Question

5 coins are tossed 128 times. the result are tabulated as below:

Answer 1

Let x be the number heads obtained is a Binomial variable with the parameters n = 5 and p = probability of getting head = \(\frac{1}{2}\)

∴ q =\(\frac{1}{2}\)

The p.m.f is –

p (x) = nC_n p^xq^n-x ; x = 0, 1, 2 ….. n. 

= 5C_x (0.5)^x (0.5)^5-x ; x = 0, 1,2 …….. 5 

Theoretical/expected frequency T_x = p(x) × N 

∴ T(x = 0) = p(x = 0) 

∴ 128 = 5C₀(0.5)°(0.5)^5-0 × 128 T (0) = 4

Using recurrence relation for theoretical frequency 

Tx = \(\frac{n+1-x}{x}\) x \(\frac{P}{q}\) T_x-1

T(x – 1) = \(\frac{5+1}{1}\) _x \(\frac{0.5}{0.5}\) x T_x-1

= 5 × 1 × T₀

= 5 × 4 T(1) 

= 20 similarly,

∴ The fitted observed theoretical frequency distribution is :

Chi-square Test is : 

H₀: Binomial distribution is good fit 

H₁ : Binomial distribution is not a good fit The 

Test statistic is –

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

Here first two and last two expected frequencies are less than 4 and are pooled with adjacent frequencies :

∴ χ²_cal = 100.932 For (n – 1) = (4 – 1) = 3 d.f at 5% level of significance the upper Tail critical value K/K₂ = 7.81 

Here χ²_cal = 100.932 is in rejection region/ χ²_cal > K₂ 

∴ H₀ is rejected and H₁ is accepted. 

Conclusion : Binomial distribution is not a good fit.