Question

Fewer young people are driving. In 1995, 63.9% of people under 20 years old who were eligible had a driver's license. Bloomberg reported that percentage dropped to 41.7% in 2016. Suppose these results are based on a random sample of 1,200 people under 20 years old who were eligible to have a driver's lic 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 ir 1995? Margin of error = (to four decimal places) Interval estimate = to (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 ir 2016? Margin of error = (to four decimal places) Interval estimate = to (to four decimal places) c. Is the margin of error the same in parts (a) and (b)? - Select your answer - V Why, or why not?

Answer 1

a. To find the margin of error and interval estimate for the number of eligible people under 20 years old who had a driver's license in 1995, we can use the formula for a confidence interval for a proportion:

margin of error = zsqrt(p(1-p)/n)

where z is the z-score corresponding to the desired level of confidence (95% confidence corresponds to a z-score of 1.96), p is the sample proportion (0.639), and n is the sample size (1200).

Plugging in the values, we get:

margin of error = 1.96sqrt(0.639(1-0.639)/1200) = 0.0261 (rounded to four decimal places)

To find the interval estimate, we can use the formula:

interval estimate = p ± margin of error

Plugging in the values, we get:

interval estimate = 0.639 ± 0.0261 = (0.6129, 0.6651) (rounded to four decimal places)

b. To find the margin of error and interval estimate for the number of eligible people under 20 years old who had a driver's license in 2016, we can use the same formulas with p = 0.417 (the sample proportion for 2016) and n = 1200:

margin of error = 1.96sqrt(0.417(1-0.417)/1200) = 0.0294 (rounded to four decimal places)

interval estimate = 0.417 ± 0.0294 = (0.3876, 0.4464) (rounded to four decimal places)

c. The margin of error is not the same in parts (a) and (b) because the sample proportion for 2016 (0.417) is smaller than the sample proportion for 1995 (0.639). As a result, the standard error (which is the square root of p*(1-p)/n) is larger for the 2016 sample than for the 1995 sample, which leads to a larger margin of error for the 2016 estimate.

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