Elite X-Naut Posted February 6, 2016 Report Share Posted February 6, 2016 Hey guys so it's time to calculate uncertainties for the IA which is due in two days. My teacher has told us that there is new uncertainties calculations for the new 2016 syllabus, although I searched videos/syllabus/forum posts and could not find any evidence of this. So what she said was: Old syllabus: For adding or subtracting A and B, you add or subtract uncertainties of A and B.Multiplying or dividing, you calculate percent uncertainties. New syllabus (what we should be doing):Adding and subtracting A and B: Square uncertainties of A, square uncertainties of B, add them together, and square root the whole thing.Multiplying and dividing A and B: Take the standard deviation of uncertainties, or divide uncertainty of A by value of A, square it, divide uncertainty of B by value of B, add them, and square root the whole thing. My calculations:10.0 +/- 0.05 (volume in a pipette) divided by 10.70 +/- 0.01 (time using stopwatch app on the phone)10.0/10.91 and 10.0/11.46 using the same uncertainties. I then need to take an average of these three results. Which way should I be calculating uncertainties? Are there actually changes in calculating them in the new syllabus?I am inclined to think I should be doing the 'old syllabus' way because otherwise, taking an average of uncertainties requires more tedious work (a lot of squaring and square roots). Thank you! Reply Link to post Share on other sites More sharing options...
Elite X-Naut Posted February 6, 2016 Author Report Share Posted February 6, 2016 Thank you, I have sent my teacher an email. I just checked my textbook and my study guide, you're right; I should be using the easier method. We're doing it early since this was the only time we got booked for the labs, however I'm unsure about the official due dates for when they are sent off. I think this is because the teacher wants to get it fully marked before officially submitting it. But since we've only covered topics 1-5, our experiments are basic ones like titrations, rate of reactions, etc, nothing complicated. Reply Link to post Share on other sites More sharing options...
Elite X-Naut Posted February 7, 2016 Author Report Share Posted February 7, 2016 (edited) Also, I'm currently stumped on the following calculation of uncertainty, and I've roamed the internet in search for an equation but I didn't get anything. I'm taking an average of the following numbers and I have the average, but not its uncertainty.0.935 +/- 0.0470.917 +/- 0.0460.873 +/- 0.044 I found ways to calculate the uncertainty of the average IF the uncertainties were the same, but they are all different. How would I do this? Edited February 7, 2016 by Elite X-Naut Reply Link to post Share on other sites More sharing options...
Vioh Posted February 7, 2016 Report Share Posted February 7, 2016 Also, I'm currently stumped on the following calculation of uncertainty, and I've roamed the internet in search for an equation but I didn't get anything. I'm taking an average of the following numbers and I have the average, but not its uncertainty.0.935 +/- 0.0470.917 +/- 0.0460.873 +/- 0.044 I found ways to calculate the uncertainty of the average IF the uncertainties were the same, but they are all different. How would I do this? Can you tell me what the values 0.047, 0.046, & 0.044 stand for? Are they the propagation of errors that originate from the uncertainties of the measuring equipments themselves? Because if they are, then they represent the so-called reading error. There's also another way to calculate the uncertainties, namely the Range/2 method, which will give you the random error for the experiment (& this gives you a measure to how spread your results are). Now, the general rule of thumb is that if the reading error is higher than random error, you use the reading error. But if the random error is higher, then you use the random error instead. So in this case, the reading error for the average result is:(0.047 + 0.046 + 0.044) / 3 = 0.046And the random error is:Range / 2 = (0.935 - 0.873) / 2 = 0.031, which is less than the reading error.So the result can be quoted as: 0.908 +/- 0.406 1 Reply Link to post Share on other sites More sharing options...
Elite X-Naut Posted February 7, 2016 Author Report Share Posted February 7, 2016 Yes, 0.047 comes from division of (10.0 +/- 0.5 cm3) / (10.70 +/- 0.01) s, where 0.5 is of the pipette, and 0.01 is my timer. So I assume this is where I use the reading error. Do both those calculations apply to propagating errors from uncertainties of the equipment? For example, should I calculate both reading and random error for the rest of my calculations and then choose which one is lower? Also, since I show sample calculations outside of a table (using the numbers above), if I use random error for another calculation inside a table should I also mention/show this work? (Different method of calculation right?) Reply Link to post Share on other sites More sharing options...
Vioh Posted February 8, 2016 Report Share Posted February 8, 2016 (edited) Do both those calculations apply to propagating errors from uncertainties of the equipment? For example, should I calculate both reading and random error for the rest of my calculations and then choose which one is lower?Random error should only be calculated if you want to know the uncertainty for an average. This is because random error is a measure of how spread your results are around an average.For most other cases, you can use reading errors (i.e. uncertainty of equipment) for propagation-of-error calculations. Also, since I show sample calculations outside of a table (using the numbers above), if I use random error for another calculation inside a table should I also mention/show this work? (Different method of calculation right?)I'm not quite sure what you mean here. Can you clarify a bit more on what you're trying to do? Edited February 8, 2016 by Vioh Reply Link to post Share on other sites More sharing options...
Elite X-Naut Posted February 11, 2016 Author Report Share Posted February 11, 2016 Sorry for the late response, I've already submitted my IA. Yeah I just needed uncertainties for the average and I used random error for all my calculations for that. But regarding the information given above, thank you so much for your help! It's really appreciated. Reply Link to post Share on other sites More sharing options...
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