Answer to Question #171676 in Statistics and Probability for Mandy

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.