= Solution
Assume independent <normally distributed> <angle> errors with known <standard deviation> $\sigma_y=0.002$. The relevant <chi-squared test> uses the sum of squared standardized residuals,
$$
\chi^2=\sum_i\frac{(y_i-y_{\mathrm{fit},i})^2}{\sigma_y^2}.
$$
For the model through the origin, the predictions and residuals, expressed in milliradians, are
$$
\begin{array}{c|rrrr}
r\ (\mathrm{mm})&16&18&22&32\\\hline
\hat c x\ (\mathrm{mrad})&70.6715&55.8392&37.3800&17.6679\\
y-\hat c x\ (\mathrm{mrad})&-4.6715&-0.8392&6.6200&7.3321
\end{array}
$$
Consequently $\chi_0^2=30.0280$. There are $4-1=3$ <statistical degrees of freedom>. The upper-tail <P-value> is approximately $1.36\times10^{-6}$; equivalently the value greatly exceeds the $5\%$ critical value $7.815$. The zero-intercept model is strongly rejected under the stated errors.
For the alternative <linear regression>, minimize $\sum_i(y_i-c'x_i-d)^2$. Setting both partial <derivatives> to zero gives the <normal equations>, whose solution is
$$
\hat c'=\frac{\sum_i(x_i-\bar x)(y_i-\bar y)}{S_{xx}},
\qquad \hat d=\bar y-\hat c'\bar x,
\qquad S_{xx}=\sum_i(x_i-\bar x)^2.
$$
For these data $\bar x=2508.83699\,\mathrm{m^{-2}}$, $\bar y=0.0475$, and $S_{xx}=4.830232\times10^6\,\mathrm{m^{-4}}$. This yields
$$
\boxed{\hat c'=1.37074\times10^{-5}\,\mathrm{m^2},\qquad
\hat d=0.0131105=13.1105\,\mathrm{mrad}.}
$$
The new predictions are $(66.6548,55.4171,41.4315,26.4966)\,\mathrm{mrad}$, with residuals $(-0.6548,-0.4171,2.5685,-1.4966)\,\mathrm{mrad}$. Therefore
$$
\boxed{\chi_1^2=2.35998,\qquad \nu=4-2=2,\qquad p=0.307.}
$$
The affine model is compatible with the data. Allowing for its additional fitted parameter, the decrease $\Delta\chi^2=27.668$ has a <chi-squared distribution> with one <statistical degree of freedom> under the nested zero-intercept hypothesis, corresponding to $p\simeq1.44\times10^{-7}$. Thus \b[the model with a fitted intercept provides the better description], rather than merely lowering the residual by adding a parameter.
With the measurement variance treated as known, the coefficient uncertainties are
$$
\sigma_{c'}=\frac{\sigma_y}{\sqrt{S_{xx}}}=9.10\times10^{-7}\,\mathrm{m^2},
\qquad
\sigma_d=\sigma_y\sqrt{\frac14+\frac{\bar x^2}{S_{xx}}}=0.0024925.
$$
Hence \b[$d=(13.11\pm2.49)\,\mathrm{mrad}$], about $5.26$ <standard deviations> above zero. A positive constant <systematic error> in the angular zero, or an additional nearly constant <torque>, could account for it. Its presence is evidence against the ideal zero-intercept model; four points alone do not identify the physical cause uniquely. A calibration error in $\kappa$ only rescales the inferred <gravitational constant> and cannot remove this residual pattern.
Using the fitted slope after allowing for this offset would give $G'=\kappa\hat c'/(mML)=(7.27\pm0.51)\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$, with the period and slope errors combined in quadrature. This is conditional on the constant-offset model.
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