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29 April, 15:02

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.

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  1. 29 April, 15:18
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    (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)
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