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 243 children on xylitol, 61 got ear infection, while of 227 children on placebo, 56 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.

*Please Show Work!*

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

2. Rejection region: We reject H0 at 5% level of significance if:
z < ?1.960.
|z| > 1.645.
|z| > 1.960.
z > 1.645.
z < ?1.645.
None of the above

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

4. If ? = 0.05, what will be your conclusion?
Reject H0.
There is not enough information to conclude.
Do not reject H0.

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

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

Explanation / Answer

2.
As, the alternative hypothesis does not have unequal sign, this is one-tail (left tail for less than sign) hypothesis test.
z value at 0.05 is -1.64
For left tail test , at 5% significance level, the rejection region is z < ?1.645.
The correct answer is z < ?1.645

3.
Pooled sample proportion is,
p = (number of infections in treatment group + number of infections in placebo group ) / Total number of observations
= (61 + 56) / (243 + 227)
= 0.2489

Standard error of proportion, SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

where n1, n2 are number of observations for treatment and placebo group.

SE = sqrt{ 0.2489 * ( 1 - 0.2489 ) * [ (1/243) + (1/227) ] } = 0.0399

Let p1, p2 are proportions of ear infections for treatment and placebo group.

p1 = 61/243 = 0.2510

p2 = 56 / 227 = 0.2467

Difference in proportions = p1 - p2 = 0.2510 - 0.2467 = 0.0043

z = (p1 - p2) / SE = 0.0043 / 0.0399 = 0.11

4.

P-value = P(z < 0.11) = 0.5438

As, the p-value is greater than 0.05, we

Do not reject H0.

5.

The p-value of the test is 0.5438

6.

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

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