Questions about the error in the slope of a graph

[toothpaste666]

Homework Statement

I am doing a lab report and I have to find the slope of a graph as well as the max and min lines of the graph. I also have to find the error in the slope. I have calculated the slope. My first problem is in calculating the error lines. My teacher said i must find the slope using just my graph and not my data, meaning when i use the slope formula i can't use my data as the points I use but must use other points on that line. So I did that and got the slope. But to find the slope of the max and min lines I am confused because due to the way i obtained the data, the error in x is the same for all points and the error in y is the same for all points. wouldn't this mean that the slope of the max and min lines will be both be the same as the slope? (because let's say the points are (3 +/-1, 4 +/-2) and (5+/-1, 6 +/- 2) then the slope of the line would be (6-4)/(5-3) = 2/2 = 1 and the slope of the max line would be (8-6)/(6-4) = 2/2 = 1 . I was going to use the following formula for the error of my slope : (max slope - min slope)/2 but now I am confused.

 

[Chestermiller]

Let's see the graph.

Chet

 

[toothpaste666]

The problem is that my error in y was .005 mA and my error in x was .005 V which are both too small to include on the graph.

 

[toothpaste666]

the two points marked are the ones i used to calculate the slope they are (.6, .6) and (4.4, 4.6) for the slope i got m = 1.05 mA/V

 

[toothpaste666]

I am trying to find the error in that value of m

 

[Chestermiller]

OK. Put error bars on each of the points (both horizontal and vertical if necessary), and find the steepest line that passes through the span of all the error bars on all the points simultaneously; this is the maximum slope. Then find the least steep line that passes through the span of all the error bars on all the points simultaneously; this is the minimum slope.

Chet

 

[toothpaste666]

But my errors are too small for the scale of my graph I tried to include them on an earlier draft of the graph but from looking at the graph it appeared that the errors were much larger than they were when really they were only .005 for every x and y

 

[BvU]

Teachers will is law. Your example is no good (*).
Better show your (unprocessed) data and the plot you made.

There are formulas for errors in linear least squares (we had a very thorough thread here), but the graphical approach is a good substitute:

• find the center of gravity of the data points (you need the data for that, but that's allowed I would say).
• take a ruler and wiggle around while keeping the ruler on the center of gravity. Gives you the error of the slope: when you don't believe that such a line could represent the data any more, you're about a factor of three sigma away from the most probable line. Take the worst of the two (higher slope/lower slope). (three sigma is about one in a thousand)
• draw the most probable line
• move the ruler up and down, parallel to the most likely line. Same recipe: when you don't believe it any more, ... etc. Gives the error in vertical direction. Error in y intercept is combination of that and error in slope: $\sigma_{\rm intercept}^2 = \sigma_{\rm y}^2 + (\sigma_{slope} * x_{\rm average})^2$

You can do this with points without error bars and also with points with error bars. Make sure the error bars don't include systematic errors (things like calibration errors that are common to all points). Such errors belong in the slope.

(*) at least not for my purposes.

 

[toothpaste666]

the points of my graph were (1.00, 1.05) , (2.00, 2.09) , (3.00, 3.14), (4.00, 4.17) and (5.00, 5.22) the error in all x was +/- .005 V and the error in all y was +/- .005 mA (half of the smallest division on the ammeter and voltmeter. my teacher confirmed that this was fine)