Could you be more clear in how you calculate the error for an individual data point?

@Mew I think both are accurate but the second one is a better indicator of % error. If you have 2 outliers in measurements, they can offset each other and thus provide inaccurate results eg an error % of -70% and 70% would be 0% total, but should be a sum of 140% error. The first method does not account for this. I use this method for each point: calculator.net/…

so for each data point, do you have an associated error, and if so, how was it calculated?

It was calculated using calculator.net/… (ie % error formula)

given that new information, method 2 makes no sense and method 1 is the way to go. Method 2 would have some merit if for each measurement you had a +- value due to the limitations of equipment etc, but if you are just comparing an experimental result to an actual result, you should just compare your final answer, that is your average.

Also, the theoretical value 'g' used was 9.82. Each measurement did have a positive or negative value relative to the theoretical benchmark. % errors can sometimes be negative eg -3.2% ie 3.2% away from 9.82 on the negative side ie 10.13. I hope I make sense

Hello

Normally you have an error associated with your devices, eg you might only be able to measure to the nearest 1mm, so your answer is 5 +1mm

and for such errors, your method 2 is permissible

Yes

but the so called "errors" you describe aren't experimental errors per se, but rather you are just comparing your obtained value to the true value

Ok

and if you just want to check how far your estimated value of g is to the actual value of g, it makes most sense to first obtain your average g, and then see how close it is to the actual g

^ the above applies if your aim is to estimate a value of g and compare with the known value

if your aim however is to test the valiidty of your experimental set up, then averaging the errors in method 2 may be of some use

I think if your teacher hasn't rejected method 1, continue to use it because method 2 may be contravertial

I can see where you are coming from. My original through process was that if my experiment showed a value of g as 700, and another value of g as -700 and only produced an error % of say, 2% it would not seem right. I will keep on using method 1 I suppose for my purposes. Thanks for the help

yes it depends on your aim. If you measure g to be 709.8 and also as -691.2 you're average is 9.8

so your error in g value is actulaly 0

But the experiment had to be invalid with results that large, so method 2 would be more applicable?

if however you are interested in how precise your equpiment is, then you would look at the errors in each experiment as per method 2. So what error you use depends on your aim of experiment

I understand now, thank you for your time

Method 2 is ok, provided you explain your results very clearly. If you're teacher sees 10 with an error of 5%, he will be confused when it is only 0.2 away from 9.8

np

laterz

Cya