Answer to Question #136470 in Statistics and Probability for Ayesha

Question

Apply Chi-square test to check whether the data given below may be regarded as conforming to a Poisson distribution.
X 0,1,2,3,4,5,6,7 and
Y 305,363,211,81,28,9,2,1

Answer 1

$\lambda$ =(305*0+363*1+211*2+81*3+28*4+9*5+2*6+1*7)/(305+363+211+81+28+9+2+1)=1204/1000=1.2

SIGNIFICANCE LEVEL = 0.05

to find expected values for Poisson distribution we should use formula:

P(x=k)=$(e^{-\lambda}*\lambda^k)/k!$

P(x=0)=0.302

E(x=0)=1000*0.302=302

P(x=1)=0.362

E(x=1)=1000*0.362=362

P(x=2)=0.216

E(x=2)=1000*0.216=216

P(x=3)=0.086

E(x=3)=1000*0.86=86

P(x=4)=0.026

E(x=4)=1000*0.026=26

P(x=5)=0.006

E(x=5)=1000*0.006=6

P(x=6)=0.001

E(x=6)=1000*0.001=1

E(x=7)=1000-E(x=0)-E(x=1)-E(x=2)-E(x=3)-E(x=4)-E(x=5)-E(x=6)=1

$\chi^2_0=\sum^6_{i=0}((O_i-E_i)^2/E_i)$ , where Oi - observed values

using table, we prove that this is the Poisson distribution if our sample data gives a value $\chi^2_0<\chi^2_{7,0.04}$ =14.06 ( 7 degrees of freedom )

Chi squared equals 3.093 with 7 degrees of freedom, so this is a Poisson distribution