Question

Fewer young people are driving. In 1983, of -year-olds had a driver's license. Twenty-five years later that percentage had dropped to (University of Michigan Transportation Research Institute website, April 7, 2012). Suppose these results are based on a random sample of -year-olds in 1983 and again in 2008. a. At confidence, what is the margin of error and the interval estimate of the number of -year-old drivers in 1983

Answer 1

Answer:

Hence the values for MOE and are interval estimate of 19 yr old age drivers in 1983  are 1%(0.01) and 19±0.0196

Step-by-step explanation:

Given: Refer the paper university of Michigan transportation research  Institute website ,April,2012.

It say that 19 year old group age people ,about 87% has driving license .

And of 1200 has a random sample space ,in 1983.

To Find : MOE(margin of error ) and   Interval Estimate  in 1983.

Solution:

Given that ,there is sample space of 1200 19 yr old people  and of which 87% has driving license.

hence , we get that

87% of 1200 = sample  size.

sample size =1044 members has driving license out of 1200.

Consider 95 % of confidence level,

for that Z-score is required.

calculating the Alpha for that=1-confidence level.

=1-0.95=0.05

therefore Z-alpha=Z(0.05)=1.96

1)MOE=margin of error is given by ,

=$Z-alpha*\sqrt{\frac{p(1-p)}{N} }$

here p=0.87 and N=total size =1200.

MOE=1.96*$\sqrt{\frac{0.87(1-0.87)}{1200} }$  =1 %.

2) Interval estimate is given by ,

For that we should know mean,standard deviation and sample size,

we are calculating for 19 year old age of entire age of population,

Hence mean will be 19 yr-old age.

Mean=19.

Sample size=1044.

standard deviation given by,

=$p*(\sqrt{(1-p)}$

=0.87*$\sqrt{0.13}$=0.3136.

Hence now calculating the interval estimate ,with 95% confidence level,

μ=M±Z(Standard error )

standard error=standard deviation/sqrt(sample size)

=0.3136/32.31=0.01.

μ=19±1.96*0.01

μ=19±0.0196.