Question

In: Statistics and Probability

The number of emails arriving at a server per minute is claimed to follow a Poisson distribution.

 

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.

Solutions

Expert Solution

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

orchestra answered 2 years ago
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