at 11:25 I think the min points should be (103,190), and would in turn affect the slope and uncertainties. Also, thanks for posting all these amazing videos.
In the topic uncertainty in the Y intercept, shouldn't the equations be like this? min slope line y = 3.40x + 21.2 max slope line y = 6,76x + 0.41 Thanks
Gabriel271294 Since we know 2 points on say the min. line, we can use these points to determine the slope and y-intercept. There are different methods for this but typically you determine the slope first, and then you plug in one of the points and the slope into y=mx+b and solve algebraically for b, the y-intercept. Excel will do all of this for you if you make a plot of your two points and then select to see the equation of the trend line. Note that the equation is y=3.40x+0.41 not y-int=3.40+0.41. That is, 0.41 is the y-intercept. The minimum slope line will always have the highest y-intercept. The maximum slope line has the lowest y-intercept, and the true y-intercept should lie between these values.
What about the fact your min slope doesnt go through the second last point? How can it be a line of worst fit if it doesnt fit/go through all the points?
+Evultz Good point. This method of choosing the first and last points isn't the best but it is simple. It is up to you to interpret what it means. Remember uncertainties are rough estimates, and using the first-last point method will likely give you overestimates of your uncertainties. Using all of your error boxes will give you smaller estimates of your uncertainties, but at the expense of added complexity.
We know 2 points on a line. We can use this to determine the slope of the line. We can then substitute the slope and a point into y=mx+b to find the y-intercept, b.
This isn't totally on the level, might work in some cases as a quick heuristic (like an approximation) but this is not how you calculate the propagation of uncertainty for linear regression...
This is high school physics, and this method was suggested by IB teachers as a useful tool and effective way to introduce the topic. All of the videos in this series are meant for high school students. Please feel free to use more advanced methods if that makes you happy... and besides, the very nature of uncertainties is rough estimates.
Use your gumption. In some cases, you might anchor your line of best fit, and maximum and minimum fit lines to the origin because your uncertainty for this point is so small. Ask yourself if (0,0) is based on a measurement, or is it an assumption. Generally, yes, use the first point even if it is (0,0)
Chris Doner thank you so much!! Very kind of you to reply. Yes it was a measured coordinate; it was for a vt graph in which all y uncertainties were constant as well as all x uncertainties (eg 2 m/s for all y and 0.2 seconds for all x). Putting an error rectangle at 0,0 would imply one of the coordinates would have a negative time or a negative velocity (when calculating m1 and m2) and I was unsure either would be possible since it’s motion down a ramp from a stationary point. In other words, a negative velocity wouldn’t have been theoretically possible
It isn't necessarily impossible to get a negative value for a positive quantity because of potential systematic error. If you are convinced there is no systematic error than it is reasonable to anchor the max and min slope lines to the origin.
From your video here (ruclips.net/video/1gEKoJsFkVI/видео.html) you describe the use of the LINEST function to get the absolute uncertainties in the slope and y-intercept of the linear regression parameters for the best-fit line. When applying this LINEST function to the last example problem given for this video I find the uncertainty in the slope to be 0.048 cm/cm and the uncertainty in the y-intercept to be 2.939 cm. These values are very different from the values given as the answer (0.26 cm/cm and 10.9 cm, respectively). Can you please provide comment as to which is the more correct answer and why? Also a comment as to why the two methods give such different answers? Thank you!
LINEST uses all the points (but does not use the uncertainty of any of the points) so it will give a smaller uncertainty than a method that only uses uncertainty on the first and last points. The two point method is a very rough approximation that should always overestimate. Provided you have enough data, LINEST should give a very good value for the two uncertainties.
@@donerphysics What is considered the more appropriate method?: Use the LINEST function to determine uncertainties OR use the "max" and "min" best-fit lines that take into account the uncertainties of all points plotted?
@@donerphysics Thank you for quick and very helpful responses! I would say the 'most correct' procedure would be to honor the data's uncertainties, regardless of how small they may be.
is there a way to calculate the uncertainty of the slope of a tangent of a curve at any specific point ? Lets say you have a quadratic curve of best fit and you want to find the uncertainty of the slope of the tangent at 0 point. The slope can be calculated as f'(0). What about the uncertainty of that ? Thank you
at 11:25 I think the min points should be (103,190), and would in turn affect the slope and uncertainties. Also, thanks for posting all these amazing videos.
Thanks. The graphs were made with Microsoft Excel.
In the topic uncertainty in the Y intercept, shouldn't the equations be like this?
min slope line y = 3.40x + 21.2
max slope line y = 6,76x + 0.41
Thanks
You are right. I interchanged them. I will make a correction when I get a chance.
Thanks. Your videos help a lot in understanding the concepts clearly. Thanks a lot for putting them together.
Mr. Doner for the last question aren't the min x coordinate suppose to be 103 and the max x coordinate suppose to be 97?
+Abbas S
You are right. The x uncertainty was 3 not 5.
This was really helpful for me. Thank you for your videos!
Glad it was helpful!
Jeez, thank you very much! My physics practicals are giving me a hard time with all those weird uncertainties and graphs.
Thanks for the comment Gabriel. I'm glad it helped.
Chris Doner
Just one question, please. How did you calculate the y-intercept and its uncertainty?
"Y-int= 3.40x + 0.41", for example.
