Fewer young people are driving. In 1995, 63.9% 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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jordanbb6698 jordanbb6698

04-03-2021
Mathematics

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Fewer young people are driving. In 1995, 63.9% 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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abiolataiwo2015 abiolataiwo2015 · Answer 1 · 2021-03-05T19:29:41+01:00

Answer:

A. 0.0272

B. 0.0278

C. No

Step-by-step explanation:

a. Calculation for the margin of error and the interval estimate of the number of eligible people who had a driver’s license in 1995u

First step is to calculate the Standard error using this formula

Standard error(SE)= √(pcap * (1 - pcap)/n)

Where,

pcap represent sample proportion = 0.639

n represent sample size = 1200

Let plug in the formula

Standard error (SE)= √(0.639 * (1 - 0.639)/1200)

Standard error(SE)= 0.0139

Second step is to find Zc using this formula

Zc = Z(α/2)

Where,

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

Zc = Z(α/2) = 1.96

Now let calculate the Margin of Error using this formula

Margin of Error= zc * SE

Let plug in the formula

Margin of Error = 1.96 * 0.0139

Margin of Error = 0.0272

Therefore the Margin of Error is 0.0272

b. Calculation for the margin of error and the interval estimate of the number of eligible people who had a driver’s license in 2016

First step is to calculate the Standard error using this formula

Standard error(SE)= √(pcap * (1 - pcap)/n)

Where,

pcap represent sample proportion = 0.417

n represent sample size = 1200

Let plug in the formula

Standard error (SE)= √(0.417 * (1 - 0.417)/1200)

Standard error(SE)= 0.0142

Second step is to find Zc using this formula

Zc = Z(α/2)

Where,

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

Zc = Z(α/2) = 1.96

Now let calculate the Margin of Error using this formula

Margin of Error= zc * SE

Let plug in the formula

Margin of Error = 1.96 * 0.0142

Margin of Error = 0.0278

Therefore the Margin of Error is 0.0278

c. Based on the above calculation for both (a) and (b) the MARGIN OF ERROR are NOT the same reason been that (a) Margin of Error is 0.0272 while (b) Margin of Error is 0.0278 which simply shows that the proportion in parts (a) and are different.