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.
![](https://www.sarthaks.com/?qa=blob&qa_blobid=3363044968137722771)
![](https://www.sarthaks.com/?qa=blob&qa_blobid=10946474762630754598)
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) = nCx pxqn – x , x = 0,1, 2-n
p(x) = 5CX (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:
![](https://www.sarthaks.com/?qa=blob&qa_blobid=11967306289035334149)
The fitted observed theoretical frequency distribution is:
![](https://www.sarthaks.com/?qa=blob&qa_blobid=17557392424195263125)
χ2 -TEST: H0: Binomial distribution is good fit (i.e; 0i = Ei)
H1: Binomial distribution is not a good fit (i.e; 0i ≠ Ei ) upper tail test Under H0, the χ2 -test statistic is:-
![](https://www.sarthaks.com/?qa=blob&qa_blobid=12698536828172078685)
Here ‘p’ is estimated from the data and of freedom will be (n – 1 – 1) = (n – 2)
![](https://www.sarthaks.com/?qa=blob&qa_blobid=12535544108016075121)
∴ χcal2 = 9.743
At α = 5% for (n – 2) = (5 – 2) = 3d.f
The upper tail critical value k2 = 7.31
Here χcal lies in rejection region
∴ H is rejected and H1 is accepted Conclusion: Binomial distribution is not good fit (0i ≠ Ei)
![](https://www.sarthaks.com/?qa=blob&qa_blobid=8420647671806404327)