Question

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid.
a) Use a 90% confidence interval to estimate the true proportion of students who receive financial aid. Explain/show how you obtain your answer.

b) If the dean wanted to estimate the proportion of all students receiving financial aid to within 3% with 99% reliability, how many students would need to be sampled? Explain/show how you obtain your answer.

Answer 1

Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     $0.533 < p < 0.64$

b

   $n = 1789$

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is $k = 118$

Generally the sample proportion is  

      $\^ p = \frac{k}{n}$

=>   $\^ p = \frac{118}{200}$

=>   $\^ p = 0.59$

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 margin of error is mathematically represented as  

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

 =>$E = 1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }$

=>  $E = 0.057$

Generally 90% confidence interval is mathematically represented as  

      $\^ p -E < p < \^ p +E$

  =>  $0.533 < p < 0.64$  

Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      $\alpha = (100 - 99 ) \%$

=>   $\alpha = 0.01$

Generally from the normal distribution table the critical value  of   is  

   $Z_{\frac{\alpha }{2} } = 2.58$

Generally the sample size is mathematically represented as      

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

=>      $n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )$

=>      $n = 1789$