Given:
ME = 1.5% = 0.015
P = 0.5 ( maximises P)
Answer and Step-by-step explanation:
Now,
Margin of Error (ME) is given by the formula:
[tex]ME = t\times \ sigma[/tex]
where,
t = critical value = 1.645
[tex]\sigma[/tex] = standard deviation
So, squaring both sides and then rearranging:
[tex]ME^{2} = t^{2}\times {\frac{P(1 - P)}{n}}[/tex]
[tex]n = t^{2}\times {\frac{P(1 - P)}{ME^{2}}}[/tex]
n = (1.645)^{2}\times {\frac{0.5(1 - 0.5)}{0.015^{2}}}
n = 3006.69
n = 3007 (approx)
