Question

Die A is rolled 50 times and a 6 is scored 4 times, whilea 6 is obtained 10 times when die B is rolled 50 times.

Construct a two-sided 98% confidence interval for the difference in the probabilities of scoring a 6 on the two dice.

		(-0.2074, -0.0325)
		(-0.2324, -0.0075)
		(-0.2792, 0.0392)
		(-0.2960, 0.0560)

Answer 1

A) Given data

Two dies are rolled (A and B)-50 times(n=50)

For Die A:

6 is scored 4 times,so x=4

Probability (pA)=4/50=0.08

qA=1-0.08=0.92

For Die B

6 is scored 10 times,so x=10

Probability (pB)=10/50=0.2

qB=1-0.2=0.8

We need to determine the 98% confidence interval for the difference in the probabilities of scoring 6 on the two dies

Formula used is

CI =(pA-pB)±Z×√[((pA×qA)/n)+(pB×qB)/n]

Z value for 98% confidence level is

Z Value=2.33

=(0.08-0.2) ±2.33× √[((0.08×0.92)/50)+(0.2×0.8)/50]

=(-0.12) ± 2.33×0.06835

=(-0.12) ± 0.1592

Lower limit=-0.12-0.1592=-0.2792

Upper limit=-0.12+0.1592=0.0392

So, finally 98% confidence interval is

(-0.2792,0.0392)