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]
=> [tex]\^ p = \frac{91}{582}[/tex]
=> [tex]\^ p = 0.1564[/tex]
From the question we are told the confidence level is 90% , hence the level of significance is
[tex]\alpha = (100 - 90 ) \%[/tex]
=> [tex]\alpha = 0.10[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.645[/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 = 1.645 * \sqrt{\frac{0.1564 (1- 0.1564)}{582} } [/tex]
=> [tex]E = 0.0248 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\^ p -E < p < \^ p +E[/tex]
=> [tex]0.1564 -0.0248 < p < 0.1564 + 0.0248[/tex]
=> [tex]0.1316 < p < 0.1812[/tex]
