Score: 0.08 of 1 pt HW Score: 58 65%, 235 of 4 pts 9.2.11-T Question Help Rhino
ID: 3051001 • Letter: S
Question
Score: 0.08 of 1 pt HW Score: 58 65%, 235 of 4 pts 9.2.11-T Question Help Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 36 of the 42 subjects treated with echinacea developed thinovirus infections In a placebo group, 96 of the 112 subjects developed rhinovirus infections Use a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (c) below a. Test the claim using a hypothesis test Consider the first sample to be the sample of subjects treated with echinacea and the second sample to be the sample of subjects treated with a placebo. What are the null and alternative hypotheses for the hypothesis test? B. Ho Pi 2p2 H1 P1,P2 D. Ho P1 P2 H1 P1 P2 Round to two decimal places as needed.)Explanation / Answer
1) p1 = 36/42 = 0.857
p2 = 96/112 = 0.857
pooled sample proportion (p) = (p1 * n1 + p2 * n2)/(n1 + n2)
= (0.857 * 42 + 0.857 * 112)/(42 + 112)
= 0.857
SE = sqrt(p * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.857 * (1 - 0.857) * (1/42 + 1/112))
= 0.063
the test statistic z = (p1 - p2)/SE
= (0.857 - 0.857)/0.063 = 0
2) p1 = 24/(24 + 208) = 0.103
p2 = 55/(55 + 463) = 0.106
pooled sample proportion (p) = (p1 * n1 + p2 * n2)/(n1 + n2)
= (0.103 * 232 + 0.106 * 518)/(232 + 518)
= 0.105
SE = sqrt(p * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.105 * (1 - 0.105) * (1/232 + 1/518))
= 0.024
The 90% confidence interval is
(p1 - p2) +/- z0.95 * SE
= 0.103 - 0.106 +/- 1.645 * 0.024
= -0.003 +/- 0.039
= -0.042, 0.036
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