Error on slope using LSQ method

AI Thread Summary
The discussion revolves around calculating the error on the slope using the least squares (LSQ) method with three given points that include errors in the y-coordinates. The initial attempt at the solution led to confusion regarding the correct formula for σy, which was miswritten, causing incorrect calculations. Clarification was provided that the correct formula should be √(1/(n-2)Σ(yk - y(xk))^2), not the version initially used. Additionally, the calculation of Δ was found to be incorrect, as it resulted in a negative number, which is not valid in this context. Proper understanding and application of the formulas are essential for accurately determining the error on the slope.
Matt21
Messages
14
Reaction score
0

Homework Statement


Three points are given with errors on y coordinates.
(2.00,4.70±0.36); (4.00,6.8&±0.45); (5.00, 9.10±0.50)
Using LSQ method, find the error on slope.

Homework Equations


σm = σy√(n/Δ), where σy = √(1/n-2(Σ(yk - y(xk))^2)) and Δ = nΣxk^2 - (Σxk)^2

The Attempt at a Solution


I do not know if this is the correct equation or if I made an error in my calculations but what I got was:
σy = 1/3-2((4.7-4.7(2))^2+(6.8-6.8(4))^2+(9.1-9.1(5))^2) = 42 and
Δ = (2^2+4^2+5^2)-(2+4+5)^2 = -76
n = 3
Obviously I can't use Δ since you can't get a root of a negative number. Can anyone see what I'm doing wrong? Any help would be much appreciated.
 
Physics news on Phys.org
Matt21 said:
1/n-2(Σ(yk - y(xk))^2))
That's a very strange looking expression. Are you sure about the minus sign after 1/n?
 
Yes I'm positive. The formula given in my notes was √((1/n-2)(Σ(yk - y(xk))^2)
 
Matt21 said:
Yes I'm positive. The formula given in my notes was √((1/n-2)(Σ(yk - y(xk))^2)
Ah, you mean √(1/(n-2)(Σ(yk - y(xk))^2)), i.e. ##\sqrt{\frac 1{n-2}\Sigma( y_k-y(x_k))^2}##? That would make more sense. Originally you had √((1/n)-2Σ(yk - y(xk))^2)), i.e. ##\sqrt{\frac 1n-2\Sigma(y_k-y(x_k))^2}##.
By the way, please clarify what you mean by y(xk).
 
Yes that was the equation was trying to convey. In regards to y(xk) I would assume that means y*(xk) where xk = x1, x2, x3...xk
 
Last edited:
Δ = (3*(2^2+4^2+5^2))-(2+4+5)^2 = 14
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top