In: Math
The following counts of vehicles arriving at a toll station, over 1-minute intervals, were made by an engineering student: 2,0,4,1,2,4,2,1,6,2,5,4,4,3,3,2,2,1,2,3,2,5,2,0,1,2,0,3,1,1, 2,3,5,6,3,4,3,4,0,6,3,2,1,4,4,0,1,4,3,7,0,0,2,3,2,4,3,2,4,5 (a) Assuming the vehicle arrivals were generated by a Poisson process, compute the maximum likelihood estimate of the arrival rate. (b) Test the hypothesis that the arrival counts follow a Poisson distribution, at the α=0.05 significance level.
a)
sample mean is unbiased estimator of lambda
sample mean = sum of all terms / number of terms
b)
this is chi-square goodness of fit test
p | Oi | Ei | (Oi-Ei)^2/Ei | |
0 | 0.0695 | 7 | 4.1690 | 1.9224 |
1 | 0.1853 | 8 | 11.1174 | 0.8741 |
2 | 0.2471 | 15 | 14.8231 | 0.0021 |
3 | 0.2196 | 11 | 13.1761 | 0.3594 |
4 | 0.1464 | 11 | 8.7841 | 0.5590 |
5 | 0.0781 | 4 | 4.6848 | 0.1001 |
>5 | 0.0541 | 4 | 3.2455 | 0.1754 |
1 | 60 | 60.0000 | 3.9926 | |
TS | 3.9926 | |||
critical value | 12.5916 | |||
p-value | 0.6777 |
TS = 3.9926
critical value = 12.5916
since TS < critical value
we fail to reject the null hypothesis
we conclude that there is not sufficient evidence that the arrival counts does not follow a Poisson distribution
Formulas
p | Oi | Ei | (Oi-Ei)^2/Ei | |
0 | =POISSON.DIST(A2,8/3,0) | 7 | =$C$10*B2 | =(C2-D2)^2/D2 |
1 | =POISSON.DIST(A3,8/3,0) | 8 | =$C$10*B3 | =(C3-D3)^2/D3 |
2 | =POISSON.DIST(A4,8/3,0) | 15 | =$C$10*B4 | =(C4-D4)^2/D4 |
3 | =POISSON.DIST(A5,8/3,0) | 11 | =$C$10*B5 | =(C5-D5)^2/D5 |
4 | =POISSON.DIST(A6,8/3,0) | 11 | =$C$10*B6 | =(C6-D6)^2/D6 |
5 | =POISSON.DIST(A7,8/3,0) | 4 | =$C$10*B7 | =(C7-D7)^2/D7 |
>5 | =1-SUM(B2:B7) | 4 | =$C$10*B8 | =(C8-D8)^2/D8 |
=SUM(B2:B8) | =SUM(C2:C8) | =SUM(D2:D8) | =SUM(E2:E8) | |
TS | =SUM(E2:E8) | |||
critical value | =CHISQ.INV(0.95,6) | |||
p-value | =1-CHISQ.DIST(E10,6,1) |