Question

6. Rhino viruses typically cause common colds. In a test of the effectiveness of​ echinacea, 35 of the 41 subjects treated with echinacea developed rhinovirus infections. In a placebo​ group, 76 of the 92 subjects developed rhinovirus infections. Use a 0.01 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.H 0​: p 1= p 2

H 1​: p 1 ≠ p2

B.H 0​: p 1 ≠ p 2

H 1​: p 1 = p 2

C.H 0​: p 1 = p 2

H 1​: p 1 > p 2

D.H 0​: p 1 ≥ p 2

H 1​: p 1 ≠ p 2

E.H 0​: p 1 ≤ p 2

H 1​: p 1 ≠ p 2

F: H 0​: p 1 = p 2

H 1​: p 1 < 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 (greater than/less than) the significance level of a=0.01​, so (fail to reject/reject) the null hypothesis. There (is/is not) sufficient evidence to support the claim that echinacea treatment has an effect.

b. Test the claim by constructing an appropriate confidence interval.

The 99​% confidence interval is ____<(p1-p2)<____

​(Round to three decimal places as​ needed.)

What is the conclusion based on the confidence​ interval?

Because the confidence interval limits (do not include/include) 0, there (does not/does) appear to be a significant difference between the two proportions. There (is not/is) evidence to support the claim that echinacea treatment has an effect.

c. Based on the​ results, does echinacea appear to have any effect on the infection​ rate?

A.Echinacea does not appear to have a significant effect on the infection rate.

B.Echinacea does appear to have a significant effect on the infection rate. There is evidence that it lowers the infection rate.

C.Echinacea does appear to have a significant effect on the infection rate. There is evidence that it increases the infection rate.

D.The results are inconclusive.

Answer 1

A.H 0​: p 1= p 2

H 1​: p 1 ≠ p2

first sample size,     n1=   41
number of successes, sample 1 =     x1=   35
proportion success of sample 1 , p̂1=   x1/n1=   0.854
      
second sample size,     n2 =    92
number of successes, sample 2 =     x2 =    76
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.8261

      
difference in sample proportions,    p̂1 - p̂2 =     0.0276
      
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.8346
      
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0698

      
Z-statistic =    (p̂1 - p̂2)/SE =     0.40

      
p-value =        0.693
decision :    p-value>α,Don't reject null hypothesis

the​ P-value is (greater than/) the significance level of a=0.01​, so (fail to reject/) the null hypothesis. There (is not) sufficient evidence to support the claim that echinacea treatment has an effect.   

b)

Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0679  
Z critical value =   Z α/2 =    2.5758   [excel function: =normsinv(α/2)
          
margin of error , E =   Z*SE =    0.1749  
          
confidence interval is           
lower limit =    (p̂1 - p̂2) - E =    -0.147  
upper limit =   (p̂1 - p̂2) + E =    0.202  

Because the confidence interval limits (include) 0, there (does not) appear to be a significant difference between the two proportions. There (is not) evidence to support the claim that echinacea treatment has an effect.

c)

A.Echinacea does not appear to have a significant effect on the infection rate.