4. A researcher counts the number of cars that pass through an intersection each
ID: 3220724 • Letter: 4
Question
4. A researcher counts the number of cars that pass through an intersection each minute for 120 minutes. He summarizes his data in the following table Cars per minute Count 49 46 2 20 3 Assuming the number of cars that pass through the intersection each minute is inde- pendent, the researcher would like to test if a Poisson distribution is a good fit for the data. Use significance level o 0.1 (a) (1 mark) What is the maximum likelihood estimate of A? (b) (6 marks) Use Pearson's chi-square test to test the hypothesis that the Poisson distribution is a good fit for the data. Make sure to state the null and alternative hypotheses fully. (c) (3 marks) What is the likelihood ratio test statistic? Does the likelihood ratio test give the same conclusion as (a)Explanation / Answer
cars per min(mi)
count (freq fi)
mi*fi
0
49
0
1
46
46
2
20
40
3
3
9
4
2
8
103
(a) maximum likelihood estimate of lambda = 103/120=0.858333333
(b)
Ho = number of arrivals per minute follows a Poisson distribution
H1 = number of arrivals per minute does not follow a poisson distribution
1
2
3
4
5
6
7
8
cars per min(mi)
count (freq fi)
mi*fi
prob for poisson
theretical freq(fe)
fi-fe
(fi-fe)squared
(fi-fe)sqr/fe
0
49
0
0.423867941
50.86415286
-1.86415
3.475065888
0.068321
1
46
46
0.363819982
43.65839787
2.341602
5.483100525
0.125591
2
20
40
0.156139409
18.73672909
1.263271
1.5958534
0.085172
3
3
9
0.04467322
5.360786378
-2.36079
5.573312321
1.039645
4
2
8
0.009586128
1.15033541
0.849665
0.721929915
0.627582
103
1.946311
column 4 was calculated using =POISSON(row1,0.858333333,FALSE) in excel ..false= noncumulative
column 5 = 120*(column 4)
degrees of freedom = 5-1-1=3
=number of categories – number of parameters – 1
chitest value=CHIINV(0.1,3) = 6.251388457 …. for 0.1 significance and 3 degrees of freedom
...can also be obtained from table
reject H0 if chisquare > CHIINV(0.1,3)
since CHIINV(0.1,3) > 1.946311 we accept the null hypothesis
(c) likelihood ratio test statistic is 1.9463
cars per min(mi)
count (freq fi)
mi*fi
0
49
0
1
46
46
2
20
40
3
3
9
4
2
8
103
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