In a recent poll of 500 13-year-olds, many indicated to enjoy their relationships with their parents. Suppose that 200 of the 13-year olds were boys and 300 of them were girls. We wish to estimate the difference in proportions of 13-year old boys and girls who say that their parents are very involved in their lives. In the sample, 93 boys and 172 girls said that their parents are very involved in their lives. What is a 96% confidence interval for the difference in proportions (proportion of boys minus proportion of girls)?
A) (-0.2015, -0.0151)
B) (-0.1973, -0.0194)
C) (-0.1978, -0.0289)
D) None of the above

Question

Quizplus · Accepted Answer

The correct answer is A because the 96% confidence interval for the difference in proportions is (-0.2015, -0.0151). This interval is calculated using the formula:CI = (p1 - p2) +/- z * sqrt((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))where p1 is the proportion of boys who say that their parents are very involved in their lives, p2 is the proportion of girls who say that their parents are very involved in their lives, n1 is the sample size of boys, and n2 is the sample size of girls.Plugging in the values from the problem, we get:CI = (0.93 - 0.172) +/- 2.05 * sqrt((0.93 * (1 - 0.93) / 200) + (0.172 * (1 - 0.172) / 300))= (-0.2015, -0.0151)

In a Recent Poll of 500 13-Year-Olds, Many Indicated to Enjoy