Question

Fit a Binomial distribution for the following data and test at 5% level of significance that binomial distribution is a good fit.

Answer 1

Let x be the number of defective items is a Binominal variate with the n = 5, and ‘p’ is obtained as below : let f be the no of samples.

5p = \(\frac{300}{100}\) = 3

p = 0.6 and q= 1 – p = 1- 0.6 = 0.4

Then the p.m.f is: p(x) = ⁿC_x p^xq^{n – x} , x = 0,1, 2-n

p(x) = ⁵C_X (0.6)x (0.4)^{5 – x} , x = 0,1, 2 5

Theoretical frequency / Expected frequency : Tx =p(x) N

T₀ = p (x = 0) 100 = 510 (0.6)⁰ (0.4)^5-0 × 100 = 1.024 = 1

Using recurrence relation: 

The fitted observed theoretical frequency distribution is:

χ² -TEST: H₀: Binomial distribution is good fit (i.e; 0_i = E_i)

H₁: Binomial distribution is not a good fit (i.e; 0_i ≠ E_i ) upper tail test Under H₀, the χ² -test statistic is:-

Here ‘p’ is estimated from the data and of freedom will be (n – 1 – 1) = (n – 2)

∴ χ_cal² = 9.743 

At α = 5% for (n – 2) = (5 – 2) = 3d.f

The upper tail critical value k₂ = 7.31

Here χ_cal lies in rejection region

∴ H is rejected and H₁ is accepted Conclusion: Binomial distribution is not good fit (0_i ≠ E_i)