Finding the x-Intercept For the line y = mx + b, the x-intercept occurs when y = 0, or x = −b/m. Let’s designate the x-intercept as the function F = −b/m. Find the x-intercept and its uncertainty for the least-squares line in Figure 4-11.

Question

Finding the x-Intercept

For the line y = mx + b, the x-intercept occurs when y = 0, or x = −b/m.
Let’s designate the x-intercept as the function F = −b/m. Find the
x-intercept and its uncertainty for the least-squares line in Figure 4-11.

Accepted Answer

The following quantities are computed in Section 4-7:

m = 0.615 38 [latex]s^{2}_{m}[/latex] = 0.002 958 6 [latex]s^{2}_{y}[/latex] = 0.038 462 ∑([latex]x_{i}[/latex]) = 14

b = 1.346 15 [latex]s^{2}_{b}[/latex] = 0.045 859 D = 52

The covariance in Equation C-3 is therefore

[latex]s_{mb} = \frac{−s^{2}_{y}∑(x_{i})}{D} = \frac{−(0.038 462)(14)}{52}[/latex] = −0.010 355

[latex]e_{F} = \sqrt{\underbrace{(\frac{\delta F}{\delta m})² s^{2}_{m} + (\frac{\delta F}{\delta b})² s^{2}_{b}}_{\begin{matrix} Variance terms from \ Equation C-1 \end{matrix} } + \underbrace{2(\frac{\delta F}{\delta m})(\frac{\delta F}{\delta b})s_{mb}}_{\begin{matrix} Covariance accounts for \ correlation of m and b \end{matrix} } }[/latex] (C_2)

The x-intercept is just F = −b/m = −(1.346 15)/(0.615 38) = −2.187 5.
To find the uncertainty in F, we use Equation C-2. The derivatives in
C-2 are

[latex]\frac{\delta F}{\delta m} = \frac{\delta(−b/m)}{\delta m} = \frac{b}{m²} = \frac{1.346 15}{0.615 38²}[/latex] = 3.554 7

[latex]\frac{\delta F}{\delta b} = \frac{\delta(−b/m)}{\delta b} = \frac{−1}{m} = \frac{−1}{0.615 38}[/latex] = − 1.625 0

Now we can evaluate the uncertainty with Equation C-2:

[latex]e_{F} = \sqrt{(\frac{\delta F}{\delta m})² s^{2}_{m} + (\frac{\delta F}{\delta b})² s^{2}_{b} + 2(\frac{\delta F}{\delta m})(\frac{\delta F}{\delta b})s_{mb}}[/latex]

= [latex]\sqrt{(3.554 7)²(0.002 958 6) + (−1.625 0)²(0.045 859) + 2(3.554 7)(−1.625 0)(−0.010 355)}[/latex]

=0.527 36

The final answer can now be written with a reasonable number of digits:

F = −2.187 5 ± 0.527 36 = −[latex]2.1_{9} ± 0.5_{3}[/latex]

[latex]e_{F} = \sqrt{(\frac{\delta F}{\delta x})² e^{2}_{x} + (\frac{\delta F}{\delta y})² e^{2}_{y} + (\frac{\delta F}{\delta \mathcal{z}})² e^{2}_{\mathcal{z}} + . . .}[/latex] (C-1)

If we had used Equation C-1 and ignored the covariance term in Equation C-2, we would have computed an uncertainty of ±[latex]0.4_{0}[/latex].
To learn how to compute variance and covariance and to see how to
include weighting factors in least-squares curve fitting, see J. Tellinghuisen,
“Understanding Least Squares Through Monte Carlo Calculations,” J. Chem. Ed. 2005, 82, 157.

Question AC.1: Finding the x-Intercept For the line y = mx + b, the x-inter...