Question

According to a poll taken last year, 45% of the cities' residents get most of their news from the Internet. To conduct a follow-up study that would provide 90% confidence that the point estimate is correct to within 0.04 of the population proportion, how large a sample size is required

Answer 1

Answer:

The sample size is  $n =419$

Step-by-step explanation:

From the question we are told that

    The population proportion is  $p = 0.45$

     The  margin of error is  $E = 0.04$

Given that the confidence level is  90%

  Then the level of significance can be mathematically represented as

          $\alpha = 100 -90$

          $\alpha = 10\%$

         $\alpha = 0.10$

Next we obtain the level of significance from the normal distribution table the value is    

             $Z_{\frac{\alpha }{2} } = 1.645$

   Generally the sample size is mathematically represented as  

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

substituting values

         $n = [ \frac{1.645 }{0.04} ]^2 * 0.45(1- 0.45 )$

        $n =419$