Question

1. a national health organization warns that 30% of the middle school students nationwide have been drunk. concerned, a local health agency randomly and anonymously surveys 110 of the middle 1212 middle school students in its city. only 21 of them report having been drunk.

a) what proportion of the sample reported having been drunk?
b) does this mean that this city’s youth are not drinking as much as the national data would indicate?
c) create a 95% confidence interval for the proportion of the city’s middle school students who have been drunk.
d) is there any reason to believe that the national level of 30% is not true of the middle school students in the city?
e) to keep the margin of error at most 5%, how many middle school students do we need to survey?

Explanation:

Step1: Calculate sample proportion

The formula for sample proportion $\hat{p}$ is $\hat{p}=\frac{x}{n}$, where $x$ is the number of successes (students who reported being drunk) and $n$ is the sample size. Here, $x = 21$ and $n=110$. So, $\hat{p}=\frac{21}{110}\approx0.191$.

Step2: Analyze if city's youth drink less

A single - sample proportion is just an estimate. We need to consider sampling variability. A confidence interval (step3) will give a range of plausible values. Just because the sample proportion ($\hat{p}\approx0.191$) is less than the national proportion ($p = 0.3$) does not immediately mean the city's youth drink less.

Step3: Create 95% confidence interval

The formula for a confidence interval for a proportion is $\hat{p}\pm z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. For a 95% confidence interval, $z = 1.96$.
First, calculate the standard error $SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.191\times(1 - 0.191)}{110}}\approx\sqrt{\frac{0.191\times0.809}{110}}\approx\sqrt{\frac{0.154}{110}}\approx0.0375$.
Then the confidence interval is $0.191\pm1.96\times0.0375$.
Lower limit: $0.191-1.96\times0.0375=0.191 - 0.073=0.118$.
Upper limit: $0.191 + 1.96\times0.0375=0.191+0.073 = 0.264$. So the 95% confidence interval is $(0.118,0.264)$.

Step4: Check if national level is not true

Since the national proportion ($p = 0.3$) is not in the 95% confidence interval $(0.118,0.264)$, there is evidence to suggest that the national level of 30% is not true for the city's middle - school students.

Step5: Calculate sample size for margin of error

The formula for sample size $n$ is $n=\frac{z^{2}\hat{p}(1 - \hat{p})}{ME^{2}}$. When no prior estimate ($\hat{p}$) is available, we use $\hat{p}=0.5$ (maximizes $\hat{p}(1 - \hat{p})$). $z = 1.96$ (for 95% confidence) and $ME = 0.05$.
$n=\frac{1.96^{2}\times0.5\times(1 - 0.5)}{0.05^{2}}=\frac{3.8416\times0.25}{0.0025}=\frac{0.9604}{0.0025}=384.16$. Since we can't have a fraction of a sample, $n = 385$.

Answer:

a) The proportion of the sample that reported having been drunk is approximately $0.191$.
b) No, a single - sample proportion does not account for sampling variability. We need to consider a confidence interval.
c) The 95% confidence interval is $(0.118,0.264)$.
d) Yes, since the national proportion ($30\%$ or $0.3$) is not in the 95% confidence interval.
e) We need to survey $385$ middle - school students.