Question

Five unbiased dice were thrown 96 times and the number of times 4, 5 or 6 was 

obtained, is given in the following table:

 

No. of dice  0. 1. 2. 3. 4. 5.

showing 4, 

5 or 6 

Frequency 1 10 24 35 18 8

At 5% level of significance test whether this data comes from a binomial distribution. 

You may like to use the following values. 

 [( x₅²10.05)=11.07 , x₆²(0.05)=12.59 ,x₇²(0.05)=14.07.]⁽

Answer 1

Let H$_o$: The given data comes from Binomial distribution

Total number of dice is thrown N=96 times

Probability of obtaining 4,5 or 6 $p=\dfrac{1}{2}$

$q=1-p=1-\dfrac{1}{2}=\dfrac{1}{2}$

For binomial distribution:

$E_i=Np_i=N/f_i$

where p_i is the probability of success,

f_i is the frequency of success.

The table for Chi-square is given by,-

The value of $\chi^2=\sum \dfrac{(O_i-E_i)^2}{E_i}=14.01$

The tabulate value of $\chi^2$ at 5% level of significance is 14.07.

Conclusion: As the calculated value of $\chi^2$ is less than the tabulated value So H_o is accepted i.e Data comes from binomial distribution.