Answer: 709
Step-by-step explanation:
The formulas we use to find the required sample size :-
1. [tex]n=(\dfrac{z^*\cdot \sigma}{E})^2[/tex]
, where [tex]\sigma[/tex] = population standard deviation,
E = Margin of error .
z* = Critical value
2. [tex]n=p(1-p)(\dfrac{z^*}{E})^2[/tex] , where p= prior estimate of population proportion.
3. If prior estimate of population proportion is unavailable , then we take p= 0.5 and the formula becomes
[tex]n=0.25(\dfrac{z^*}{E})^2[/tex]
Given : Margin of error : E= 3% =0.03
Critical value for 95% confidence interval = z*= 1.96
A study conducted several years ago revealed that the percent of junior executives leaving within three years was 21%.
i.e. p=0.21
Then by formula 2., the required sample size will be :
[tex]n=0.21(1-0.21)(\dfrac{1.96}{0.03})^2[/tex]
[tex]n=0.21(0.79)(65.3333)^2[/tex]
[tex]n=(0.1659)(4268.44008889)\\\\ n=708.134933333\approx709[/tex] [Round to the next integer.]
Hence, the required sample size of junior executives should be studied = 709
