Do World War II Bomb Hits Fit a Poisson Distribution? In analyzing hits by V-1 buzz bombs in World War II, South London was subdivided into regions, each with an area of 0.25\(k{m^2}\). Shown below is a table of actual frequencies of hits and the frequencies expected with the Poisson distribution. (The Poisson distribution is described in Section 5-3.) Use the values listed and a 0.05 significance level to test the claim that the actual frequencies fit a Poisson distribution. Does the result prove that the data conform to the Poisson distribution?

Number of Bomb Hits	0	1	2	3	4
Actual Number of Regions	229	211	93	35	8
Expected Number of Regions (from Poisson Distribution)	227.5	211.4	97.9	30.5	8.7

The observed and expected frequencies of bomb hits in different regions are provided.

Let O denote the observed frequencies of hits.

The following values are obtained:

\(\begin{aligned}{l}{O_0} = 229\\{O_1} = 211\\{O_2} = 93\\{O_3} = 35\\{O_4} = 8\end{aligned}\)

Let E denote the expected frequencies.

The expected frequencies are noted below:

\(\begin{aligned}{l}{E_0} = 227.5\\{E_1} = 211.4\\{E_2} = 97.9\\{E_3} = 30.5\\{E_4} = 8.7\end{aligned}\)

It is observed that each of the frequencies is greater than 5.

Thus, the requirements for the test are satisfied, assuming the sampling is done randomly.

The hypotheses are,

\({H_0}:\)The observed frequencies of bomb hits fit well with the Poisson distribution.

\({H_a}:\)The observed frequencies of bomb hits do not fit well with the Poisson distribution.

The test is right-tailed.

The table below shows the necessary calculations:

The value of the test statistic is equal to:

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \;\\ = 0.00989 + 0.000757 + ...... + 0.056322\\ = 0.976\end{aligned}\)

Thus,\({\chi ^2} = 0.976\).

Let k be the different number of bomb hits, which is equal to 5.

The degrees of freedom for\({\chi ^2}\)is computed below:

\(\begin{aligned}{c}df = k - 1\\ = 5 - 1\\ = 4\end{aligned}\)

The critical value of\({\chi ^2}\)at\(\alpha = 0.05\)with 4 degrees of freedom is equal to 9.4877.

The p-value is equal to,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 0.976} \right)\\ = 0.913\end{aligned}\)

Since the test statistic value is less than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to be rejected.

There is not enough evidence to conclude that theobserved frequencies of bomb hits do not fit well with the Poisson distribution.

Therefore, the given data conforms to the Poisson distribution.