Suppose we observe the following random sample of

Chi-Square Test: In statistical hypothesis test, a chi-square test, or specifically Pearson's chi-square test is performed when the test statistic is distributed as chi-square under the null hypothesis. This test is used to determine whether there is a significant difference between the observed and the expected frequencies in one or more categories of a contingency table.
In the standard applications of this test, the observations are classified into mutually exclusive classes. If the null hypothesis is true i.e. there are no significant differences between the classes in the population, the test statistic which is computed from the observations follows a chi-square frequency distribution. The purpose of the test is to determine how likely the observed frequencies would be assuming the null hypothesis is true.
If ${\displaystyle O_i}$ and ${\displaystyle E_i}$ be the observed and expected frequency for the ${\displaystyle i^{th}}$ category, then the chi-square test statistic is given by,
${\displaystyle {\chi }_c^2=\sum _i\frac{{(O_i-E_i)}^2}{E_i}}$
where c is the degrees of freedom of the chi-square distribution.
Here, the null hypothesis is that the data comes from a Poisson distribution.
So, from the given data, we find the mean of the values.
$\begin{array}{|ccc|}\hline \text{ Value (x) }& \text{ Count (f) }& \text{ xf }\\ 0& 32& 0\\ 1& 36& 36\\ 2& 21& 42\\ 3& 18& 54\\ 4& 6& 24\\ 5& 2& 10\\ 6& 2& 12\\ \text{ Total }& 117& 178\\ \hline\end{array}$
Therefore,
$\overline{\chi }=\frac{1}{N}\sum _{i=1}^n{\chi }_i{f}_i=\frac{178}{117}=1.52$ (rounded to 2 decimal places)
Now, the parameter of the population is ${\displaystyle \lambda }$ i.e.
Average = ${\displaystyle \lambda }$ = 1.52
Then the probability mass function of the random variable of the value is given by,
$P(X=x)=(\frac{e^{-1.52}(1.52{)}^{\chi }}{\chi !}),(0):$, for ${\displaystyle \chi =0,1,2,\dots \mathrm{\infty }}$
Using this, we calculate the expected frequencies multiplying each probability by the total frequency i.e. 117.
$\begin{array}{|cccc|}\hline \text{ Value (x) }& \text{ Observed frequency }(O_i)& \text{ Expected frequency }(Ei)& (O_i-E_i{)}^2/E_i\\ 0& 32& 25.6& 1.6060\\ 1& 36& 38.9& 0.2156\\ 2& 21& 29.6& 2.4792\\ 3& 18& 15.0& 0.6100\\ 4& 6& 5.7& 0.0167\\ 5& 2& 1.7& 0.0421\\ 6& 2& 0.4& 5.5642\\ \text{ Total }& 117& 117& 10.5337\\ \hline\end{array}$
We have,
${\displaystyle {\chi }_c^2=\sum _i\frac{{(O_i-E_i)}^2}{E_i}}$
=10.5337 (rounded to 4 decimal places)
The value of the test statistic is 10.534 (rounded to 3 decimal places).
From the chi-square value, we calculate p-value for this test using MS-Excel,
p-value=CHISQ.DIST(chi-square value,n-1,TRUE)
=CHISQ.DIST(10.534,6,TRUE)
=0.896(rounded to 3 decimal places)
Answer: The p-value of the observed test statistic is 0.896.