Question

In randomized, double-blind clinical trials of a new vaccine, monkeys were randomly divided into two groups. Subjects in group 1 received the new vaccine while subjects in group 2 received a control vaccine. After the second dose, 128 of 695 subjects in the experimental group (group 1) experienced drowsiness as a side effect. After the second dose, 85 of 576 of the subjects in the control group (group 2) experienced drowsiness as a side effect. Does the evidence suggest that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the a= 0.01 level of significance? Verify the model requirements. Select all that apply. A. The samples are independent. B. The sample size is more than 5% of the population size for each sample. C. The sample size is less than 5% of the population size for each sample. D. The data come from a population that is normally distributed. E. The samples are dependent. F. na f1 (1- @1 ) 2 10 and n2f2 (1 - f2) = 10 Determine the null and alternative hypotheses. Ho: P1 P2 H1:21 P2 Find the test statistic for this hypothesis test. (Round to two decimal places as needed.) Determine the P-value for this hypothesis test.

Answer 1

Sample 1:

n1 = 695, x1 = 128

p̂1 = x1/n1 = 0.1842

Sample 2:

n2 = 576, x2 = 85

p̂2 = x2/n2 = 0.1476

α = 0.01

All that apply:

• The samples are independent.
• The sample size is less than 5% of the population size for each sample.
• n1p̂1(1-p̂1) >= 10 and n2p̂2(1-p̂2) >= 10

Null and Alternative hypothesis:

Ho : p1 = p2

H1 : p1 > p2

Pooled proportion:

p̄ = (x1+x2)/(n1+n2) = (128+85)/(695+576) = 0.167585

Test statistic:

z = (p̂1 - p̂2)/√ [p̄*(1-p̄)*(1/n1+1/n2)] = (0.1842 - 0.1476)/√[0.1676*0.8324*(1/695+1/576)] = 1.74

p-value = 1- NORM.S.DIST(1.7393, 1) = 0.041

Conclusion:

p-value > α, Do not reject the null hypothesis. There is not enough evidence