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	20	40	20
B	40	30	15
C	45	50	30

Determine the marginal probability P ( Y = L 1 ) =  (round to the third decimal place)

Determine the conditional probability P ( X = A | Y = L 1 ) =  (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

The marginal probability, P(Y=L₁) = Number of occurances corresponding to L₁/ Total number of occurances = (20+40+45)/ (20+40+20+40+30+15+45+50+30) = 105/ 290 = 0.362

Now, P(X=A | Y = L₁) = P(X=A , Y = L₁) / P(Y = L₁) = Number of simultaneous occurance of A and L₁/ Number of occurances of L₁ = 20/(20+40+45) = 0.190

x<-c(rep("A",80),rep("B",85),rep("C",125))
y<-c(rep("L1",20),rep("L2",40),rep("L3",20),rep("L1",40),rep("L2",30),rep("L3",15),
     rep("L1",45),rep("L2",50),rep("L3",30))

tbl<-table(x,y)
tbl

chi<-chisq.test(tbl)
chi

Result of chi-square test :-

Degrees of freedom = 4

Chi-square = 9.04

p-value = 0.06

At 5% level of significance our conclusion would be, B