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 6% margin of error at a 90% confidence level, what size of sample is needed? Be sure to round accordingly.

Answer 1

Answer: The size of sample of 193 is needed.

Step-by-step explanation:

Since we have given that

Margin of error = 6%

At 90% confidence level, z = 1.645

Let p = 0.5

and q = 0.5

( because it is the most conservative estimation).

So, using formula, we get :

$e=z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\0.06= 1.645\times \sqrt{\dfrac{0.5\times 0.5}{n}}\\\\\dfrac{0.06}{1.645}=\sqrt{\dfrac{0.25}{n}}\\\\(0.036)^2=\dfrac{0.25}{n}\\\\n=\dfrac{0.25}{0.036^2}\\\\n=192.9\\\\n=193$

Hence, the size of sample of 193 is needed.