Question

Consider the following data:

Fit a Poisson distribution to the data and test the goodness of fit. (Use 5% L.O.S.)

Answer 1

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., O_i = E_i}

H₁: Poisson distribution is not good fit {i.e., 0_i ≠ E_i}

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

∴ χ²_cal= 11.0979; n = 4

At α = 5% for (n – 2) = 4 – 2 = 2 d.f the upper tail critical value K² = 9.49. Here χ²_cal lies in R.R. 

∴ H₀ is rejected and H₁ is accepted.

OR

Conclusion: 0_i ≠ E_i; i.e., P.D. is not a good fit.