At 95% confidence, what is the margin of error of the number of eligible people under 20 years old who had a driver's license in year A? (Round your answer to four decimal places.

(0.6231 , 0.6749)

Step-by-step explanation:

With the information we have, it is impossible to solve the exercise, therefore I was looking for information to complete it and we have to:

the sample proportion is 64.9%, or 0.649 plus the sample size is 1300 (n)

Now, we have that the standard error is given by:

SE = (p * (1 - p) / n) ^ (1/2)

replacing

SE = (0.649 * (1 - 0.649) / 1300) ^ (1/2)

SE = 0.0132

Now we have that confidence level is 95%, hence α = 1 - 0.95 = 0.05

α / 2 = 0.05 / 2 = 0.025, Zc = Z (α / 2) = 1.96

With this we can calculate margin of error like so:

ME = z * SE

ME = 1.96 * 0.0132

ME = 0.0259

Finally the interval would be:

CI = (p - ME, p + ME)

CI = (0.649 - 0.0259, 0.649 + 0.0259)

CI = (0.6231, 0.6749)

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x