Question

 

The number of emails arriving at a server per minute is claimed to follow a Poisson distribution. To test this claim, the number of emails arriving in 70 randomly chosen 1-minute intervals is recorded. The table below summarizes the result.

Number of emails	0	1	2	≥ 3
Observed Frequency	13	22	23	12

Test the hypothesis that the number of emails per minute follows a Poisson distribution? Use a significance level of α = 0.05.

Answer 1

H0: Null Hypothesis: X follows Poisson Distribution

HA: Alternative Hypothesis: X does not follow Poisson Distribution

Mean of Poisson Distribution is got as follows:

So,

E₀= 0.2263 X 70 = 15.841

So,

E₁= 0.3363 X 70 = 23.541

So,

E₂= 0.2498 X 70 = 17.485

So,

E₃= 0.1876 X 70 = 13.132

Test Statistic is got as follows:

Observed (O)	Expected (E)	(O - E)2/E
13	15.841	0.5095
22	23.541	0.1009
23	17.485	1.7395
12	13.133	0.0977
Total = =		2.448

Nuber of Degrees of Freedom = 4 - 1 - 1 = 2

By Technology, P - Value = 0.2941

Since p value is greater than , the difference is not significant. Fail to reject null hypothesis.

Conclusion:

the number of emails per minute follows a Poisson distribution