Question

Q - Die A is rolled 50 times and a 6 is scored 4 times,

while a 6 is obtained 10 times when die B is rolled

50 times.

A). Construct a two-sided 98% confidence interval

for the difference in the probabilities of scoring

a 6 on the two dice.

B). Calculate the test statistics for the two- sided null

hypothesis that the two dice have equal

probabilities of scoring a 6.

C). Calculate a p-value for the two-sided null

hypothesis that the two dice have equal

probabilities of scoring a 6.

Please Answer and Explain

Thank you....

Answer 1

(A)

n1 = 50

n2 = 50

p1 = 0.08

p2 = 0.2

% = 98

Pooled Proportion, p = (n1 p1 + n2 p2)/(n1 + n2) = (50 * 0.08 + 50 * 0.2)/(50 + 50) = 0.14

q = 1 - p = 1 - 0.14 = 0.86

SE = √(pq * ((1/n1) + (1/n2))) = √(0.14 * 0.86 * ((1/50) + (1/50))) = 0.069397406

z- score = 2.326347874

Width of the confidence interval = z * SE = 2.32634787404085 * 0.0693974062915899 = 0.161442509

Lower Limit of the confidence interval = (p1 - p2) - width = -0.12 - 0.161442508590389 = -0.281442509

Upper Limit of the confidence interval = (p1 - p2) + width = -0.12 + 0.161442508590389 = 0.041442509

The confidence interval is [-0.2814, 0.0414]

(B)

Data:    

n1 = 50   

n2 = 50   

p1 = 0.08   

p2 = 0.2   

Hypotheses:    

Ho: p1 = p2    

Ha: p1 ≠ p2    

Decision Rule:    

α = 0.02   

Lower Critical z- score =   -2.326347874

Upper Critical z- score = 2.326347874

Reject Ho if |z| >   2.326347874

Test Statistic:    

Average proportion, p = (n1p1 + n2p2)/(n1 + n2) = (50 * 0.08 + 50 * 0.2)/(50 + 50) = 0.14

q = 1 - p = 1 - 0.14 = 0.86

SE = √[pq * {(1/n1) + (1/n2)}] = √(0.14 * 0.86 * ((1/50) + (1/50))) = 0.069397406

z = (p1 - p2)/SE = (0.08 - 0.2)/0.0693974062915899 = -1.729171253

p- value = 0.08377845   

Decision (in terms of the hypotheses):    

Since 1.729171253 < 2.326347874 we fail to reject Ho

Conclusion (in terms of the problem):    

There is no sufficient evidence that the two dice give different proportions of a 6

(C)

p- value = 0.08377845