Question

Exercise 6.11 presents the results of a poll evaluating support for a generic "National Health Plan" in the US in 2019, reporting that 55% of Independents are supportive. If we wanted to estimate this number to within 1% with 90% confidence, what would be an appropriate sample size?

Answer 1

Answer:

The sample size is   $n =6697$

Step-by-step explanation:

From the question we are told that

    The  population proportion is  p =  0.55

    The  margin of error is  E  =  0.01

 

Given that the confidence level is  then the level of significance is mathematically represented as

      $\alpha = (100 -90)\%$

=>   $\alpha = 0.10$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.645$

Gnerally the sample size is mathematically represented as

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

=>   $n = \frac{1.645^2 * 0.55(1-0.55)}{0.01^2}$

=>    $n = \frac{1.645^2 * 0.55(1-0.55)}{0.01^2}$

=>    $n =6697$