= Solution
For a candidate attenuation coefficient, the one-second expected count is $\mu_n=2018.5e^{-bn}$. The model predictions are
$$
\begin{array}{c|rrrrrrrr}
n&1&2&3&4&5&6&7&8\\ \hline
b=1&742.56&273.17&100.50&36.97&13.60&5.00&1.84&0.677\\
b=1.1&671.90&223.66&74.45&24.78&8.25&2.75&0.914&0.304
\end{array}
$$
Use first the conservative signal-dominated range $n=1,\ldots,4$, where the count approximation is reliable and background is a small correction. Using the estimated errors from part (c), the weighted residual statistic is
$$
Q_{\mathrm{obs}}(b)=\sum_{n=1}^4\frac{(y_n-\mu_n)^2}{y_n}.
$$
It gives $\boxed{Q_{\mathrm{obs}}(1)=4.45,\qquad Q_{\mathrm{obs}}(1.1)=12.74}$. Since both $b$ candidates are fixed and $N_0$ was calibrated independently, neither was fitted to these four counts: the approximate reference has $\nu=4$, not three. The upper-tail probabilities are approximately $0.349$ and $0.0126$, respectively. Thus $b=1$ is compatible with the scatter, whereas $b=1.1$ is disfavored at the usual $5\%$ level in this approximate test.
A <Pearson chi-squared statistic>, using model variances instead of estimated observed variances, gives the same preference:
$$
Q_{\mathrm P}(1)=4.04,\quad Q_{\mathrm P}(1.1)=15.62,\qquad p\simeq0.401,\ 0.00357.
$$
Including the fifth point also leaves the decision unchanged: the estimated-error statistics are $8.37$ and $12.74$ on five degrees of freedom, with approximate upper-tail probabilities $0.137$ and $0.0259$; the Pearson statistics are $6.34$ and $15.63$. Hence
$$
\boxed{b=1\text{ provides the better exponential fit}.}
$$
The small uncertainty in $N_0$ contributes correlated prediction uncertainty, as in <chi-squared comparison of a calibrated attenuation model>, but is negligible compared with the one-second counting errors here. Choosing the test range after inspecting the observations limits a strict interpretation of its nominal significance; the numerical comparison and its stated range are the relevant evidence.
If all eight observations are inserted into the estimated-error formula, the totals are $14.99$ and $22.48$. Their nominal eight-degree-of-freedom calibration is unreliable because the final expected counts are much smaller than one for some predictions. In fact the full-range Pearson statistics are $43.34$ and $118.25$, exposing the late excess much more strongly. The full <Poisson deviance> values are $24.09$ and $47.03$, again favoring $b=1$ but signaling that a background-free exponential is inadequate over the complete range. These sparse-count comparisons should not be used to claim that either unmodified model explains all eight measurements.
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