Question

a. Poisson distribution was used to model the number of arrivals per minute at a bank located in the central business district of a city. Suppose that the actual arrivals per minute were observed in 200 01e. minute periods over the course of a week are given below. Determine whether the number of arrivals per minute follows a Poisson distribution Frequency 14 31 47 41 29 21 10 Arrivals 2 6 8 Total 1 -8 CI- b. In the z test, a hypothesized distribution more likely to accepted at a- 005 than at (1-E 200 -001? Explain your answer

Answer 1

Answer:

x	f	f*x
0	14	0
1	31	31
2	47	94
3	41	123
4	29	116
5	21	105
6	10	60
7	5	35
8	2	16
Total	200	580
	mean = 580/200=2.9

λ= 2.9

P(X=x) = e^-λ λ^x / x!    for x = 0, 1, ....8

P(X = 0) = 0.0550

POISSON.DIST Probabilities Table
	X	P(X)
	0	0.0550
	1	0.1596
	2	0.2314
	3	0.2237
	4	0.1622
	5	0.0940
	6	0.0455
	7	0.0188
	8	0.0068

x	f	f*x	p	200*p= expected value
0	14	0	0.0550	11
1	31	31	0.1596	31.92
2	47	94	0.2314	46.28
3	41	123	0.2237	44.74
4	29	116	0.1622	32.44
5	21	105	0.0940	18.8
6	10	60	0.0455	9.1
7	5	35	0.0188	3.76
8	2	16	0.0068	1.36
Total	200	580		199.4

Ho: number of arrivals follows Poisson distribution

H1: number of arrivals not follows Poisson distribution

Goodness of Fit Test
	observed	expected	O - E	(O - E)² / E
	14	11.000	3.000	0.818
	31	31.920	-0.920	0.027
	47	46.280	0.720	0.011
	41	44.740	-3.740	0.313
	29	32.440	-3.440	0.365
	21	18.800	2.200	0.257
	10	9.100	0.900	0.089
	5	3.760	1.240	0.409
	2	1.360	0.640	0.301
Total	200	199.400	0.600	2.590
Chi square	2.590
DF	8

Calculated chi square = 2.590 is < critical chi square at 0.05 level of significance value 15.51.

Ho is not rejected at 0.05 level.

Therefore Ho is not rejected at 0.01 level also.( critical value for 0.01 level is 20.09)