The proportion of junior executives leaving large manufacturing companies within three years is to be estimated within 3 percent. the 0.95 degree of confidence is to be used. a study conducted several years ago revealed that the percent of junior executives leaving within three years was 21. to update this study, the files of how many junior executives should be studied?

709

Step-by-step explanation:

Sample size for an Interval estimate of population proportion is

n = (zα/2)^2 p (1-p) / E^2

given:

proportion, p = 0.21

margin of error, E = 0.03

Confidence level of 95%, that means the the siginficance level α is 1 – p

α = 1 – 0.95 = 0.05

Z(α /2) = Z(0.05/2) = Z (0.025)

Using a z table Z = 1.96

n = (zα/2)^2 p (1-p) / E^2

n = 1.96^2*0.21 (1-.21)/0.03^2

n = 708.13 ≈ 709

sorry, i can't understand your question. will later if you describe more clearly!

: )