Gabriel271294
Since we know 2 points on say the min. line, we can use these points to determine the slope and y-intercept. There are different methods for this but typically you determine the slope first, and then you plug in one of the points and the slope into y=mx+b and solve algebraically for b, the y-intercept. Excel will do all of this for you if you make a plot of your two points and then select to see the equation of the trend line.
Note that the equation is y=3.40x+0.41 not y-int=3.40+0.41. That is, 0.41 is the y-intercept.
The minimum slope line will always have the highest y-intercept. The maximum slope line has the lowest y-intercept, and the true y-intercept should lie between these values.
Chris Doner
Thank you very much for the reply. Yeah, my bad, I meant just "y=". This was very helpful.
To find uncertainty in y-int, would you also do (max-min)/2 just like for slope?
Good, yes.
What about the fact your min slope doesnt go through the second last point? How can it be a line of worst fit if it doesnt fit/go through all the points?
+Evultz
Good point. This method of choosing the first and last points isn't the best but it is simple. It is up to you to interpret what it means. Remember uncertainties are rough estimates, and using the first-last point method will likely give you overestimates of your uncertainties. Using all of your error boxes will give you smaller estimates of your uncertainties, but at the expense of added complexity.
what method should i use for young's modulus graph? Its not straight line ...
+Š П ł P Σ Г
Watch my video on linearization to find out what to plot to transform your data into a linear graph.
Can you still calculate the slope uncertainty with error bars that are too small to be considered graphically significant?
Sure....you do not need to be able to see the uncertainty bars.
Thanks Alot.
Always welcome
Hi there. How would i calculate the percentage uncertainties in gradient and y intercept from the absolute uncertainties?
Percentage uncertainty is always equal to the absolute uncertainty divided by the value x100%
thank you so much!
Glad it helped!
Hi just have a question, at 8:02, how did you know the y intercept is 0.41? Thank you
We know 2 points on a line. We can use this to determine the slope of the line. We can then substitute the slope and a point into y=mx+b to find the y-intercept, b.
Where do you get the uncertainties from in the table from the beginning?
It is made up data.
how are you getting the y-intercepts?
At what time in the video?
@@donerphysics 7:53
Standard procedure is to use the two points to find the slope, m, and then plug one of the points into y=mx+b to find the y-intercept, b.
@@donerphysics thank you
thank you!!!!
This isn't totally on the level, might work in some cases as a quick heuristic (like an approximation) but this is not how you calculate the propagation of uncertainty for linear regression...
This is high school physics, and this method was suggested by IB teachers as a useful tool and effective way to introduce the topic. All of the videos in this series are meant for high school students. Please feel free to use more advanced methods if that makes you happy... and besides, the very nature of uncertainties is rough estimates.
what if my first coordinate is (0,0)? Do I still draw the error rectangle on the 0,0 point
Use your gumption. In some cases, you might anchor your line of best fit, and maximum and minimum fit lines to the origin because your uncertainty for this point is so small. Ask yourself if (0,0) is based on a measurement, or is it an assumption. Generally, yes, use the first point even if it is (0,0)
Chris Doner thank you so much!! Very kind of you to reply. Yes it was a measured coordinate; it was for a vt graph in which all y uncertainties were constant as well as all x uncertainties (eg 2 m/s for all y and 0.2 seconds for all x). Putting an error rectangle at 0,0 would imply one of the coordinates would have a negative time or a negative velocity (when calculating m1 and m2) and I was unsure either would be possible since it’s motion down a ramp from a stationary point. In other words, a negative velocity wouldn’t have been theoretically possible
It isn't necessarily impossible to get a negative value for a positive quantity because of potential systematic error. If you are convinced there is no systematic error than it is reasonable to anchor the max and min slope lines to the origin.
From your video here (ruclips.net/video/1gEKoJsFkVI/видео.html) you describe the use of the LINEST function to get the absolute uncertainties in the slope and y-intercept of the linear regression parameters for the best-fit line. When applying this LINEST function to the last example problem given for this video I find the uncertainty in the slope to be 0.048 cm/cm and the uncertainty in the y-intercept to be 2.939 cm. These values are very different from the values given as the answer (0.26 cm/cm and 10.9 cm, respectively). Can you please provide comment as to which is the more correct answer and why? Also a comment as to why the two methods give such different answers? Thank you!
LINEST uses all the points (but does not use the uncertainty of any of the points) so it will give a smaller uncertainty than a method that only uses uncertainty on the first and last points. The two point method is a very rough approximation that should always overestimate. Provided you have enough data, LINEST should give a very good value for the two uncertainties.
@@donerphysics What is considered the more appropriate method?: Use the LINEST function to determine uncertainties OR use the "max" and "min" best-fit lines that take into account the uncertainties of all points plotted?
That is for you to decide. For your data, which is the more appropriate method?
@@donerphysics Thank you for quick and very helpful responses! I would say the 'most correct' procedure would be to honor the data's uncertainties, regardless of how small they may be.
is there a way to calculate the uncertainty of the slope of a tangent of a curve at any specific point ? Lets say you have a quadratic curve of best fit and you want to find the uncertainty of the slope of the tangent at 0 point. The slope can be calculated as f'(0). What about the uncertainty of that ? Thank you
The best way would be to perform linearization first, then use the procedures from the video.