Question

A die is rolled 300 times; it lands five 58 times. Is this evidence significant enough to conclude that the die is not fairly balanced (alpha is 0.05)? ....Continuing the previous die problem, suppose that the true chance of landing five is 0.25. Draw the picture to state the region of , which is the probability of making the Type II error.

Answer 1

Ans:

1)sample proportion=58/300=0.1933

Test statistic:

z=(0.1933-0.1667)/sqrt(0.1667*(1-0.1667)/300)

z=1.239

critical z value=+/-1.96

Fail to reject the null hypothesis.

There is not sufficient evidnece to conclude that the die is not fairly balanced.

2)

lower cut off=0.1667-1.96*SQRT(0.1667*(1-0.1667)/300)=0.1245

upper cut off=0.1667-1.96*SQRT(0.1667*(1-0.1667)/300)=0.2089

when true p=0.25

z(0.1245)=(0.1245-0.25)/sqrt(0.25*(1-0.25)/300)=-5.02

z(0.2089)=(0.2089-0.25)/sqrt(0.25*(1-0.25)/300)=-1.645

P(type II error)=P(-5.02<z<-1.646)=P(z<-1.646)-P(z<-5.02)

=0.0498-0.0000

=0.0498