Question

Fewer young people are driving. In , of people under years old who were eligible had a driver's license. Bloomberg reported that percentage had dropped to in . Suppose these results are based on a random sample of people under years old who were eligible to have a driver's license in and again in . a. At confidence, what is the margin of error and the interval estimate of the number of eligible people under years old who had a driver's license in ?

Answer 1

Complete Question

Fewer young people are driving. In 1995, 63.9% of people under years 20 old who were eligible had a driver's license. Bloomberg reported that percentage had dropped to 41.7% in 2016. Suppose these results are based on a random sample 1,200 of people under 20 years old who were eligible to have a driver's license in 1995 and again in 2016.

a. At 95% confidence, what is the margin of error and the interval estimate of the number of eligible people under 20 years old who had a driver's license in 1995?

Margin of error(to four decimal places)

Interval estimate (to four decimal places)

b. At 95% confidence, what is the margin of error and the interval estimate of the number of eligible people under 20 years old who had a driver's license in 2016?

Margin of error(to four decimal places)

Interval estimate    to   (to four decimal places)

Answer:

a

  $0.6120 < p < 0.639 + 0.6670$

b

  $0.3900 < p < 0.4440$

Step-by-step explanation:

Considering question a

   The sample proportion is 1995 is  $\^ p_1 = 0.639$

    The sample size is  $n = 1200$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

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

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as  

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

=>   $E = 1.96 * \sqrt{\frac{0.639 (1- 0.639)}{1200} }$

=>   $E = 0.027$

Generally 95% Interval estimate is mathematically represented as  

      $\^ p -E < p < \^ p +E$

=>    $0.639 -0.027 < p < 0.639 + 0.027$

=>    $0.6120 < p < 0.639 + 0.6670$

Considering question b

   The sample proportion is 1995 is  $\^ p_2 = 0.417$

    The sample size is  $n = 1200$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

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

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as  

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

=>   $E = 1.96 * \sqrt{\frac{0.417 (1- 0.417)}{1200} }$

=>   $E = 0.027$

Generally 95% Interval estimate is mathematically represented as  

      $\^ p -E < p < \^ p +E$

=>    $0.417 -0.027 < p < 0.417 + 0.027$

=>    $0.3900 < p < 0.4440$