Here is a least-squares problem that you can do by hand with a calculator. Find the slope and intercept and their standard deviations for the straight line drawn through the points$\left(x,y\right)\mathbf{=}\left(0,1\right)\mathbf{,}\left(2,2\right)$and $\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{3}\mathbf{)}$. Make a graph showing the three points and the line. Place error bars $\left(\pm s_y\right)$on the points.

The equation of a straight line is:

$$\begin{gathered} \mathbf{y}\mathbf{=}\mathbf{mx}\mathbf{+}\mathbf{b} \\ \mathbf{y}\mathbf{=}\left[m\left(\mp u_m\right)\right]\mathbf{x}\mathbf{+}\left[b\left(\pm u_b\right)\right] \end{gathered}$$

The standard deviation of $\mathbf{y}\mathbf{:}{\mathbf{s}}_{\mathbf{y}}\mathbf{:}$

The standard deviation of y is mathematically presented as ${\mathbf{\sigma }}_{\mathbf{y}}\mathbf{\approx }{\mathbf{s}}_{\mathbf{y}}\mathbf{=}\sqrt{\frac{\mathbf{\sum }_{\mathbf{}}\left(d_i^2\right)}{\mathbf{n}\mathbf{-}\mathbf{2}}}$

The standard uncertainty of slope and intercept:

The standard uncertainty of slope is:

${\mathbf{u}}_{\mathbf{m}}^{\mathbf{2}}\mathbf{=}\frac{{\mathbf{s}}_{\mathbf{y}}^{\mathbf{2}}\mathbf{n}}{\mathbf{D}}$

The standard uncertainty of intercept is mathematically presented as ${\mathbf{u}}_{\mathbf{b}}^{\mathbf{2}}\mathbf{=}\frac{{\mathbf{s}}_{\mathbf{y}}^{\mathbf{2}}\mathbf{\sum }_{\mathbf{}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{i}}^{\mathbf{2}}\mathbf{\right)}}{\mathbf{D}}$

Given data:

The given points of a straight line are:$\left(\mathrm{x},\mathrm{y}\right)-\left(0,1\right),\left(2,2\right),\mathrm{and}\left(3,3\right)$.

The equation of the least-squares straight line:

The equation of the straight line will be $\mathrm{y}\left(\mp {\mathrm{s}}_{\mathrm{y}}\right)=\left[\mathrm{m}\left(\mp {\mathrm{u}}_{\mathrm{m}}\right)\right]\mathrm{x}+\left[\mathrm{b}\left(\pm {\mathrm{u}}_{\mathrm{b}}\right)\right]$

From the given data, the value of $x_{\mathrm{i}}^2,{\mathrm{d}}_{\mathrm{i}}$and ${\mathrm{d}}_{\mathrm{i}}^2$are tabulated as follows:

$$\begin{gathered} D=\left|\begin{array}{cc}\sum x_i^2& \sum x_i\\ \sum x_i& n\end{array}\right| \\ m=\frac{\left|\begin{array}{cc}\sum \left(x_i{y}_i\right)& \sum x_i\\ y_i& n\end{array}\right|}{D} \\ =\frac{\left(3\right)\left(13\right)-\left(5\right)\left(6\right)}{14} \\ =\frac{9}{14} \\ =0.64286 \\ \mathrm{b}=\frac{\left|\sum \left({\mathrm{x}}_{\mathrm{i}}^2\right)\sum \left({\mathrm{x}}_{\mathrm{i}}{\mathrm{y}}_{\mathrm{i}}\right)\right|\sum {\mathrm{y}}_{\mathrm{i}}\mid }{\mathrm{D}} \\ =13\times 6-13\times 5÷14 \\ =0.92857. \\ {\mathrm{s}}_{\mathrm{y}}=\sqrt{\frac{\sum \left({\mathrm{d}}_{\mathrm{i}}^2\right)}{\mathrm{n}-2}} \\ =\sqrt{\frac{0.07143}{3-2}} \\ =0.26726 \\ {\mathrm{s}}_{\mathrm{m}}={\mathrm{s}}_{\mathrm{y}}\sqrt{\frac{\mathrm{n}}{\mathrm{D}}} \\ =\left(0.26726\right)\sqrt{\frac{3}{14}} \\ =0.12371. \\ {\mathrm{s}}_{\mathrm{b}}={\mathrm{s}}_{\mathrm{y}}\sqrt{\frac{\mathrm{n}}{\mathrm{D}}} \\ =\left(0.26726\right)\sqrt{\frac{13}{14}} \\ =025754 \end{gathered}$$

Therefore, the slope is$0.6\pm 0.1$and the intercept is$0.93\pm 0.2$.

The graph shows the three given points along with the error bars.