Answer:
The size of the sample needed = 67.65
Explanation:
Given that :
The margin of error = 10% = 0.10
The confidence level = 90% = 0.90
Level of significance = 1 - 0.90 = 0.10
At ∝ = 0.10
[tex]Z_{\alpha/2}= Z_{0.10/2} = 1.645[/tex]
Since the proportion of people that supported him/her is not given, we assumed p = 0.50
The margin of error formula can be expressed as:
[tex]M.O.E = Z_{\alpha/2} \times \sqrt{\dfrac{p(1-p)}{n}}[/tex]
[tex]0.10 = 1.645 \times \sqrt{{\dfrac{0.5(1-0.5)}{n}[/tex]
Squaring both sides; we have:
[tex]0.10^2 = 1.645^2 \times {\dfrac{0.5(1-0.5)}{n}[/tex]
[tex]0.01 = 2.706025 \times \dfrac{0.25}{n}[/tex]
[tex]n= 2.706025 \times \dfrac{0.25}{0.01 }[/tex]
n = 67.65
Therefore, the required sample size = 67.65
