 Solution

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.