Question

A research firm needs to estimate within 3% the proportion of junior executives leaving large manufacturing companies within three years. A 0.95 degree of confidence is to be used. Several years ago, a study revealed that 35% of junior executives left their company within three years. To update this study, how many junior executives should be surveyed

Answer 1

Answer:

972 junior executives should be surveyed.

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 of:

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

35% of junior executives left their company within three years.

This means that $\pi = 0.35$

0.95 = 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 $Z = 1.96$.

To update this study, how many junior executives should be surveyed?

Within 3% of the proportion, which means that this is n for which M = 0.03. So

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

$0.03 = 1.96\sqrt{\frac{0.35*0.65}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.35*0.65}$

$\sqrt{n} = \frac{1.96\sqrt{0.35*0.65}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.35*0.65}}{0.03})^2$

$n = 971.2$

Rounding up:

972 junior executives should be surveyed.