Respuesta :
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]
