Question

Students at a certain school were​ surveyed, and it was estimated that 20​% of college students abstain from drinking alcohol. To estimate this proportion in your​ school, how large a random sample would you need to estimate it to within 0.08 with probability 0.95​, if before conducting the study​ (a) you are unwilling to predict the proportion value at your school and​ (b) you use the results from the surveyed school as a guideline.

Answer 1

Answer:

a) A sample size of at least 251 students is needed.

b) A sample of at least 97 students is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

(a) you are unwilling to predict the proportion value at your school

We need a sample size of at least n.

n is found when M = 0.08.

We wont predict a proportion value for the school, so we use $\pi = 0.5$

So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.08 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.08\sqrt{n} = 1.96\sqrt{0.5*0.5}$

$\sqrt{n} = \frac{1.96\sqrt{0.5*0.5}}{0.08}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.5*0.5}}{0.08})^{2}$

$n = 150.1$

Rounding up

A sample size of at least 251 is needed.

(b) you use the results from the surveyed school as a guideline.

Now we have that $\pi = 0.2$

So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.08 = 1.96\sqrt{\frac{0.2*0.8}{n}}$

$0.08\sqrt{n} = 1.96\sqrt{0.2*0.8}$

$\sqrt{n} = \frac{1.96\sqrt{0.2*0.8}}{0.08}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.2*0.8}}{0.08})^{2}$

$n = 96.04$

Rounding up

A sample of at least 97 students is needed.