Solutions HWB12

7.8: This is the six-step procedure to test the hypothesis North location has greater median cash receipts than the South location, using the sign test.

Let pi be the proportion of times that the North location has higher receipts than the South location.

Step 1: H0: pi = .5

Step 2: HA: pi > .5

Step 3: alpha = .10

Step 4: Look at the sample of 9 daily receipts. Assign (+) if North location receipts exceeds South location, assign (-) is it’s the other way around, and assign (0) if they are tied. We have N0 = 0, N+ = 6, and N- = 3. . This makes n = 9 – 0 = 9.

Test statistic is ks = 3., n = 9. Having only three 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 <= 3) = (9 take 0)(.5)^0(1-.5)^9 + (9 take 1)(.5)^2(1-.5)^8 + (9 take 2)(.5)^2(1-.5)^7 + (9 take 3)(.5)^3(1-.5)^6 = binomcdf(9 , .5 , 3) = .2539

Step 6: Conclusion in English. There is insufficient evidence (P-value = .2539) to conclude that the North location has higher median receipts than the South location.

7.15

	Headache	No Headache	Totals
DrugTreatment	9	111	120
Placebo	3	117	120
Totals	12	228	240

Among subjects with headaches, let pi be the proportion of those who get the drug treatment.

Step 1: H0: The drug group has headaches with probability equal to the proportion of subjects in this group: pi = 120/2140= .5

Step 2: HA: Those taking drug have headaches with probability higher than .5: pi > .5

Step 3: alpha = .05

Step 4: In the sample of 12 subjects complaining of headaches, let ks be the sample count of in the drug treatment group. ks = 9

Step 5: P-value = Pr(K >= ks) = Pr(K >= 9) = 1 – Pr(K <= 8) = 1 - binomcdf(12 , .5 , 8) = .0730

Step 6: Conclusion in English. There is insufficient evidence (P-value = .0730) to conclude that the drug treatment increases incidence of headaches.

7.16

	Headache	No Headache	Totals
DrugTreatment	60	840	900
Placebo	4	96	100
Totals	64	936	1000

Among subjects with headaches, let pi be the proportion of those who get the drug treatment.

Step 1: H0: The drug group has headaches with probability equal to the proportion of subjects in this group: pi = 900/1000 = .9

Step 2: HA: Those taking drug have headaches with probability higher than .9: pi > .9

Step 3: alpha = .05

Step 4: In the sample of 64 subjects complaining of headaches, let ks be the sample count of in the drug treatment group (We use the sample count in the drug treatment group because pi measures the probability for this group. If we had use considered the probability of headache in the placebo group, pi = 100/1000 = .10, then ks has to be the count in the placebo group.). Here ks = 60

Step 5: P-value = Pr(K >= ks) = Pr(K >= 60) = 1 – Pr(K <= 59) = 1 - binomcdf(64 , .9 , 59) = .2205

Step 6: Conclusion in English. There is insufficient evidence (P-value = .2205) to conclude that the drug treatment increases incidence of headaches.