the correct formula for the Poisson probability mass functio

the correct formula for the Poisson probability mass function is
P(X = x) = ? x exp( ??) / x!

Let X(s) = {0, 1, 2, ...}. The random variable X Is said to haw a Poisson distribution with intensity parameter mu (0, infinity) if X has probability mass function f(x) = P(X = x) = lambda^x e^-x/X!. The Poisson distribution frequently arises when counting arrivals in a fixed time interval. If X_1, ..., X_n ~ Poisson(mu), then EX_i= mu and the maximum likelihood estimator of mu is X_n. In 1910. E. Rutherford and M. Geiger counted the numbers of alpha-particle scintillations observed in each of n = 2608 72-second intervals. The following table^7 reports the frequency of each count: Use the above data to compute x, the average observed count. Now we partition X(S) by setting E_j = {j - 1} for j = 1, ..., 10 and E_11 = {10, 11, 12, ...}. Test the null hypothesis that counts of alpha-particle scintillations follow a Poisson distribution.

Solution

We are already given that Xi follows Poison (µ).

We can compute xbar by usin the given data.

Formula is,

xbar = x*f / f

For this we make a table of x*f and x.

xbar = average observed count = 3.8715

Here we have have to test the hypothesis that,

H0 : Counts of alpha-particle scintillations follow a Poison distribution.

H1 : Counts of alpha-particle scintillationsis not follow a Poison distribution.

Now here we have to calculate probability for each x = 0,1,2,..........,14 for finding expected frequencies.

We can find probability by using formula ,

P(X=x) = ( e^- * ^x) / x!

using = 3.8715

So the tabe is as follow,

The test statistic for testing is,

² = (O-E)² / E

where O is the observed frequency.

E is the expected frequency.

Now next step is the table of observed frequency and expected frequency foe calculating test statistic.

² = 22.7834

alpha = level of significance = 0.05

The critical value we can calculate by using EXCEL.

syntax :

chiinv(probability,d.f.)

where probability = alpha

d.f. = n-1

n = total number of counts = 15

d.f. = 15 - 1 = 14

critical value = 23.6848

Test statistic < critical value

Accept H0 at 5% level of significance.

Conclusion : Counts of alpha-particle scintillations follow a Poison distribution.

x	f	xf
0	57	0
1	203	203
2	383	766
3	525	1575
4	532	2128
5	408	2040
6	273	1638
7	139	973
8	45	360
9	27	243
10	10	100
11	4	44
12	0	0
13	1	13
14	1	14
total	2608	10097