Question

In: Statistics and Probability

Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver’s license. Twentyfive years...

Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver’s license. Twentyfive years later that percentage had dropped to 75%. Suppose the results are based on a random sample of 1200 19-year-olds in 1983 and again in 2008, and we are interested in estimating the population proportion of 19-year-old drivers in 2008.
a. At 95% confidence, what is the margin of error?
b. Develop a 95% confidence interval for the proportion of 19-year-old drivers in 2008.

Expert Solution

Solution :

Given that,

n = 1200

Point estimate = sample proportion = = 75% = 0.75

1 - = 1 - 0.75 = 0.25

a) At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z_/2 = 1.96

Margin of error = E = Z_{/ 2} * $\sqrt$ (( * (1 - )) / n)

= 1.96 * ( $\sqrt$ ((0.75 * 0.25 ) / 1200 )

= 0.025

b) A 95% confidence interval for population proportion p is ,

- E < p < + E

0.75 - 0.025 < p < 0.75 + 0.025

(0.725 < p < 0.775)

orchestra answered 2 years ago

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