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

972 junior executives should be surveyed.

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is of:

`M = z\sqrt{(\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 - (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{(\pi(1-\pi))/(n)}`

`0.03 = 1.96\sqrt{(0.35*0.65)/(n)}`

`0.03√(n) = 1.96√(0.35*0.65)`

`√(n) = (1.96√(0.35*0.65))/(0.03)`

`(√(n))^2 = ((1.96√(0.35*0.65))/(0.03))^2`

`n = 971.2`

Rounding up:

972 junior executives should be surveyed.