Question

For the following data fit a Poisson distribution and test whether it is a good fit (test at 5% level of significance)

Answer 1

Let x be the no. of T. V. Sold is a Poisson variate. 

The parameter λ can be obtained as below: 

Let f be the no. of days.

We know that; mean 

i.e., λ

Theoretical/expected frequency T_x = P(x) × N.

∴ T(x = 0) = p(x= 0) × 150

 

= 0.1496 × 150

T₀ = 22.44 

using recurring relation for theoretical frequency

∴ The fitted poisson distribution is :

Chi- square test: 

H₀: Poisson distribution is good fit 

H₁ : Poisson distribution is not a good fit. 

The test statistic is –

Since the parameter X is estimated so, one more d.f is lessened. 

Let ‘O’ and ‘E’ be the observed and expected frequencies.

∴ χ²_cal = 1.938 

For (n – 2) = (6 – 2) = 4 d.f at 5% level of significance the upper tail critical value K₂ = 9.49 

Here χ²_cal = 1.938 is in acceptance region. (χ²_cal < K) 

∴ H₀ is accepted. 

Conclusion : Poisson distributions is good fit.