Hypothesis Testing involving Poisson Distribution

[marsuconn]

Hi, Could anybody help me solving it? I am sure how to do a hypothesis test of Poisson distribution with small sample.

A person will decide to have her telephone disconnected if her
average number of calls per day is lsess than 2. On five randomly selected days,
her recorded number of calls were 0, 2, 1, 1, 1.
a) Briefly state why the Poisson(λ) assumption is suitable as a probability
distribution for the number of calls.
b) At the 5% level of significance, and based on information from this random
sample, discuss whether she should decide to disconnect her phone.

Any help would be much appreciated!

Thanks!

[BGM]

a) I guess Poisson distribution is suitable for modelling the frequency of rare events

Not quite sure about the discrete test
b) Model: Xi ~ Poission(λ), i = 1, 2, 3, 4, 5
Test H0: λ = 2 vs H1: λ < 2
Test Statistics = (X1 + X2 + X3 + X4 + X5) ~ Poission(5λ = 10) under H0
Observed Test Statistics = 5
Pr{(X1 + X2 + X3 + X4 + X5) ≤ 5} ≈ 0.06708596 > 0.05
Therefore, H0 is rejected
She should disconnect her phone

[TheEcologist]

Originally Posted by BGM

a) I guess Poisson distribution is suitable for modelling the frequency of rare events

The Poisson distribution is a discrete probability distribution that expresses the probability of a number of discrete events occurring in a fixed period of time. When these events occur independently with a known average rate.
http://en.wikipedia.org/wiki/Poisson_distribution

Thus perfect for such problems.

I wouldn't make any statements about rare events, though the Poisson has a "tail", other distributions would be better suited for 'rare' event modeling e.g. negative binomial. At least in my opinion.

Originally Posted by BGM

Not quite sure about the discrete test
b) Model: Xi ~ Poission(λ), i = 1, 2, 3, 4, 5
Test H0: λ = 2 vs H1: λ < 2
Test Statistics = (X1 + X2 + X3 + X4 + X5) ~ Poission(5λ = 10) under H0
Observed Test Statistics = 5
Pr{(X1 + X2 + X3 + X4 + X5) ≤ 5} ≈ 0.06708596 > 0.05
Therefore, H0 is rejected
She should disconnect her phone

The probability of obtaining a total of 5 calls or less in 5 days (under H0, thus when the actual λ = 2 calls per day, totaling 10 calls in 5 days) = 0.067 as you stated. Your calculations are spot on. However here: P > 0.05, convention specifies rejection at P < 0.05.

Is there any particular reason why you rejected H0 here?

[BGM]

Sorry I have made that silly mistake, twisting the things.
Thanks for the correction