Question

11. The following 30 datapoints were generated from one of our discrete distributions. 8, 12,7,7,7,12,1,5,1,5,6,5,4,7,3,3,9,4,5,4,8,9,3,6,8,4,7, 4, 11,8 In this problem, we will use various x2 tests to try to figure out which distribution it was.

Answer 1

solution:

Given data:

Summary

Distribution	Parameter	DF	Test statistics	p_value	Conclusion
Uniform	a=1,b=12	4	10.23333333	0.036675	Reject Ho
Geometric	p=1/6.1	5	19.12003525	0.001826	Reject Ho
Poission	lambda=6.1	5	2.561647242	0.767184	Do not Reject Ho

Conclusion:

Underline data is coming from poission distribution

a)

For Uniform[a,b] most likely choce is

a= Xmin

b=Xmax

Hence here a=1 and b=12

b)

p(X=xi)= 1/12

Ei=N*p(X=xi) N=30

We have merged class to have observed frequency >4 thats why first class=<=3 and last >=9

x	Oi=Observed frequency	P(X=xi)	EiUniform[1,12]	(oi-Ei)^2/Ei
<=3	5	0.25	7.5	0.833333
4	5	0.083333333	2.5	2.5
5	4	0.083333333	2.5	0.9
6	2	0.083333333	2.5	0.1
7	5	0.083333333	2.5	2.5
8	4	0.083333333	2.5	0.9
>=9	5	0.333333333	10	2.5
			X^2	10.23333

df= No. of class-No.of parameter-1=7-2-1=4

p_value =p($\chi2$(n=4)>10.233)=0.037

P-value < 0.05 hence at 5% level of significance we reject null hypothesis underline data comning from Uniform distribution

c)

for Geo(p)

Most likely choce for p=1/E(X)

E(X)=6.1

p=1./E(x)=0.164

d)

WE need to calculate probability using geometric distribution

p(X=xi)=p*(1-p)^(xi-1)'

p(X<=3)=p(X=1)+p(X=2)+p(X=3)

=(1/6.1)*((5.1/6.1)^0+(5.1/6.1)^1+(5.1/6.1)^2)

=0.4155

p(X>=9)=1-p(X<=8)

x	Oi=Observed frequency	P(X=xi)	EGeometric[1/6.1]	(oi-Ei)^2/Ei
<=3	5	0.415585	12.46756	4.472767
4	5	0.095806	2.87417	1.572333
5	4	0.0801	2.402995	1.061353
6	2	0.066969	2.009061	4.09E-05
7	5	0.05599	1.679707	6.563257
8	4	0.046811	1.404345	4.797557
>=9	5	0.238739	7.162159	0.652727
				19.12004

df= No. of class-No.of parameter-1=7-1-1=5

p_value =p($\chi2$(n=5)>19.12)=0.002

P-value < 0.05 hence at 5% level of significance we reject null hypothesis underline data comning from geometric distribution

e)

Most likely chice for lambade(X)=6.1

f)

p(X=xi)=exp($\lambda$)*$\lambda$^xi/xi!

we have calculated probability using excell formulae detail provided in last colum

x	Oi=Observed frequency	P(X=xi)	EiUniform[1,12]	(oi-Ei)^2/Ei	Formulae to calculate
<=3	5	0.142501	4.275029	0.122942461	"=POISSON.DIST(3,6.1,1)"
4	5	0.129393	3.881799	0.322112034	=POISSON.DIST(4,6.1,0)
5	4	0.15786	4.735794	0.114319461	=POISSON.DIST(5,6.1,0)
6	2	0.160491	4.814724	1.645509195	=POISSON.DIST(6,6.1,0)
7	5	0.139856	4.195688	0.15418619	=POISSON.DIST(7,6.1,0)
8	4	0.10664	3.199212	0.200443339	=POISSON.DIST(8,6.1,0)
>=9	5	0.163258	4.897752	0.002134561	=1-POISSON.DIST(8,6.1,1)
				2.561647242

df= No. of class-No.of parameter-1=7-1-1=5

p_value =p($\chi2$(n=5)>2.56)=0.767

P-value > 0.05 hence at 5% level of significance we do not reject null hypothesis i.e underline data comning from possion distribution

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