Respuesta :
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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
35% of junior executives left their company within three years.
This means that [tex]\pi = 0.35[/tex]
0.95 = 95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
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
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.35*0.65}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96\sqrt{0.35*0.65}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.35*0.65}}{0.03}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.35*0.65}}{0.03})^2[/tex]
[tex]n = 971.2[/tex]
Rounding up:
972 junior executives should be surveyed.
