Question

A political candidate has asked his/her assistant to conduct a poll to determine the percentage of people in the community that supports him/her. If the candidate wants a 10% margin of error at a 90% confidence level, what size of sample is needed? Be sure to round accordingly. The candidate would need to survey people in the community in order to be within a 10% margin of error at a 90% confidence level.

Answer 1

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

$Z_{\alpha/2}= Z_{0.10/2} = 1.645$

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:

$M.O.E = Z_{\alpha/2} \times \sqrt{\dfrac{p(1-p)}{n}}$

$0.10 = 1.645 \times \sqrt{{\dfrac{0.5(1-0.5)}{n}$

Squaring both sides; we have:

$0.10^2 = 1.645^2 \times {\dfrac{0.5(1-0.5)}{n}$

$0.01 = 2.706025 \times \dfrac{0.25}{n}$

$n= 2.706025 \times \dfrac{0.25}{0.01 }$

n = 67.65

Therefore, the required sample size = 67.65