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 210 children on xylitol, 53 got ear infection, while of 292 children on placebo, 73 got ear infection.
Suppose p1p1 and p2p2 represent population proportions with ear infections on xylitol and placebo, respectively. Again p̂ 1p^1 and p̂ 2p^2 represent sample proportions with ear infections on xylitol and placebo, respectively.

What are the appropriate hypotheses one should test?
H0:p1=p2H0:p1=p2   against   Ha:p1>p2Ha:p1>p2.
H0:p̂ 1=p̂ 2H0:p^1=p^2   against   Ha:p̂ 1<p̂ 2Ha:p^1<p^2.
H0:p̂ 1=p̂ 2H0:p^1=p^2   against   Ha:p̂ 1>p̂ 2Ha:p^1>p^2.
H0:p1=p2H0:p1=p2   against   Ha:p1<p2Ha:p1<p2.
H0:p̂ 1=p̂ 2H0:p^1=p^2   against   Ha:p̂ 1≠p̂ 2Ha:p^1≠p^2.
H0:p1=p2H0:p1=p2   against   Ha:p1≠p2Ha:p1≠p2.

Tries 0/3

Rejection region: We reject H0H0 at 1% level of significance if:
z<−2.326z<−2.326.
|z|>2.576|z|>2.576.
z<−2.576z<−2.576.
|z|>2.326|z|>2.326.
z>2.326z>2.326.
None of the above

Tries 0/3

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

Tries 0/5

If α=0.01α=0.01, what will be your conclusion?
Reject H0H0.
Do not reject H0H0.
There is not enough information to conclude.

Tries 0/3

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

Tries 0/5

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

Tries 0/3

Answer 1

we have to test whether the use of sweetener xylitol reduce the incidence of ear infections. So, it is a one tailed hypothesis test

z critical value for left tailed hypothesis at 0.01 significance level is -2.326

So, we will reject the null hypothesis when z statistic is less than -2.326

option A is correct (z<−2.326)

Using TI 84 calculator

press stat then tests then 2-propZtest

enter the data

x1= 53, n1 = 210, x2 =73, n2 = 292

p1< p2

press enter, we get

z statistic = 0.06

z statistic is not less than critical value. So, at 0.01 significance level, we failed to reject the null hypothesis

p value= 0.4761

it is clear that the p value is larger than the alpha level of 0.01

So, we failed to reject the null hypothesis

p value =