Question

Two four-sided dice are tossed and summed 50 times. The resulting frequencies of the sums are given below:

Value 2 3 4 5 6 7 8

Frequency 2 7 12 14 7 3 4

Compute the chi-squared statistic assuming both dice are fair.

Answer 1

sol:

(i) The null hypothesis H0: each face is equally likely to be the outcome of a single roll.means die is fair.

(ii) The alternative hypothesis, Ha: the null hypothesis is false.

(iii) α = 0.05.

(iv) The degrees of freedom: k − 1 = 8 − 1 = 7.

(v) The test statistic can be calculated using a table:

expected value: sum of 2 on two die can be come in one way i.e. (1,1) = (1/16)*50 = 3.125

sum of 3 = (1,2), (2,1) = two ways (2/16)*50 = 6.25

sum of 4 = (1,3), (3,1), (2,2) = (3/16)*50 = 9.375

sum of 5 = (1,4), (2,3), (3,2), (4,1) = (4/16)*50 = 12.5

sum of 6 = (2,4 ), (3,3), (4,2) = (3/16)*50 = 9.375

sum of 7 = (4,3) (3,4) = (2/16)*50 = 6.25

sum of 8 = (4,4) = (1/16)*50 = 3.125

sum E O O-E (O-E)^2 {(E-O)^2}/E

2	3.125	2	1.125	1.266	0.405
3	6.25	7	-0.75	0.563	0.09
4	9.375	12	-2.625	6.891	0.735
5	12.5	14	-1.5	2.250	0.18
6	9.375	7	2.375	5.641	0.602
7	6.25	3	3.25	10.563	1.69
8	3.125	4	-0.875	0.766	0.245

(vi) From α = 0.01 and 8 − 1 = 7, the critical value is 18.475

(vii) Is there enough evidence to reject H0? Since χ 2 ≈ 3.947 < 18.475, we fail to reject the null hypothesis, that the die is fair.