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For the Poisson variance function $V(\mu )=\mu$, the code computes the Pearson chi-squared statisticand the Pearson dispersion estimatorThe first quantity measures goodness of fit; the second estimates the dispersion parameter, which equals one in a correctly specified Poisson regression. A standard rough calculation substitutes the residual deviance for the Pearson statistic and givesIf the reported upper-tail probability $4.908651\times 10^{-11}$ is inverted numerically, the actual Pearson statistic used by the code is about $82.93$, giving $\hat{\varphi }\simeq 5.18$. Either calculation reveals severe overdispersion.

The quasi-Poisson standard errors are the Poisson standard errors multiplied by $\sqrt{\hat{\varphi }}$, and the Pearson estimator divides the Pearson chi-squared statistic by the residual degrees of freedom $40-3=37$. Therefore the requested sum of squared Pearson residuals isup to the rounding in the printed standard errors.

Fit the Poisson distribution using the estimated daily mean. For $k=0,\dots ,8$, its expected frequency isand $E_{\ge 9}=N\left(1-{\sum }_{k=0}^8{e}^{-\hat{\mu }}{\hat{\mu }}^k/k!\right)$. Each row's contribution to the Pearson chi-squared statistic is $(O_k-E_k{)}^2/E_k$. For example the zero-count class contributes $(162-126.784{)}^2/126.784\approx 9.781$. Summing the class contributions gives the test statistic; expected frequencies depend on the fitted model, while observed frequencies come directly from counting days.

A chi-squared test compares residuals with the random scatter predicted by a model. For independent measurements $y_i$ with known Gaussian standard deviations ${\sigma }_i$ and predictions ${\mu }_i(\theta )$, formIf the mean model and error assumptions are correct, a full-rank linear fit of $p$ parameters gives a chi-squared distribution with $\nu =N-p$ statistical degrees of freedom. With a fully specified prediction and no fitted parameters, $\nu =N$. Nonlinear fits usually make this an approximation. Independent external calibration is not the same as fitting a parameter to these test data.

Under the reference distribution, the mean statistic is $\nu$ and its variance is $2\nu$. The upper-tail p-value $\mathrm{Pr}({\chi }_{\nu }^2\ge Q_{\mathrm{obs}})$ quantifies how unusual a discrepancy at least this large would be under the assumed model. A small value suggests model failure or incorrect uncertainties; it is not the probability that the model is true. An unusually small statistic can also reveal inflated uncertainties or correlations. A rejection threshold should be specified in advance.

For independent Poisson distribution counts, the model variance is its expected count ${\mu }_i$. The Pearson chi-squared statistic is $\sum (y_i-{\mu }_i{)}^2/{\mu }_i$. The estimated-error alternative ${\sigma }_i\simeq \sqrt{y_i}$ is common for large counts but can differ substantially when counts are small. In that regime the Poisson deviance or a direct likelihood calculation is preferable, with simulation if an accurately calibrated goodness-of-fit probability is required.

For a candidate attenuation coefficient, the one-second expected count is ${\mu }_n=2018.5e^{-bn}$. The model predictions areUse first the conservative signal-dominated range $n=1,\dots ,4$, where the count approximation is reliable and background is a small correction. Using the estimated errors from part (c), the weighted residual statistic isIt gives $\boxed{Q_{\mathrm{obs}}(1)=4.45,Q_{\mathrm{obs}}(1.1)=12.74}$. 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: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 $15.63$. HenceThe 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.

For independent Poisson random variables with known expected counts ${\mu }_i$, the model variance is ${\mu }_i$. The Pearson chi-squared statistic is thereforeThere are six independently specified bins and no parameters fitted from them, giving approximately six degrees of freedom. This is far above its expected value six; the approximate chi-squared distribution tail probability is $9.5\times 10^{-6}$. The discrepancies are much too large for a good Poisson-noise fit. Replacing model variances by the observed-count estimates gives the alternative statistic $30.92$, with the same qualitative conclusion. There is no fitted normalization to justify subtracting a degree of freedom or rescaling the model to the observed total.

Let $N_{ij}$ be the observed cell counts, $N_{i+}$ the row totals, $N_{+j}$ the column totals and $N=33009$. The fitted expected frequencies under independence are $E_{ij}=N_{i+}{N}_{+j}/N$. For example the three-child, one-boy cell has expected count $7541\cdot 16472/33009=3763.1$.

The Pearson chi-squared statistic isUnder $H_0$ it has asymptotically a chi-squared distribution with $(2-1)(3-1)=2$ degrees of freedom, assuming the multinomial sampling model and sufficiently large expected counts. All expected counts here are large. The distribution is an asymptotic approximation, rather than an exact finite-sample identity.

Write $y_i=e_i+r_i$. For $|r_i|/e_i$ small, the cell contribution to the Poisson deviance expands asSumming gives the Pearson chi-squared statistic as the quadratic approximation. The approximation can be inaccurate for sparse expected counts or cells with relative residuals near minus one.

For a two-by-two contingency table, subtracting one half from each absolute observed-minus-expected count difference, with truncation at zero, gives the displayed corrected Pearson chi-squared statistic. Expected counts use the independence model $E_{ij}=O_{i+}{O}_{+j}/N$. The subtraction compensates approximately for using a continuous chi-squared distribution for discrete counts. It can make the test conservative. The corrected statistic should be distinguished from the ordinary uncorrected Pearson chi-squared statistic.