Question

Records pertaining to the monthly number of job-related injuries at an underground coal mine were being studied by a federal agency. The values for the past 100 months are given (see file H9.xlsx, sheet P3).

Apply the chi-square test to the data to test the hypothesis that the underlying distribution is Poisson. Use a level of significance of a = 0.05.

Apply the chi-square test to the data to test the hypothesis that the underlying distribution is Poisson with mean 1.0. Use a level of significance of a = 0.05.

Note that in (a) you have to estimate mean and then perform the hypothesis test, whereas in (b) you are being asked to test the hypothesis with the given mean, which may be different from the sample mean. Think carefully about the number of degrees of freedom that you will be using in (a) and (b).

Injuries per month	Frequency of Occurrence
0	35
1	40
2	13
3	6
4	4
5	1
6	1

Answer 1

Part A) H_0: poisson distribution vs H_a: not poisson distribution

x = 0:6
f = c(35,40,13,6,4,1,1)
p = c(.35,.4,.13,0.06,0.04,0.01,0.01)
part.a = chisq.test(x,p)
part.a

Pearson's Chi-squared test

data: x and p
X-squared = 35, df = 30, p-value = 0.2426

so failed to reject H_0

part B)

Ho:poisson dist

H_a: not poisson dist

mean = lambda = xbar

prob.poisson = function(x){
100*((exp(-m)*m^x)/factorial(x))
}
p_0=prob.poisson(0)
p_1=prob.poisson(1)
p_2 =prob.poisson(2)
p_3 =prob.poisson(3)
p_4 =prob.poisson(4)
p_5=prob.poisson(5)
p_6 =prob.poisson(6)
pnew=c(p_0,p_1,p_2,p_3,p_4,p_5,p_6)
part.b = chisq.test(x,pnew)
part.b

Pearson's Chi-squared test

data: x and pnew

X-squared = 42, df = 36, p-value = 0.227

accept Ho

x follow poisson distribution with mean 1.11

Note: degrees of freedom

in part a we have df=30 and part b df =36