Question

The Investment department of the Zambia Development Agency randomly selected 100 SMEs and examined them for profitability on their investment. The following results were obtained:

Number of Profits

1

2

3

Number of SMEs

36

40

19

5

Test at 5% significance level whether the distribution of profits follows a Poison distribution.

Answer 1

Denote X as number of SMEs.

Estimation of parameter $\lambda$ of Poisson distribution:

$\lambda = EX = \frac{1}{n}\sum_{i=0}^{\infty}\nu_iX_i = \frac{36*0 + 40 * 1 + 19 * 2 + 5 * 3}{100} = 0.93$

Distribution function:

$p_k = Pr(X = k) = e^{-\lambda} \frac{\lambda^k}{k!}$

Let's consider Pearson's test for parametric distribution with null hypothesis

$H_0$ : X has poisson distribution.

Statistics:

$T = \sum_{i=0}^3\frac{(\nu_i - p_i n)^2}{np_i} = \frac{36 - e^{-0.93} * 100}{e^{-0.93} * 100} + ... = 0.836$

T should be distributed as $\Chi^2(k - 1 - l) = \Chi^2(4 - 1 - 1) = \Chi^2(2)$ , where $l$ is parameters count.

At 5% significance level for this distribution we have critical value 5.991 > 0.836.

Hence, at 5% significance level the distribution of profits follows a Poison distribution.