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 (0) Expected Frequency (E) 5 6.767 18 13.534 2 10 13.534 3 12 9.022 24 5 7.144 Ho: The distribution is Poisson Hy: The distribution is not Poisson Part a (5 points): 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. Part b (5 points): Using a =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) Part c (5 points): Calculate the P-value for this test. State the conclusion for this test.

Answer 1

Step -1: State the hypothesis:

H0: The distribution is Poisson

H1: The distribution is not Poisson

Number of plants	Observed Frequency fo	Expected Frequency fe	( fo - fe )	( fo - fe )²	( fo - fe )² / fe
0	5	6.766	-1.766	3.118756	0.460945315
1	18	13.534	4.466	19.94516	1.473707404
2	10	13.534	-3.534	12.48916	0.922798581
3	12	9.022	2.978	8.868484	0.982984261
≥ 4	5	7.144	-2.144	4.596736	0.64344009
Total	50	50			4.48387565

Step-2: Compute Chi-square:

χ2 = ∑ [ (foi - fei)² / fei ]

χ2 = 4.48387565 -------------------------(Calculated value / test statistics)

Step-3: Compute the degrees of freedom (df):
df = n – 1 = 50 – 1 = 49

df = 49
Level of significance = 0.01

From chi square table we get;

χ2 = 76.15       -------------------------(Tabulated value/ Critical value)

P value = 1 – ( α/2)

              = 1 – ( 0.01 /2)

              = 1 – 0.005

P value = 0.995

Step-4: Decision:

Calculated value < Tabulated value

   4.48387565 < 76.15

Step-5: Conclusion:

Null hypothesis is accepted.

“The distribution is Poisson”

99% a/2 (Two tail) Rejection region - 76.15 Critical value 4.48387565 Test Statistics 76.15 (Critical Value)