Question

A civil engineer has been studying the frequency of vehicle accidents on a certain stretch of interstate highway. Longterm history indicates that there has been an average of 1.74 accidents per day on this section of the interstate. Let r be a random variable that represents number of accidents per day. Let O represent the number of observed accidents per day based on local highway patrol reports. A random sample of 90 days gave the following information.

r	0	1	2	3	4 or more
O	21	19	17	18	15

(a) The civil engineer wants to use a Poisson distribution to represent the probability of r, the number of accidents per day. The Poisson distribution is given below.

P(r) =

e^???^r

r!

Here ? = 1.74 is the average number of accidents per day. Compute P(r) for r = 0, 1, 2, 3, and 4 or more. (Round your answers to three decimal places.)

P(0) =
P(1) =
P(2) =
P(3) =
P(4 or more) =

(b) Compute the expected number of accidents E = 90P(r) for r = 0, 1, 2, 3, and 4 or more. (Round your answers to two decimal places.)

E(0) =
E(1) =
E(2) =
E(3) =
E(4 or more) =

(c) Compute the sample statistic ?² = ?((O – E)²/E) and the degrees of freedom. (Round your sample statistic to three decimal places.)

df =
?2 =

(d) Test the statement that the Poisson distribution fits the sample data. Use a 1% level of significance.

Fail to reject the null hypothesis, there is sufficient evidence to conclude the Poisson distribution does not fit.

Reject the null hypothesis, there is insufficient evidence to conclude the Poisson distribution does not fit.    

Reject the null hypothesis, there is sufficient evidence to conclude the Poisson distribution does not fit.

Fail to reject the null hypothesis, there is insufficient evidence to conclude the Poisson distribution does not fit.

Answer 1

	p	Oi	Ei	(Oi-Ei)^2/Ei
0	0.175520401	21	15.79683606	1.713818827
1	0.305405497	19	27.48649474	2.620217442
2	0.265702782	17	23.91325042	1.99860038
3	0.154107614	18	13.86968524	1.229984652
4 or more	0.099263706	15	8.933733543	4.11917241
	1	90	90
			TS	11.68179371
			p-value	0.019881254

a) columns with p

b) column with Ei

c) df = r-1 = 4

TS = 11.68179371

d )

p-value =0.019881254

p-value > 0.01

we fail to reject the null

option D)

Fail to reject the null hypothesis, there is insufficient evidence to conclude the Poisson distribution does not fit.