Math  /  Data & Statistics

QuestionQuestion 3, 9.1.15-T HW Score: 26.17\%, 3.92 of 15 points Part 2 of 7 Points: 0 of 1 Save
Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 34 of the 39 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 81 of the 98 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? A. H0:p1=p2H_{0}: p_{1}=p_{2} B. H0:p1p2H_{0}: p_{1} \leq p_{2} C. H0:p1=p2H_{0}: p_{1}=p_{2} H1:p1>p2H_{1}: p_{1}>p_{2} H1:p1p2H_{1}: p_{1} \neq p_{2} H1:p1p2H_{1}: p_{1} \neq p_{2} D. H0:p1p2H_{0}: p_{1} \geq p_{2} E. H0:p1p2H_{0}: p_{1} \neq p_{2} F. H0:p1=p2\mathrm{H}_{0}: \mathrm{p}_{1}=\mathrm{p}_{2} H1:p1p2H_{1}: p_{1} \neq p_{2} H1:p1=p2H_{1}: p_{1}=p_{2} H1:p1<p2H_{1}: p_{1}<p_{2}
Identify the test statistic. z=\mathrm{z}=\square (Round to two decimal places as needed.)

Studdy Solution
Make a decision based on the test statistic and significance level:
- Compare the calculated z-value to the critical z-value for α=0.05\alpha = 0.05 (two-tailed). - Critical z-value for α=0.05\alpha = 0.05 is approximately ±1.96\pm 1.96.
Since 0.6180.618 is within the range 1.96-1.96 to 1.961.96, we fail to reject the null hypothesis.
The test statistic is:
z=0.62 z = 0.62

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