Answer:
The size of the sample = 1088.90
Step-by-step explanation:
Given that:
The confidence interval level = 0.95
The margin of error = 0.03
Suppose the proportion of the boys and the girls are p₁ and p₂ respectively.
The Standard error S.E of difference is:
[tex]S.E (p_1-P_2) = \sqrt{ \dfrac{p_1(1-p_1) }{n} + \dfrac{p_2(1-p_2)}{n} }[/tex]
Assume p₁ = p₂ = 0.5
[tex]S.E (p_1-P_2) = \sqrt{ \dfrac{0.5(1-0.5) }{n} + \dfrac{0.5(1-0.5)}{n} }[/tex]
[tex]S.E (p_1-P_2) =\dfrac{0.70711}{\sqrt{n}}[/tex]
The z-critical value at 95% C.I = 1.96
Margin of error = Z_{critical} × S.E
∴
[tex]0.03 = 1.96 \times \dfrac{0.70711}{\sqrt{n}}[/tex]
[tex]n =\dfrac{\left(1.96\ \cdot\ 0.70711^{2}\right)}{0.03^{2}}[/tex]
n ≅ 1088.90
