Answer:
The P-value (0.159) is higher than the significance level (0.1), so we can not reject the null hypothesis.
The claim of the anti-tobacco campaign could be right about their estimation.
Step-by-step explanation:
This is a problem of hypothesis testing. In this case, about the teenage population's proportion of smokers.
<u><em>Hypothesis</em></u>
The null hypothesis, that needs to be tested, is that the proportion of teenage that use tobacco is less than 20%:
![H_0: \mu](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%3C0.2)
The alternative hypothesis is that this proportion is higher than 20%
![H_0: \mu>0.2](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%3E0.2)
The significance level is 0.10 and it is used a one-sample z-test.
Analysis
The proportion for the sample is:
![p=207/1100=0.188](https://tex.z-dn.net/?f=p%3D207%2F1100%3D0.188)
Then, we calculate the standard deviation of the sample:
![\sigma=\sqrt{\frac{p(1-p)}{n} } =\sqrt{\frac{0.188(1-0.188)}{1100} }= 0.012](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D%20%3D%5Csqrt%7B%5Cfrac%7B0.188%281-0.188%29%7D%7B1100%7D%20%7D%3D%200.012)
The z-value can now be calculated as:
![z=\frac{p-\mu}{\sigma}=\frac{0.188-0.2}{0.012} =-1](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bp-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B0.188-0.2%7D%7B0.012%7D%20%20%3D-1)
The P-value for
is
![P(z](https://tex.z-dn.net/?f=P%28z%3C-1%29%3D0.15866)
The P-value (0.159) is higher than the significance level (0.1), so we can not reject the null hypothesis.