Answer:
96
Step-by-step explanation:
From the given information:
At 95% Confidence interval level,Level of significance [tex]\alpha[/tex] 0.05, the value of Z from the standard normal tables = 1.96
Margin of Error = 0.10
Let assume that the estimated proportion = 0.5
therefore; the sample size n can be determined by using the formula: [tex]n =(\dfrac{Z}{E})^2 \times p\times (1-p)[/tex]
[tex]n =(\dfrac{1.96}{0.1})^2 \times 0.5\times (1-0.5)[/tex]
[tex]n =(19.6)^2 \times 0.5\times (0.5)[/tex]
n = 96.04
n [tex]\approx[/tex] 96
