Question

a political candidate has asked you to conduct a poll to determine what percentage of people support her. if the candidate only wants a 8% margin of error at a 95% cnofidence level, what size of sample is needed

Answer 1

Answer:

The  sample needed is  $n =150$

Step-by-step explanation:

From the question we are told that

   The margin of error is  $E = 0.08$

   The confidence level is  $C = 95 \% = 0.95$

Given that the confidence level is  95% the level of significance is mathematically represented as

          $\alpha = 1 - 0.95$

           $\alpha = 0.05$

Next we obtain the critical value of  $\frac{\alpha }{2}$ from the z-table , the value is $Z_{\frac{\alpha }{2} } = 1.96$

The reason we are obtaining critical value of    $\frac{\alpha }{2}$ instead of    $\alpha$ is because  

$\alpha$ represents the area under the normal curve where the confidence level interval (   $1-\alpha$ ) did not cover which include both the left and right tail while  

$\frac{\alpha }{2}$ is just the area of one tail which what we required to calculate the margin of error

The sample size is mathematically represented as    

            $n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \r p[1-\r p]$

Here $\r p$ is sample proportion of people that supported her and we will assume this to be 50% =  0.5

So

            $n = [\frac{1.96}{ 0.08} ]^2 * [0.5 (1- 0.5)]$

           $n =150$