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mathxl.com Do Homework Nathalie Morales How to take a screenshot on your Mac - Apple Support MA 121 3285 Nathalie Morales | 11/8/17 10:03 PM Homework: Section 11.1 Homework Save Score: 0 of 1 pt 301702 complete) Hw Score: 28.57%, 2 of 7 pts Question Help 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, 116 of 664 subjects in the experimental group (group 1) experienced drowsiness as a side effect. After the second dose, 54 of 557 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 fever as a side effect than subjects in group 2 at the = 0.10 level of significance? Verify the model requirements. Select all that apply. A. The samples are dependent. CA n, .)210and naP2(1 ofthe populationszetreachsempie. C. D. The samples are independent. The sample size is less than 5% of the population size for each sample. E. F. The sample size is more than 5% of the population size for each sample. The data come from a population that is normally distributed. Click to select your answer(s) and then click Check Answer. Clear All Check Answer remainingExplanation / Answer
Solution:-
(B) n1p1(1 - p1) > 10 and n2p2(1 - p2) > 10
(D) The samples are independent.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P1< P2
Alternative hypothesis: P1 > P2
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the proportion of women catching cold (p1) is sufficiently smaller than the proportion of men catching cold (p2).
Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.1392
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.0199
z = (p1 - p2) / SE
z = 3.91
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
Since we have a one-tailed test, the P-value is the probability that the z-score is more than 3.91.
Thus, the P-value = less than 0.001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.10), we have to reject the null hypothesis.
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