Answer: 1068
Step-by-step explanation:
[tex]n= p(1-p)(\dfrac{z_c}{E})^2[/tex] , where p is the prior population proportion , E is the margin of error and [tex]z_c[/tex] is the z-value at a certain confidence level.
As per given , we have
Margin of error at 95% confidence interval : E= half width of the confidence interval = 0.03
z-value for 95% confidence interval : [tex]z_c=1.96[/tex] [using z-value table.]
Since no prior population proportion is given , so we take p= 0.5
[∵ the half-width of a CI for a population proportion is maximized when the sample proportion is set to p = 0.5. ]
Then , the sample size would be :-
[tex]n= 0.5(1-0.5)(\dfrac{1.96}{0.03})^2=1067.11111111\approx1068[/tex] [Rounded to the next integer.]
∴ They must sample 1068 individuals.