According to a report in USA Today (February 1,2012),more and more parents are helping their young adult children buy homes.You would like to construct a 90% confidence interval of the proportion of all young adults in Louisville,Kentucky,who got help from their parents in buying a home.How large a sample should you take so that the margin of error is no more than 0.06?
A)n = 114
B)n = 188
C)n = 267
D)Cannot be determined

Question

Quizplus · Accepted Answer

We need to find the sample size to construct a 90% confidence interval with a margin of error no more than 0.06.

Using the formula for margin of error:

Margin of error = z*sqrt(p*(1-p)/n)

We need to find n.

The critical value for a 90% confidence interval is 1.645.

We don't know the value of p, so we use 0.5 as a conservative estimate to get the largest sample size:

0.06 = 1.645*sqrt(0.5*(1-0.5)/n)

Solving for n:

n = (1.645/0.06)^2 * 0.5*(1-0.5)

n = 187.8

Rounding up to the nearest whole number, we get

n = 188

Therefore, the best choice is B.

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