According to a report in USAToday (February 1,2012),more and more parents are helping their young adult children get homes.Suppose eight persons in a random sample of 40 young adults who recently purchased a home in Kentucky received help from their parents.You have been asked to construct a 95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents.What is the margin of error for a 95% confidence interval for the population proportion?
A) 1.645(0.0040)
B) 1.645(0.0632)
C) 1.96(0.0040)
D) 1.96(0.0632)

Question

Quizplus · Accepted Answer

We are asked to construct a confidence interval for the population proportion, so we can use the formula:

margin of error = z*(standard error)

where z is the z-score corresponding to the desired confidence level (95% corresponds to 1.96), and the standard error is given by:

standard error = sqrt((p*(1-p))/n)

where p is the sample proportion (8/40 = 0.2) and n is the sample size (40).

Substituting the values, we get:

standard error = sqrt((0.2*(1-0.2))/40) = 0.0632

Thus, the margin of error is:

margin of error = 1.96*0.0632 = 0.124032

Therefore, the best choice is D, 1.96(0.0632), which matches our calculation.

According to a Report in USAToday (February 1,2012),more and More