Solution (source code)

= Solution

A <chi-squared test> compares the sum of squared standardized <regression residuals> with the appropriate <chi-squared distribution>. With both line parameters fitted, the data give $\chi^2=4.6537$ on $6-2=4$ degrees of freedom, with upper-tail <P-value> about $0.325$: the fitted <exponential decay> model describes the data adequately.

First consider fixing the alternative lifetime at its central value $\tau_a=0.478\,\mathrm{ps}$. Its slope is $m_a=-1/\tau_a=-2.09205\,\mathrm{ps}^{-1}$. One must refit the normalization rather than retain the earlier intercept. Minimization over the intercept gives $c_a=\bar y-m_a\bar x=7.29773$, and
$$
\chi^2(\tau_a,\widehat c_a)=19.9111
$$
on $6-1=5$ degrees of freedom, with upper-tail <P-value> about $0.00130$. Thus that fixed central lifetime gives a substantially worse fit than $\widehat\tau=0.4216\,\mathrm{ps}$. Equivalently, the increase after profiling over the intercept is
$$
\Delta\chi^2=\frac{S_{xx}}{s^2}(m_a-\widehat m)^2=15.2574.
$$
This fixed-value comparison does not yet include the uncertainty of the alternative measurement.

For the <compatibility of independent lifetime measurements>, include both errors. Under the approximately Gaussian lifetime-error model, equality of the two true lifetimes is tested by
$$
\boxed{\chi^2_{\mathrm{compat}}
=\frac{(\widehat\tau-\tau_a)^2}{\sigma_\tau^2+\sigma_a^2}
=\frac{(0.42156-0.478)^2}{0.01274^2+0.023^2}
\simeq4.61.}
$$
This has one degree of freedom and <P-value> approximately $0.032$, equivalent to a $2.15$-<standard deviation> discrepancy. The two measurements are therefore in mild tension: equality is rejected at a $5\%$ significance threshold, but not at $1\%$. There is no unique yes-or-no compatibility verdict without choosing such a threshold. Treating the alternative central value as exact would overstate the disagreement between the experiments.

The lifetime that provides the best fit to these six rates remains $\boxed{\tau=0.4216\,\mathrm{ps}}$, not $0.478\,\mathrm{ps}$. The compatibility calculation above assumes the two experiments have independent uncertainties; a shared systematic error would require its covariance.