Solutions HWB11

7.2: Decision Rule:

Reject H0 if an ace never comes up in 24 rolls

Retain H0 if an ace comes up at least once.

alpha = Pr(Reject H0 | H0 is true) = Pr(Ace never comes up | The die is good) = f(0) = (24 take 0)(1/6)^0(5/6)^24 = .0126

beta = Pr(Retain H0 | H0 is false) = Pr(At least one Ace | Die has no aces) = 0

7.4: Decision Rule:

Reject H0 is Heads comes up 11 or 12 times in 12 tosses

Retain H0 if Heads comes up fewer than 11 times.

alpha = Pr(Reject H0 | H0 is true) = Pr(Heads comes up 11 or 12 times | pi = .5) = Pr(K = 11) + Pr(K = 12) = (12 take 11)(.5)^11(1 - .5)^1 + (12 take 12)(.5)^12(1 - .5)^0 = 13/(2^12) = .0033

beta = Pr(Retain H0 | H0 is false) = Pr(Heads comes up less than 11 times | pi = .75) = 1 – {Pr(K = 11) + Pr(K = 12)} = 1 – {(12 take 11)(.75)^11(.25)^1 + (12 take 12)(.75)^12(.25)^0 = 1 - .1584 = .8416

7.6: This is the 6 step procedure to test the hypothesis Med > 25 using the sign test.

Step 1: H0: pi = .5

Step 2: HA: pi > .5

Step 3: alpha = .05

Step 4: There is one value tied with 25; so N0 = 1. This makes n = 10 – 1 = 9.

There are 7 values greater than 25; so N+ = 7.

There are 2 values less than 25; so N- = 2.

Test statistic is ks = 2., and n = 9. Having only two successes out of 9, is evidence for the alternative hypothesis. So we continue with the next step.

Step 5: P-value = Pr(K <= ks) = Pr(K <= 2) = (9 take 0)(.5)^0(1-.5)^9 + (9 take 1)(.5)^2(1-.5)^8 + (9 take 2)(.5)^2(1-.5)^7 = binomcdf(9 , .5 , 2) = .0898

Step 6: Conclusion in English. There is insufficient evidence (P-value = .0898) to conclude that the population median is greater than 25.