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Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver’s license. Twenty-five years later that percentage had dropped to 75% (University of Michigan Transportation Research Institute website, April 7, 2012). Suppose these results are based on a random sample of 1200 19-year-olds in 1983 and again in 2008.

a. At 95% confidence, what is the margin of error and the interval estimate of the number of 19-year-old drivers in 1983?
b. At 95% confidence, what is the margin of error and the interval estimate of the number of 19-year-old drivers in 2008?
c. Is the margin of error the same in parts (a) and (b)? Why or why not?

Respuesta :

contdai0955

Answer: the answer to this question is B

Step-by-step explanation: Hope This Helps

JeanaShupp

Answer with explanation:

Formula to find the margin of error :

[tex]E=z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] , where n= sample size  , [tex]\hat{p}[/tex] is the sample proportion and z*= critical z-value.

Let p be the proportion of 19-year-olds had a driver’s license.

A) As per given , In 1983

[tex]\hat{p}=0.87[/tex]

n=  1200

Critical value for 95% confidence level is 1.96 (By z-table)

So ,Margin of error : [tex]E=(1.96)\sqrt{\dfrac{0.87(1-0.87)}{1200}}\approx0.019[/tex]

Interval : [tex](\hat{p}-E , \ \hat{p}+E)=(0.87-0.019 ,\ 0.87+0.019)[/tex]

[tex]=(0.851,\ 0.889)[/tex]

B) In 2008 ,

[tex]\hat{p}=0.75[/tex]

Margin of error :    [tex]E=(1.96)\sqrt{\dfrac{0.75(1-0.75)}{1200}}\approx0.0245[/tex]

Interval : [tex](\hat{p}-E, \hat{p}+E)=(0.75-0.0245,\ 0.75+0.0245)[/tex]

[tex]=(0.7255,\ 0.7745)[/tex]

c. The margin of error is not the same in parts (a) and (b) because the sample proportion of 19-year-olds had a driver’s license are not same in both parts.

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