Question

Use a 5% significance level.

In a large city,  200  persons were selected at random and each person was asked how many tickets he purchased that week in the state lottery. The results are given in the following table. Suppose that among the  7  persons who had purchased  five  or more tickets,  3  persons had purchased exactly five tickets,  2  persons had purchased six tickets,  1  had purchased seven tickets, and  1  had purchased ten tickets. Test the hypothesis that these  200  observations form a random sample from a  Poisson distribution.                                                                                                      

Number of tickets previously purchased	Number of Persons
0	52
1	60
2	55
3	18
4	8
5 or more	7

Answer 1

ANSWER::

Ho: distribution is poisson probability distribution
Ha: distribution is not the poisson probability distribution

total observation = ΣO =200      
µ = ΣX*O/n = 291/200=1.455      
expected proportion = e-µ * µ^x / x!      
expected frequency = expected proportion * total observation      

X	observed (O)=f	X*O	expected proportion,p = e-µ * µ^x / x!	expected frequency ( E )=p*ΣO
0	52	0	0.233	46.680
1	60	60	0.340	67.920
2	55	110	0.247	49.411
3	18	54	0.120	23.965
4	8	32	0.044	8.717
5 or more	7	35	0.017	3.307

expecte dfrequency is less than 5, so we will merge last two rows.

X	observed (O)=f	X*O	expected proportion,p = e-µ * µ^x / x!	expected frequency ( E )=p*ΣO	(O-E)²/E
0	52	0	0.233	46.680	0.606
1	60	60	0.340	67.920	0.923
2	55	110	0.247	49.411	0.632
3	18	54	0.120	23.965	1.485
4 or more	15	60	0.060	12.024	0.736

chi square test statistic,X² = Σ(O-E)²/E =   4.383              
                  
level of significance, α=   0.05              
Degree of freedom=k-2 =    5   -   2   =   3
                  
P value =   0.2230   [ excel function: =chisq.dist.rt(test-stat,df) ]          
Decision: P value >α , Do not reject Ho                  

conclusion:distribution is poisson probability distribution

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