Question

Given contingency table for simultaneous occurance of two categorical random variables X (levels A,B,C) and Y (levels L1,L2,L3)

	L1	L2	L3
A	10	20	30
B	15	20	30
C	25	22	32

Determine the marginal probability P ( X = B ) =  (round to the sthird decimal place)

Determine the conditional probability P ( Y = L 2 | X = B ) =  (round to the second decimal place)

Use R to conduct a chi-square test of independence and thus determine

• Number of degrees of freedom DF=
• Ch-square, round to two decimal places=
• p-value, round to two decimal places=
• Make your conclusion at 5% significance. Enter the correct answer using the following options:  (type the corresponding capital letter, do not type the "dot" at the end)
  1. Not independent at 5% significance level.
  2. Independent with 95% confidence.
  3. Do not reject that they are independent at 5% significance level and reserve judgement. We may accept independence, but with some unknown probability.
  4. Reject both, H0 and H1, the test has failed.
  5. Accept both, H0 and H1, the test has failed.

Answer 1

Observed		L1	L2	L3	Total
	A	10	20	30	60
	B	15	20	30	65
	C	25	22	32	79
	total	50	62	92	204

from above

1) P(X=B)=65/204=0.32

2)P(Y=L2 |X=B) =20/65 =0.31

3)

degree of freedom(df) =(rows-1)*(columns-1)=

4

4)

Applying chi square test of independence:

Expected	Ei=row total*column total/grand total	L1	L2	L3	Total
	A	14.706	18.235	27.059	60
	B	15.931	19.755	29.314	65
	C	19.363	24.010	35.627	79
	total	50	62	92	204
chi square    χ2	=(Oi-Ei)2/Ei	L1	L2	L3	Total
	A	1.506	0.171	0.320	1.9964
	B	0.054	0.003	0.016	0.0736
	C	1.641	0.168	0.369	2.1788
	total	3.2016	0.3421	0.7051	4.2487
test statistic X2=		4.249

5)

p value =

0.373

6)

Do not reject that they are independent at 5% significance level and reserve judgement. We may accept independence, but with some unknown probability.