Hypothesis testing in poisson distribution

Clarkson

New Member
At a nuclear plant great care is taken to measure the employees health.These are the number of visits made by each of the 10 employees to the doctor during a calender year. 3,6,5,7,4,2,3,5,1,4

Assuming the number of visits made by employee has a poisson distribution ,test the hypothesis that the annual mean per employee is greater than 3.

I am using the graphical method and I am not sure of which p[X=x] should i consider.

X: no.of visits by each employee to the doctor.

H0:lambda=3
H1:lambda>3
X follows a Poisson(3)
Then what is the probability that I should check?

What I did was as the average of sample data is 4,calculated p[X=4] and checked if it was in the critical region. Is this the correct probability to calculate?

asterisk

New Member
The sum of poisson process is a poisson process, so we can expect the total number of visits to be a new poisson distribution with a mean of 30.

Consider the p[x>=40] of a poisson distribution with a mean of 30.

$$1- \sum_{x = 0}^{39} {\frac {30^x * e^{-30}} {x!}}$$

Clarkson

New Member
how was the mean calculated as 30.In a Poisson distribution how does the mean get calculated.As (sigma f*x)/(sigma x) or sigma(x*p[X=x]).And why is p[X>=40]
considered

asterisk

New Member
The hypothesis being tested is that the mean is 3. If the mean for 1 employee is 3, the mean for 2 employees is 6, for 10 employees the mean is 30. This gives a new Poisson distribution with mean = lambda = mu = 30.

The data for 10 employees shows 40 sick days.

I'm not sure what graphical method you are using, but I would graph a Poisson distribution with lambda = 30 and evaluate at x=40

I Hope this helps.

Clarkson

New Member
Thanks for the help.But I want to know if my Ho should be changed as Ho:lamda=30 or is it lamda=3

asterisk

New Member
$$H_0$$ is the annual mean per employee is 3

$$H_1$$ is the annual mean per employee is greater than 3

the annual mean per employee is 3 implies the annual mean per 10 employees is 30