Final answer:
To calculate the margin of error and the interval estimate for individuals under 20 with a driver's license in 1995 from a sample size of 1200, we determined the margin of error to be ±36 individuals, with an interval estimate ranging from approximately 730 to 802 individuals at a 95% confidence level.
Step-by-step explanation:
To calculate the margin of error and the interval estimate of the number of eligible people under 20 years old who had a driver's license in 1995, we will use the given sample proportion and sample size. The sample proportion is 63.9%, and the sample size is 1200. First, we find the standard error (SE) for the proportion using the formula SE = √p(p(1 - p)/n), where p is the sample proportion and n is the sample size.
For a sample proportion of 63.9%, or 0.639, and a sample size of 1200:
SE = √(0.639(1 - 0.639)/1200) = √(0.639 * 0.361 / 1200) = √(0.000230319) ≈ 0.0152
To find the margin of error (ME) at a 95% confidence level, we use the Z-score associated with 95% confidence, which is approximately 1.96.
ME = Z * SE = 1.96 * 0.0152 ≈ 0.0298 or 2.98%
To convert the margin of error to the number of individuals, we multiply by the sample size:
Number ME = 1200 * 0.0298 = 35.76
We can approximate this to 36 individuals. Thus, at 95% confidence, the margin of error in terms of individuals is ±36.
The interval estimate is calculated by taking the sample proportion and adding and subtracting the margin of error. In percentage form:
Lower limit = 63.9% - 2.98% = 60.92%
Upper limit = 63.9% + 2.98% = 66.88%
Converting these percentages into the number of individuals:
Lower limit = 1200 * 60.92% ≈ 730 individuals
Upper limit = 1200 * 66.88% ≈ 802 individuals
Therefore, the interval estimate for the number of eligible individuals under 20 years old who had a driver's license in 1995 is approximately between 730 and 802, with a margin of error of ±36 individuals based on a random sample of 1200.