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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 ?

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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

  [tex]0.6120 <  p <  0.639 + 0.6670[/tex]

b

  [tex]0.3900 <  p < 0.4440[/tex]

Step-by-step explanation:

Considering question a

   The sample proportion is 1995 is  [tex]\^ p_1 = 0.639[/tex]

    The sample size is  [tex]n = 1200[/tex]

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

      [tex]\alpha = (100 - 95 ) \%[/tex]

=>   [tex]\alpha = 0.05[/tex]

Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is  

   [tex]Z_{\frac{\alpha }{2} } =  1.96[/tex]

Generally the margin of error is mathematically represented as  

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

=>   [tex]E =  1.96  * \sqrt{\frac{0.639 (1- 0.639)}{1200} } [/tex]

=>   [tex]E =  0.027 [/tex]

Generally 95% Interval estimate is mathematically represented as  

      [tex]\^ p -E <  p <  \^ p +E[/tex]

=>    [tex]0.639 -0.027 <  p <  0.639 + 0.027[/tex]

=>    [tex]0.6120 <  p <  0.639 + 0.6670[/tex]

Considering question b

   The sample proportion is 1995 is  [tex]\^ p_2 = 0.417[/tex]

    The sample size is  [tex]n = 1200[/tex]

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

      [tex]\alpha = (100 - 95 ) \%[/tex]

=>   [tex]\alpha = 0.05[/tex]

Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is  

   [tex]Z_{\frac{\alpha }{2} } =  1.96[/tex]

Generally the margin of error is mathematically represented as  

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

=>   [tex]E =  1.96  * \sqrt{\frac{0.417 (1- 0.417)}{1200} } [/tex]

=>   [tex]E =  0.027 [/tex]

Generally 95% Interval estimate is mathematically represented as  

      [tex]\^ p -E <  p <  \^ p +E[/tex]

=>    [tex]0.417 -0.027 <  p <  0.417 + 0.027[/tex]

=>    [tex]0.3900 <  p <   0.4440[/tex]

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