Question

How large a sample is needed if we wish to be 96% confident that our sample proportion in Exercise 9.53 will be within 0.02 of the true fraction of the voting population?
A. A random sample of 200 voters in a town is selected, and 114 are found to support an annexation suit. Find the 96% confidence interval for the fraction of the voting population favoring the suit.
B. What can we assert with 96% confidence about the possible size of our error if we estimate the fraction of voters favoring the annexation suit to be 0.57?

Answer 1

Complete Question

How large a sample is needed if we wish to be 96% confident that our sample proportion in Exercise 9.53 will be within 0.02 of the true fraction of the voting population?

Exercise 9.53

A. A random sample of 200 voters in a town is selected, and 114 are found to support an annexation suit. Find the 96% confidence interval for the fraction of the voting population favoring the suit.

B. What can we assert with 96% confidence about the possible size of our error if we estimate the fraction of voters favoring the annexation suit to be 0.57?

Answer:

The First question

                    $n = 1879  $

A

    The 96% confidence interval is  

                      $0.51 <  p <  0.63$

B

   The possible size of our error if we estimate the fraction of voters favoring the annexation suit to be 0.57 is  

           $E =    0.06$

Step-by-step explanation:

From the question we are told that

    The margin of error is  $E = 0.02$

     The sample size is  n =  200

     The number that supported the annexation suit is  k = 114

Considering the first question

Generally our sample in Exercise 9.53 is  $\^ p = 0.57$

From the question we are told the confidence level is  96% , hence the level of significance is    

      $\alpha = (100 - 96 ) \%$

=>   $\alpha = 0.04$

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

   $Z_{\frac{\alpha }{2} } =  1.751$

Generally the sample size is mathematically represented as  

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

=>  $n = [\frac{1.751}{0.02 } ]^2 * \^ p (1 - \^ p )$

=>  $n = [\frac{1.751}{0.02 } ]^2 * 0.57  (1 -0.57  )$

=>  $n = 1879  $

Considering question B

Generally the margin of error is mathematically represented as  

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

=>  $E =    1.751 * \sqrt{\frac{ 0.57  (1- 0.57 )}{200} }$

=>  $E =    0.06$

Considering question A

Generally the sample proportion of the number that supported the annexation suit is mathematically represented as

        $\^ p = \frac{114}{200}$

=>      $\^ p = 0.57$

Generally 96% confidence interval is mathematically represented as  

      $\^ p -E <  p <  \^ p +E$

=>  $0.57  - 0.06 <  p <  0.57  + 0.06$

=>  $0.51 <  p <  0.63$