Question

Please help

After getting trounced by your little brother in a children's game, you suspect the die he gave you to roll may be unfair. To check, you roll it 60 times, recording the number of times each face appears. Do these results cast doubt on the die's fairness? Face 1 Count 11 8 5 Face 4 5 6 Full data set Count 6 7 23 2 3 a) If the die is fair, how many times would you expect each face to show?

Answer 1

a)
If the die is fair, how many times you expect each face to show?
## (1/6)^th time i.e. 10 times. [60*(1/6)=10]

Ir the situation the appropriate test o Goodren AL Xe tead of goodness of ft is a test of significance to that whether the experimental results support the theoretical results or not. any difference Under Hoi une actreally that if there is betoren o bollenved frequency and frequency and expected frequency. NE be number of parametus estimated for filtry the corne test statiche : 72 = (01-08) E k= number of classes cofter X2 k-pl poolfy (ef required) Total Conditions: The observations should be independent of frequency os hould be greater than so. Each expected frequency of hould be greater than or equal to 5. [if it is less than 5, then such a closs to merged onto neighbouring class by procedure of 'poolny ]

b)
To see if the results are Unusual, Goodness of fit will be tested.

c)
Hypothesis:
H0: The die is fair. [It means that P(i^th Face)=1/6 for all i, i=1,2,3,4,5,6]
H1: The die is unfair.

d)
Yes the counted data condition is met. [Total frequency be greater than 50]
Yes, the randomization condition is also met. [Die is rolled which will give random outcomes]
Yes, the expected cell frequency condition is also met. [Each expected frequency is greater than 5; Ei=10 for all i]

e)
Degrees of freedom=df=k-1=6-1=5
k=number of classes (here faces of die), p=0

f)
Computation in MS-EXCEL:

Face	Oi (Count)	Ei	(Oi-Ei)	(Oi-Ei)^2	(Oi-Ei)^2/Ei
1	11	10	1	1	0.1
2	8	10	-2	4	0.4
3	5	10	-5	25	2.5
4	6	10	-4	16	1.6
5	7	10	-3	9	0.9
6	23	10	13	169	16.9
	60	60			22.4
			p-value	CHIDIST(22.4,5)	0.000439

Calculated Chi-square=22.4
P-value=0.000439412

g)
As p-value is less than alpha(0.05), Reject the null hypothesis H0. There is sufficient evidence to support the claim that the die is unfair.