HBM1003 Applied Maths& Biostatistics s2 2017 Test:Chapter 9 Post-Test This Quest
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Question
HBM1003 Applied Maths& Biostatistics s2 2017 Test:Chapter 9 Post-Test This Question: 1 pt 12 of 1908 complete) This Test: 19 pts posb cause common colds. ln atest ofthe efectiveness of echnacea, 42 of the 47 s ects treated with echinoo deveeped mno rus nhotons n a placebo group, 89 one 10g Mpch developed thinovirus infections. Use a 0.01 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (e) 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 nul and altemative hypotheses for the hypothesis test? densty the test staistie Round to two decimail places as needed.) Round to tree decimal places as needed.) Click to select your answerta). 7 8 0 2 4Explanation / Answer
Solution:
From the given information
= 0.01
x1= 42 , n1= 47 , p1 = x1/n1= 42/47 = 0.8936
x2= 89 , n2= 103 , p2 = x2/n2= 89/103 = 0.8641
a. Test Hypothesis:
D) H0: p1 = p2
H1: p1 p2
Pooled proportion p = p1n1+ p2n2 / n1+n2
= 0.8936 * 47 + 0.8641*103 /47+103
= 0.8733
Standard error SE = sqrt(p(1-p) *(1/n1+1/n2))
= sqrt(0.8733(1-0.8733) * (1/47 + 1/103))
= 0.0586
Test statistic Z = (p1-p2)/ SE
= (0.8936- 0.8641) / 0.0586
= 0.5034
Since it is a 2-tailed test, hence
P-value = 0.6147
Conclusion:
The P-value is greater than the significance level of = 0.01, so fail to reject null hypothesis. There is not sufficient evidence to support the claim the eachinaces treatment has an effect.
b) Z/2 = Z0.005 = 2.575
Standard error SE = sqrt(p1(1-p1)/n1 + p2(1-p2)/n2)
= sqrt( 0.8936(1-0.8936)/47 + 0.8641(1-0.8936)/103)
= 0.0539
Confidence Interval = (p1 - p2) ± [Z/2 * SE]
= (0.8936-0.8641) ± [2.575 * 0.0586]
= 0.0295 ± 0.1509
= -0.1214 <(p1-p2)< 0.1804
The 99% confidence interval is -0.1214 <(p1-p2)< 0.1804
Conclusion: Because the confidence interval limits include 0, there doesn't appear to be significant difference between the two proportions. There is not evidence to support the claim that echinacea treatment has an effect.
c) B. Eachinacea does not appear to have a significant effect on the infection rate.
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