Rhino viruses typically cause common colds. In a test of the effectiveness of ec

Question

Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 43 of the 49 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 95 of the 108 subjects developed rhinovirus infections. Use a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts through below. 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? H0: P_1 notequalto P_2 Ho: P_1 = P_2 H_0: P_1 graeterthanorequaltoP_2 H_1: P_1 notequalto P_2 Identify the test statistic, z = (Round to two decimal places as needed.) Identify the P-value. P-value =) (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P-value is the significance level of a = 0.05, so the null hypothesis. There sufficient evidencee to support the claim that echinacea treatment has an effect. b. Test the claim by constructing an appropriate confidence interval. The 95% confidence interval is

Explanation / Answer

a)

Formulating the hypotheses          
Ho: p1 - p2   =   0  
Ha: p1 - p2   =/=   0   [ANSWER, B]

***************************

Here, we see that pdo =    0   , the hypothesized population proportion difference.  
          
Getting p1^ and p2^,          
          
p1^ = x1/n1 =    0.87755102      
p2 = x2/n2 =    0.87962963      
          
Also, the standard error of the difference is          
          
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] =    0.056332465      
          
Thus,          
          
z = [p1 - p2 - pdo]/sd =    -0.036898957   [ANSWER, TEST STATISTIC]

********************************

Also, the p value is, as this is two tailed,          
          
P =    0.970565571   [ANSWER]

*****************************

1. GREATER THAN
2. FAIL TO REJECT
3. IS NO [ANSWERS]

****************************  

b)

For the   95%   confidence level, then  
          
alpha/2 = (1 - confidence level)/2 =    0.025      
z(alpha/2) =    1.959963985      
Margin of error = z(alpha/2)*sd =    0.110409603      
lower bound = p1^ - p2^ - z(alpha/2) * sd =    -0.112488212      
upper bound = p1^ - p2^ + z(alpha/2) * sd =    0.108330994      
          
Thus, the confidence interval is          
          
(   -0.112488212   ,   0.108330994 ) [ANSWER]

*******************************

4. INCLUDE
5. DOES NOT
6. IS NO [ANSWER]

********************************

c)

OPTION B: Echinacea does not appear to have a significant effect on the infection rate. [ANSWER, B]
          

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.