The earlier test had two requirements:
The chi-squared test cannot therefore be used directly to test whether the following data set is a random sample from a \(\PoissonDistn(\lambda=2)\) distribution, since the \(\{E_i\}\) are all 2.
1 | 3 | 2 | 2 | 5 | 4 | 5 | 2 | 0 | 2 |
2 | 4 | 3 | 5 | 2 | 3 | 1 | 4 | 2 | 6 |
Frequency table
We first summarise the data in a frequency table.
x | 0 | 1 | 2 | 3 | 4 | 5 | 6+ |
---|---|---|---|---|---|---|---|
Freq(x) | 1 | 2 | 7 | 3 | 3 | 3 | 1 |
Treating these frequencies as our observed counts, \(\{O_i\}\), we can find expected counts from the Poisson distribution's probability function,
\[ p(x) \;\;=\;\; \frac{\lambda^x e^{-\lambda}}{x!}\]Since \(n=20\) and \(\lambda=2\) when H0 holds,
\[ E_x \;\;=\;\; 20 \times p(x) \;\;=\;\; 20 \times \frac{2^x e^{-2}}{x!}\]giving
x | 0 | 1 | 2 | 3 | 4 | 5 | 6+ |
---|---|---|---|---|---|---|---|
\(O_x\) | 1 | 2 | 7 | 3 | 3 | 3 | 1 |
\(E_x\) | 2.707 | 5.413 | 5.413 | 3.609 | 1.804 | 0.722 | 0.331 |
Combining cells
Since we still do not have all \(\{E_x\}\) ≥1 and 80% of them ≥5, cells in the table should be combined.
x | 0,1 | 2 | 3+ |
---|---|---|---|
\(O_x\) | 3 | 7 | 10 |
\(E_x\) | 8.120 | 5.413 | 6.466 |
Based on these three counts,
\[ \begin{align} X^2 \;&=\; \sum_{i=1}^{10} {\frac{\left(O_i - E_i\right)^2}{E_i}} \\ &=\; \frac{(3-8.120)^2}{8.120} + \frac{(7-5.413)^2}{5.413} + \frac{(10-6.466)^2}{6.466} \\ &=\; 5.624 \end{align} \]P-value and conclusion
There are 3 categories (after grouping) and one constraint .
\[ \sum{E_i} \;=\; \sum{O_i} \;=\; 20 \]The test should therefore be based on a chi-squared distribution with \((3-1) = 2\) degrees of freedom. The p-value here is
p-value = \(P(X^2 \ge 5.621) = 0.060\)
We conclude that there is only very weak evidence against H0 (that the original data set was a random sample from a \(\PoissonDistn(2)\) distribution).
(It is hardly surprising that a data set with only 20 values does not show up problems with a model — a larger data set would be more sensitive to any possible lack of fit of the model.)