Question

Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output (cal/cm2/min) was measured. For m 8 subjects with the syndrome, the average heat output was x = 0.64, and for n 8 nonsufferers, the average output was 2.06. Let u, and u2 denote the true average heat outputs for the sufferers and nonsufferers, respectively. Assume that the two distributions of heat output are normal with o1 = 0.2 and o2 = 0.3. (a) Consider testing Hn: u1 - i2 = -1.0 versus Hau- fi2 < -1.0 at level 0.01. Describe in words what H2 says, and then carry out the test OHa says that the average heat output for sufferers is the same as that of non-sufferers Ha says that the average heat output for sufferers is more than 1 cal/cm2/min below that of non-sufferers. OHa says that the average heat output for sufferers is less than 1 cal/cm2/min below that of non-sufferers. Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = p-value = State the conclusion in the problem context. Reject Hn. The data suggests that the average heat output for sufferers is the same as that of non-sufferers OFail to reject Hn. The data suggests that the average heat output for sufferers is the same as that of non-sufferers. Reject Ho. The data suggests that the average heat output for sufferers is more than 1 cal/cm2/min below that of non-sufferers. Fail to reject H. The data suggests that the average heat output for sufferers is less than 1 cal/cm2/min below that of non-sufferers. (b) What is the probability of a type II error when the actual difference between u and u2 is u1 - i2 = -1.2? (Round your answer to four decimal places.) (c) Assuming that m = n, what sample sizes are required to ensure that = 0.1 when u, - i2 = -1.2? (Round your answer up to the nearest whole number.) subjects

Answer 1

a)

std error of difference σ1-2=				√(σ21/n1+σ22/n2)    =	0.1275
test stat z                                       =				(x1-x2-Δo)/σ1-2     =	-3.29
p value : =					0.0005

b)

-1 Hypothesized mean difference true mean difference first sample standard deviation second sample standard deviation first sample size second sample size =n; standard error of mean =0x =(V(0,-/n1+0,-/n)) for 0.01 level and left tailed test critival value Za rejection region: X <= Ao +Za*ox or type II error =probability of not rejecting B = -1.2 0.200 0.300 8.000 8.000 0.1275 -2.33 -1.2970 A)/ox)) P(Z >-1.297- = P(Z >-0.76) 0.7764

type II error =0.7764

c)

for 0.01 level and right tailed test critival value Zα					=	2.33
for 0.1 level of type II error critival value Zβ					=	1.28
required sample size             =n			=(Zα/2+Zβ)2(σ12+σ22)/(Δo-Δ)2		=	43