Question

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?

Answer 1

Given Information:

Young adults who received help from parents  = 8

Total number of young adults = 40

Confidence level = 95%

Required Information:

Margin of error = ?

95% confidence interval = ?

Answer:

Margin of error = 0.124

95% Confidence Interval = (0.076, 0.324)

Step-by-step explanation:

The proportion of young adults who purchased a home in Kentucky and received help from their parents is given by

$p = 8/40\\p = 0.20\\p = 20 \%$

The margin of error is given by

$MoE = z\cdot \sqrt{\frac{p(1-p)}{n} }$

Where p is the proportion of young adults who purchased a home in Kentucky and received help from their parents and n is the total number of young adults.

The z-score corresponding to 95% confidence level is 1.96

$MoE = 1.96\cdot \sqrt{\frac{0.20(1-0.20)}{40} }\\MoE = 1.96\cdot 0.063\\MoE = 0.124$

Whereas the confidence interval is given by

$CI = p \pm MoE\\CI = 0.20 \pm 0.124\\CI = 0.20 - 0.124 \: and \: 0.20 + 0.124\\CI = 0.076 \: and \:0.324$

Therefore, we are 95% Confident that the young adults who purchased a home in Kentucky and received help from their parents is within the Interval of (0.076, 0.324)