Answer:
The 90% confidence interval is  [tex]0.1316 <  p <  0.1812[/tex]
Step-by-step explanation:
From the question we are told that
  The sample size is  n  =  582
  The number of teenagers at the wheel is k = 91
Generally the sample proportion is mathematically represented as
     [tex]\^ p = \frac{k}{n}[/tex]
=> Â Â Â [tex]\^ p = \frac{91}{582}[/tex]
=> Â Â Â [tex]\^ p = 0.1564[/tex]
From the question we are told the confidence level is  90% , hence the level of significance is  Â
   [tex]\alpha = (100 - 90 ) \%[/tex]
=> Â [tex]\alpha = 0.10[/tex]
Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is Â
  [tex]Z_{\frac{\alpha }{2} } =  1.645[/tex]
Generally the margin of error is mathematically represented as Â
   [tex]E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} } [/tex]
=> Â [tex]E = Â 1.645 * \sqrt{\frac{0.1564 Â (1- 0.1564)}{582} } [/tex]
=> Â [tex]E = Â 0.0248 [/tex]
Generally 95% confidence interval is mathematically represented as Â
   [tex]\^ p -E <  p <  \^ p +E[/tex]
=> Â [tex]0.1564 Â -0.0248 < Â p < Â 0.1564 Â + 0.0248[/tex]
=> Â [tex]0.1316 < Â p < Â 0.1812[/tex]
