# Adding error to a percentage change

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• rwooduk
In summary, the conversation discusses how to deal with errors in determining the percentage increase or decrease of a value with fixed errors. The standard procedures for calculating uncertainties and adding them up are mentioned, as well as the importance of using the correct values for calculating the uncertainty in the percentage change. There are also considerations for time variables and the difference between initial and final values in measurements.

#### rwooduk

This may be a really dumb question and it's been a long day and I can't get my head around this.

If I have two values with errors, say 4 ± 0.5 and 6 ± 1.5, the precentage increase from 4 to 6 is easy its 50 %.

But how do I deal with the errors? I want to write (50 ± X) % where X represents the error in the percentage increase. Do I just take the standard deviation of the two errors so it would be (50.0 ± 0.7) % or do I take an average so its (50 ± 1) %.

Not sure what to do with this, thanks for any pointers.

rwooduk said:
But how do I deal with the errors? I want to write (50 ± X) % where X represents the error in the percentage increase. Do I just take the standard deviation of the two errors so it would be (50.0 ± 0.7) % or do I take an average so its (50 ± 1) %.

No. Your starting function is of the form A/B. According to standard procedures you determine the differential of A/B which will be the uncertainty in A/B. Anything look familiar.?

gleem said:
No. Your starting function is of the form A/B. According to standard procedures you determine the differential of A/B which will be the uncertainty in A/B. Anything look familiar.?

Ahhh of course, so something like this...

Thanks!

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gleem
Try again, so like this...

Which gives me an error of 0.33, this must be too low? (50.0 ± 0.3) % ??

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OOPs not quite.get rid of the Δy it is 0.

IF Z = A/B -1

Let ΔA = uncertainty in Z associated with the uncertainty in A = δA The same with ΔBIf δA and δB are random and uncorrelated we may write that ΔZ2 = ΔA2 + ΔB2What is ΔA? it is (∂Z/∂A)δA The same with ΔB.

You have to be careful in using the correct values for δA and δB .in determining the uncertainty in the percentage change. You must calculate the actual error in Z first then determine the percentage.of ΔZ to Z to get the uncertainty of the percent change of A relative to B.

rwooduk
No, you have an uncertainty of 0.33 for a value of 0.50. 50%±33%.

Edit: Looks like there is another error somewhere, see the post above. The answer is not too far away from 33%, however.

rwooduk
Excellent I see now, will have another go tomorrow and shouldn't be a problem, many thanks for the help !

More than 50 years ago, in my first University physics course, we were taught the basics of measurement errors and how to add them up. In short:
1. Take several measurements
2. The average of the measurements is the best estimate of the "real" value of the thing you are measuring.
3. The standard deviation of the measurements is the best estimate of the absolute measurement error
4. The relative measurement error is the absolute measurement error divided by the average of the measurements
5. If your measurement involves several variables, calculate the average and the relative measurement error for each.
6. You add relative measurement errors in the same way as you add standard deviations - square them, add them and take the square root of the sum.

rwooduk
This isn't going to work for percentage decrease for fixed error values because if the percentage decrease is greater this will lead to a greater error value!

Example, here is my data:

The fixed error for each initial and final value is ± 282.40 therefore for the greatest percentage decrease where it drops to 311 (i.e. for that value it will be 311 ± 282), so the error is huge (± 27) and doesn't reflect the error in the system. The greater the percentage decrease the greater the error will be, so the error now depends on the percentage decrease and not the original standard error. Do you see the problem?

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rwooduk said:
The fixed error for each initial and final value is ± 282.40 therefore for the greatest percentage decrease where it drops to 311 (i.e. for that value it will be 311 ± 282), so the error is huge (± 27) and doesn't reflect the error in the system. The greater the percentage decrease the greater the error will be, so the error now depends on the percentage decrease and not the original standard error. Do you see the problem?
Yes, and there is something you do not tell me. Since you have "Initial" and "Final" - is there a time variable involved? What is your measurement - the final value or the decrease? If it is your final values, all you have is an estimate of 605 and an absolute error of 287.

rwooduk said:
The fixed error for each initial and final value is ± 282.40 therefore for the greatest percentage decrease where it drops to 311 (i.e. for that value it will be 311 ± 282), so the error is huge (± 27) and doesn't reflect the error in the system. The greater the percentage decrease the greater the error will be, so the error now depends on the percentage decrease and not the original standard error. Do you see the problem?

I do not know what you are doing. You have three measurements. You are looking to determine the percent decrease in the measurements You have two fixed errors for the final and initial measurements. Why didn't you calculate the average decrease and the standard deviation of the data first? You should then incorporate the fixed error to get your total error. Then determine the percentages.

Svein said:
Yes, and there is something you do not tell me. Since you have "Initial" and "Final" - is there a time variable involved? What is your measurement - the final value or the decrease? If it is your final values, all you have is an estimate of 605 and an absolute error of 287.

edit
Apologies, I see where the misunderstanding lies. Each Set is a separate experiment unrelated to the others. So I need a percentage change and associated error for each set. Edited...

So as you can see with a fixed error for each initial and final values (± 282.40), the error becomes dependant upon the amount of decrease (the greater the decrease the more error on the final value) not the standard (± 282.40) error.

For example:

At 10 W the final value is 311 ± 282.40, which will lead to a greater error value for the percentage change (27%) than say for 20 W where the final value is 620 ± 282.40 which gives (6.9%)

The error in the percentage change is dependant upon the amount of change, and does not reflect the systematic ± 282.40 error.

Apologies, hope that makes sense.

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I don't understand how you get your values.
The ratio is final/initial. Its relative uncertainty is sqrt((282.4/final)^2 + (282.4/initial)^2) which is about 90%, 46% and 32%, respectively. The absolute uncertainty for the ratio is then that number multiplied by the ratio - and it is about 0.03 in all cases. The absolute uncertainty for 1-ratio is the same. You get 3% uncertainty for the percent decrease.

rwooduk
mfb said:
I don't understand how you get your values.
The ratio is final/initial. Its relative uncertainty is sqrt((282.4/final)^2 + (282.4/initial)^2) which is about 90%, 46% and 32%, respectively. The absolute uncertainty for the ratio is then that number multiplied by the ratio - and it is about 0.03 in all cases. The absolute uncertainty for 1-ratio is the same. You get 3% uncertainty for the percent decrease.

Sorry, yes I had derived the uncertainty formula, then things have gone wrong somewhere, you are right, but I will recheck my calculations.

Thanks again for the help ! Three years of chemistry and I've completely forgotten my physics ! Thanks again

## 1. How do you add error to a percentage change?

To add error to a percentage change, you first need to determine the original percentage change. Then, you can calculate the error by multiplying the original percentage change by the percentage error. Finally, you can add the calculated error to the original percentage change to get the new percentage change with added error.

## 2. Why is it important to add error to a percentage change?

Adding error to a percentage change is important because it helps to account for any uncertainty or variability in the data. This allows for a more accurate representation of the change and can help to avoid misleading conclusions.

## 3. Can you add error to both the numerator and the denominator in a percentage change?

Yes, you can add error to both the numerator and the denominator in a percentage change. This would result in an overall added error to the final percentage change.

## 4. How does the magnitude of the error affect the final percentage change?

The magnitude of the error can greatly impact the final percentage change. A larger error will result in a larger change, while a smaller error will result in a smaller change. It is important to consider the size of the error when interpreting the final percentage change.

## 5. What are some common sources of error when calculating a percentage change?

There are several common sources of error when calculating a percentage change, including human error in measurement or data entry, sampling error, and measurement devices with limited precision. It is important to carefully consider and account for these sources of error when adding error to a percentage change.