Question

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?

Answer 1

Answer: the answer to this question is B

Step-by-step explanation: Hope This Helps

Answer 2

Answer with explanation:

Formula to find the margin of error :

`E=z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}` , where n= sample size ,
`\hat{p}` 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

`\hat{p}=0.87`

n= 1200

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

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

Interval :
`(\hat{p}-E , \ \hat{p}+E)=(0.87-0.019 ,\ 0.87+0.019)`

`=(0.851,\ 0.889)`

B) In 2008 ,

`\hat{p}=0.75`

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

Interval :
`(\hat{p}-E, \hat{p}+E)=(0.75-0.0245,\ 0.75+0.0245)`

`=(0.7255,\ 0.7745)`

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.