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.

5 0

3 years ago

X² + 15x + 14 Solve this please

Yuri [45]

Answer:

(x+14)·(x+1)

Step-by-step explanation:

I just put the equation on Photomath....

6 0

3 years ago

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