Chi-square goodness of fit: Poisson errors

Chi-square distribution: if Z₁, Z₂, …, Z_k are independent standard normal variables, then χ² = Z₁² + Z₂² + … + Z_k² follows the chi-square distribution with k degrees of freedom. It is a right-skewed continuous distribution (mean = k, variance = 2k) used for goodness-of-fit tests, tests of independence and tests of a variance.

Goodness of fit to a Poisson distribution

1. H0: the number of errors per page follows a Poisson distribution. H1: it does not follow a Poisson distribution.
2. Level of significance: α = 0.05
3. Estimate the Poisson mean from the data:

λ = Σfx/Σf = (0×275 + 1×138 + 2×75 + 3×7 + 4×4 + 5×1)/500
  = 330/500 = 0.66

4. Expected frequencies E = 500 × e^(−0.66)·0.66^x/x!:

Errors  Observed  P(x)      Expected
0       275       0.51685   258.43
1       138       0.34112   170.56
2        75       0.11257    56.29
3         7       0.02477    12.38
4         4       0.00409     2.04
5+        1       0.00060     0.30

The classes x = 4 and x = 5+ have expected frequencies below 5, and even together they stay below 5 (2.34). So they are pooled with x = 3:

Errors   O     E        (O−E)²/E
0        275   258.43   1.063
1        138   170.56   6.216
2         75    56.29   6.223
3 or more 12    14.73   0.505
Total    500   500.00   χ² = 14.007

5. Degrees of freedom = (number of classes) − 1 − (parameters estimated) = 4 − 1 − 1 = 2. Critical value: χ²(0.05; 2) = 5.991.
6. Decision: calculated χ² = 14.007 > 5.991 (p-value = 0.0009), so reject H0.

Conclusion: the errors per page do not follow a Poisson distribution at the 5% level. The observed data have too many pages with no error and too many with 2 errors compared with the Poisson model.

Note: without pooling the classes the calculated χ² is 19.33 (d.f. = 4, critical 9.488); the decision is the same.