The parameter of Poisson variate λ is:
From the distribution: Mean = x̄ = λ = \(\frac{∑fx}{N}\)
T1 = \(\frac{0.99}{1}\) T0 = 0.99 × 37.16 = 36.7884 = 37
T2 = \(\frac{0.99}{2}\) T1 = 0.495 × 36.7884 = 18.2103 = 18
T3 =\(\frac{0.99}{3}\) T2 = 0.33 × 18.2103 = 6.0094 = 6
T = \(\frac{0.99}{4}\)T3 = 0.2475 × 6.0094 = 1.4873 = 2
T5, =\(\frac{0.99}{4}\) T4 0.198 × 1.4873 = 0.2945=0
T6 =\(\frac{0.99}{5}\) T5 = 0.165 × 0.2945 = 0.0486= 0
T7 or more=N – T6= 100 – 100 = 0.
The fitted observed and theoretial frequency distribution is:
H0: Poisson distribution is good fit {i.e., Oi = Ei}
H1: Poisson distribution is not good fit {i.e., 0i ≠ Ei}
CHI – SQUARE TEST: Let 0≠ and E. be the observed (f) frequency and theoretical frequency (Tx) Then the χ2-test statistic is:
λ is estimated from the data so (n – 2) d.f
∴ χ2cal= 11.0979; n = 4
At α = 5% for (n – 2) = 4 – 2 = 2 d.f the upper tail critical value K2 = 9.49. Here χ2cal lies in R.R.
∴ H0 is rejected and H1 is accepted.
OR
Conclusion: 0i ≠ Ei; i.e., P.D. is not a good fit.