Question

At a local supermarket receiving dock, the number of truck arrivals per day is recorded for 100 days.

Arrivals per Day at a Loading Dock
	Number of Arrivals
	0	1	2	3	4	5	6	7	Total
Frequency	6	25	30	18	7	8	3	3	100

(a)	Estimate the mean from the sample. (Round your answer to 2 decimal places.)

Sample mean

(b)	Carry out the chi-square test, combining end categories as needed to ensure that all expected frequencies are at least five. (Perform a Poisson goodness-of-fit test for ? = .05, combining the last three categories into a single category for 5 or more. Round your answers to 4 decimal places.)

Chi-square
d.f.
p-value

(c)	Truck arrivals per day follow a Poisson process.
	True False

Answer 1

Question - a

X	F	F * X
0	6	0
1	25	25
2	30	60
3	18	54
4	7	28
5	8	40
6	3	18
7	3	21
Total	100	246

Sample Mean = total ( F* X ) / Total (F) = 2.46 /100 = 2.46

Question - C             Yes its a poisson distribution

Any happenings that are at the rate of random, shall be treated as poisson process

Question - B .......... E(X) = Expected frequency = e^-m - m^x / ! x   * 100

e^-m = e ^{- 2.46} = 0.08543

f(0) = 0.08543 * 1 / 1 * 100 = 8.54

f(1) = 0.08543 * 2,46 / 1 * 100 =   21.02

f(2) = 0.08543 * 6.0516 / ! 2 * 100 = 25.85

f(3) = 0.08543 * 14.8869 / ! 3 * 100 = 21.20

f(4) = 0.08543 * 36.62186256 / ! 4 * 100 = 13.04

f(5) = 0.08543 * 90.0897818/ ! 5 * 100 = 6.41

f(6) = 2.63

F(7) = 0.92

Calculating the chi square value = 10.51

OF	EF	(OF - EF)	(OF - EF)2	(OF - EF)2 /EF
6	8.54	-2.54	6.4516	0.755456674
25	21.02	3.98	15.8404	0.75358706
30	25.85	4.15	17.2225	0.666247582
18	21.2	-3.2	10.24	0.483018868
7	13.04	-6.04	36.4816	2.797668712
8	6.41	1.59	2.5281	0.394399376
3	2.63	0.37	0.1369	0.052053232
3	0.93	2.07	4.2849	4.607419355
100			Total	10.51

DOF = degrees of freedom = n - 1 = 8 - 1 = 7

p - value is the table value from chi square tables. Against 7 dof, and 5% significance level you have p - value 2.167