Question

Skipping Lunch A nutritionist wishes to determine, within 3%, the true proportion of adults who do not eat any lunch. If he wishes to be 95% confident that his estimate contains the population proportion, how large a sample will be necessary? A previous study found that 15% of the 125 people surveyed said they did not eat lunch.

Answer 1

Answer:

A sample of at least 545 adults 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 $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$\pi = 0.15$

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 $Z = 1.96$.  

If he wishes to be 95% confident that his estimate contains the population proportion, how large a sample will be necessary?

We need a sample of at least n.

n is found when M = 0.03. So

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

$0.03 = 1.96\sqrt{\frac{0.15*0.85}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.15*0.85}$

$\sqrt{n} = \frac{1.96\sqrt{0.15*0.85}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.15*0.85}}{0.03})^{2}$

$n = 544.23$

Rounding up

A sample of at least 545 adults is needed.