Does the use of sweetener xylitol reduce the incidence of ear infections? Some c

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 192 children on xylitol, 48 got ear infection, while of 201 children on placebo, 50 got ear infection.

Suppose p1p1 and p2p2 represent population proportions with ear infections on xylitol and placebo, respectively. Again p 1p1 and p 2p2 represent sample proportions with ear infections on xylitol and placebo, respectively.

1. What are the appropriate hypotheses one should test?
Incorrect H0:p 1=p 2H0p1p2   against   Ha:p 1>p 2Hap1p2.
Incorrect H0:p1=p2H0p1p2   against   Ha:p1>p2Hap1p2.
Incorrect H0:p 1=p 2H0p1p2   against   Ha:p 1p 2Hap1p2.
Incorrect H0:p1=p2H0p1p2   against   Ha:p1p2Hap1p2.
Incorrect H0:p 1=p 2H0p1p2   against   Ha:p 1<p 2Hap1p2.
Correct: H0:p1=p2H0p1p2   against   Ha:p1<p2Hap1p2.

2. Rejection region: We reject H0H0 at 1% level of significance if:
|z|>2.326z2.326.
z>2.326z2.326.
z<2.576z2.576.
z<2.326z2.326.
|z|>2.576z2.576.

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

4. If =0.010.01, what will be your conclusion?
Incorrect Reject H0H0.
Correct: Do not reject H0H0.
Incorrect There is not enough information to conclude.

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

6. We should reject H0H0 for all significance level () which are
not equal to p-value.
smaller than p-value.
larger than p-value.

Explanation / Answer

1)since we want to know if  the use of sweetener xylitol reduce the incidence of ear infections

hence

Ho : p1 = p2

Ha : p1 <p2

2) At 1 % P(Z < z*) = 0.01

z* = -2.327

hence critical region is   z<2.326

option D) is correct

3) n1 = 192 ,X1 = 48 , n2 = 201 , X2 = 50

p =(X1 +X2)/(n1+n2) = (48+50)/(192+201) = 0.249363

TS = (p1^ - p2^)/(sqrt(pq*(1/n1+1/n2))

=(48/192 - 50/201)/(sqrt(0.249363*(1-0.249363) (1/192 + 1/201))

= 0.001243 /0.04365

=0.0284765177

4)since TS does not lie in critical region ,we fail to reject the null

5)The p-value of the test is P(Z < 0.0284765177)

=0.511358

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

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