Question

A researcher wants to construct a 90% confidence interval for the proportion of elementary school students in Seward County who receive free or reduced-price school lunches. A state-wide survey indicates that the proportion is 0.50. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.1?

Answer 1

Answer:

The sample  size is  $n =68$

Step-by-step explanation:

    The population proportion is  $\^ p = 0.50$

     The margin of error is  $E = 0.1$

From the question we are told the confidence level is  90% , hence the level of significance is    

      $\alpha = (100 - 90 ) \%$

=>   $\alpha = 0.10$

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

   $Z_{\frac{\alpha }{2} } =  1.645$

Generally the sample size is mathematically represented as  

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

=>  $n = [\frac{1.645 }{0.1} ]^2 * 0.50  (1 - 0.50 )$

=>  $n =68$