Question

Does the use of sweetener xylitol reduce the incidence of ear infections? Some children are randomly allocated t xylitol treatment group and other children to the control group (who are treated with placebo). Of 214 children on xylitol, 54 got ear infection, while of 292 children on placebo, 73 got ear infection.
Suppose p₁ and p₂ represent population proportions with ear infections on xylitol and placebo, respectively. Again [^(p)]₁ and [^(p)]₂ represent sample proportions with ear infections on xylitol and placebo, respectively.

1. What are the appropriate hypotheses one should test?
H₀: p₁ = p₂   against   H_a: p₁ >  p₂.
H₀: [^(p)]₁ = [^(p)]₂   against   H_a: [^(p)]₁ ≠ [^(p)]₂.
H₀: p₁ = p₂   against   H_a: p₁ <  p₂.
H₀: p₁ = p₂   against   H_a: p₁ ≠ p₂.
H₀: [^(p)]₁ = [^(p)]₂   against   H_a: [^(p)]₁ > [^(p)]₂.
H₀: [^(p)]₁ = [^(p)]₂   against   H_a: [^(p)]₁ < [^(p)]₂.

Tries 0/3

2. Rejection region: We reject H₀ at 1% level of significance if:
z < −2.326.
|z| > 2.326.
z > 2.326.
z < −2.576.
|z| > 2.576.
None of the above

Tries 0/3

3. The value of the test-statistic is: Answer to 2 decimal places.

Tries 0/5

4. If α = 0.01, what will be your conclusion?
Reject H₀.
There is not enough information to conclude.
Do not reject H₀.

Tries 0/3

5. The p-value of the test is: Answer to 4 decimal places.

Tries 0/5

6. We should reject H₀ for all significance level (α) which are
larger than p-value.
smaller than p-value.
not equal to p-value.

Answer 1

1)

H₀: p₁ = p₂   against   H_a: p₁ <  p₂.

2)

z-critical value , Z* =        -2.3263 [Excel function =NORMSINV(α)  
We reject H₀ at 1% level of significance if: z < −2.326.

3)

sample #1   ----->              
first sample size,     n1=   214          
number of successes, sample 1 =     x1=   54          
proportion success of sample 1 , p̂1=   x1/n1=   0.2523          
                  
sample #2   ----->              
second sample size,     n2 =    292          
number of successes, sample 2 =     x2 =    73          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.2500          
                  
difference in sample proportions, p̂1 - p̂2 =     0.2523   -   0.2500   =   0.0023
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.2510          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0390          
Z-statistic = (p̂1 - p̂2)/SE = (   0.002   /   0.0390   ) =   0.06

4)

Do not reject H₀.

5)

p-value =        0.5239   [Excel function =NORMSDIST(z)  

6)

We should reject H₀ for all significance level (α) which are
smaller than p-value.