Fewer young people are driving. In year A, 67.9% of people under 20 years old who were eligible had a driver's license. Twenty years later in year B that percentage had dropped to 47.7%. Suppose these results are based on a random sample of 1,800 people under 20 years old who were eligible to have a driver's license in year A and again in year B.a. At 95% confidence, what is the margin of error and the interval estimate of the number of nineteen year old drivers in year A?b. At 95% confidence, what tis the margin of error and the interval estimate of the number of nineteen year old drivers in year B?c. Is the margin of error the same in parts (a) and (b)?

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aliakbar921853 aliakbar921853 · Answer 1 · 2020-02-07T18:25:13+01:00

Answer:

Case a Case b

margin of error 0.0216 0.0231

Interval estimate (0.7016 , 0.6795) (0.5031 , 0.4569)

margin of error is not same in both cases.

Step-by-step explanation:

a

At 95% confidence interval the interval estimate of number of 20 year old drivers in year A can be computed as

p' ± z [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]

= 0.68 ± 1.96 [tex]\sqrt{\frac{0.68(1-0.68)}{1800} }[/tex]

= 0.7016 , 0.6795

the margin of error can be written as

z [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]

= 1.96 [tex]\sqrt{\frac{0.68(1-0.68)}{1800} }[/tex]

= 0.0216

b

At 95% confidence interval the interval estimate of number of 20 year old drivers in year B can be computed as

p' ± z [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]

= 0.48 ± 1.96 [tex]\sqrt{\frac{0.48(1-0.48)}{1800} }[/tex]

= 0.5031 , 0.4569

the margin of error can be written as

z [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]

= 1.96 [tex]\sqrt{\frac{0.48(1-0.48)}{1800} }[/tex]

= 0.0231

c

Sample size is same in case A and B but proportion is different in both cases so margin of error is different in both cases