could be without being so empty that it would have induced me to notice and refill it, and measured the weight of the feed I added to bring it up to 1/3 cup. The result was an error of 35% (say 40% given the slight error in my attempt to make the scoop to exactly full).

Even with a 40% error for each scoop the error for the sum of 850 scoops is only around 18.5 scoops, or 2.18%.

12. Aug 24, 2015

### Hornbein

Not a typo. I'm saying that I bet you systematic error makes this sampling error insignificant. 1 > .34. I doubt that you measure so badly that your systematic error is greater than 10%.