Question

The following table presents the observed and expected data on the number of plants found in each of 50 sampling quadrants.

# of plants	Observed Frequency (Oi)	Expected Frequency (Ei)
0	5	6.676
1	18	13.534
2	10	13.534
3	12	9.022
$\geq$ 4	5	7.144

H0 : The distribution is Poisson

H1 : The distribution is not Poisson

a.) Justify why the assumption of the Poisson distribution seem appropriate as a probability model for this data? Find the value of the only parameter Poisson distribution has.

b.) Using α = 0.01, find the test statistic and critical region to make a conclusion about this goodness of fit test. (Hint: we did not estimate the parameter)

c.) Calculate the P-value for this test. State the conclusion for this test.

Answer 1

Solve using Excel:

D E F 1 Expected Probability (0-5) / E 2 3 А B C No. of Plants (x) | Observed Frequency (0) x*Observed Frequency 0 5 0.00 1 18 18.00 42 10 20.00 5 3 12 36.00 64 or more 5 20.00 7 Total 50 94.00 8 1.88 0.15 0.29 0.27 0.17 0.12 Expected Frequency (E) 7.63 14.34 13.48 8.45 6.09 Total 0.91 0.93 0.90 1.49 0.20 4.43 9 (Ctr) - )/E2 10 No. of Plants (x) Observed Frequency (0) x*Observed Frequency 11 0 50 =A2*B2 12 1 19 =A3*B3 13 2 7 =A4*B4 14 3 4 =A5 B5 15 4 or more 0 =4*B6 16 Total ESUM(B2:B6) =SUM(C2:06) 17 =C7/B7 18 Probability ="c*p***** =POISSON.DIST(A2,$B$ 8,FALSE) =POISSON.DIST(A3,$B$ 8,FALSE) =POISSON.DIST(A4,$B$ 8,FALSE) =POISSON.DIST(A5, $B$ 8,FALSE) =1-SUM(D2:05) Expected Frequency (E) (0-E) / E =D2*50 =D3*50 FTDS-C51"z)/E3 =D4*50 ={(B4-E4)^2)/E4 =D5*50 =((B5-E5)^2)/E5 =D6*50 =((B6-E6)^2)/E6 Total =SUM(F2:F6)

No. of Plants (x)	Observed Frequency (O)	x*Observed Frequency	Expected Probability	Expected Frequency (E)	(Oi - Ei)2 / Ei
0	5	0.00	0.15	7.63	0.91
1	18	18.00	0.29	14.34	0.93
2	10	20.00	0.27	13.48	0.90
3	12	36.00	0.17	8.45	1.49
4 or more	5	20.00	0.12	6.09	0.20
Total	50	94.00		Total	4.43
λ	1.88
No. of Plants (x)	Observed Frequency (O)	x*Observed Frequency	Probability = nCx*px*qn-x	Expected Frequency (E)	(Oi - Ei)2 / Ei
0	50	=A2*B2	=POISSON.DIST(A2,$B$8,FALSE)	=D2*50	=((B2-E2)^2)/E2
1	19	=A3*B3	=POISSON.DIST(A3,$B$8,FALSE)	=D3*50	=((B3-E3)^2)/E3
2	7	=A4*B4	=POISSON.DIST(A4,$B$8,FALSE)	=D4*50	=((B4-E4)^2)/E4
3	4	=A5*B5	=POISSON.DIST(A5,$B$8,FALSE)	=D5*50	=((B5-E5)^2)/E5
4 or more	0	=4*B6	=1-SUM(D2:D5)	=D6*50	=((B6-E6)^2)/E6
Total	=SUM(B2:B6)	=SUM(C2:C6)		Total	=SUM(F2:F6)
λ	=C7/B7

H₀ : The distribution is Poisson

H₁ : The distribution is not Poisson

a) Only parameter: λ = 1.88

b) Test statistic = 4.43

Level of significance = 0.01

Degrees of freedom: df = n-k-1 (k is no. of parameters) = 50-1-1 = 48

Critical value (Using Excel function CHISQ.INV.RT(probability,df)) = CHISQ.INV.RT(0.01,48) = 73.68

c) p-value (Using Excel function CHISQ.DIST.RT(test statistic,df)) = CHISQ.DIST.RT(4.43,48) = 1

Since p-value is more than 0.01, we do not reject the null hypothesis.

So, the distribution is Poisson.