Mathematics, 02.12.2020 17:10 shambrayperkins

Fewer young people are driving. In , of people under years old who were eligible had a driver's license. Bloomberg reported that percentage had dropped to in . Suppose these results are based on a random sample of people under years old who were eligible to have a driver's license in and again in . a. At confidence, what is the margin of error and the interval estimate of the number of eligible people under years old who had a driver's license in ?

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Answer from: perfect32

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)

0.6120 < p < 0.639 + 0.6670

0.3900 < p < 0.4440

Step-by-step explanation:

Considering question a

The sample proportion is 1995 is $\^ p_1 = 0.639$

The sample size is n = 1200

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{0.639 (1- 0.639)}{1200} }$

=> E = 0.027

Generally 95% Interval estimate is mathematically represented as

$\^ p -E < p < \^ p +E$

=> 0.639 -0.027 < p < 0.639 + 0.027

=> 0.6120 < p < 0.639 + 0.6670

Considering question b

The sample proportion is 1995 is $\^ p_2 = 0.417$

The sample size is n = 1200

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{0.417 (1- 0.417)}{1200} }$

=> E = 0.027

Generally 95% Interval estimate is mathematically represented as

$\^ p -E < p < \^ p +E$

=> 0.417 -0.027 < p < 0.417 + 0.027

=> 0.3900 < p < 0.4440

Answer from: Quest

Do the inverse operation

Answer from: Quest

Ithink the right answer could be 50%

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