A study determined that in 2009, 38% of those in a random sample of 100 adult Americans had accounts with some type of social web site. In a similar survey conducted in 2012, 53% reported having accounts. Suppose that the 2012 figure was also based on a random sample of size 100. The 95% confidence interval for the difference in population proportions, P₂₀₁₂-P₂₀₀₉ , is: A)0.15 ± 0.088 B)0.29 ± 0.136 C)0.15 ± 0.136 D)0.29 ± 0.88 E)0.00 ± 0.136

Question

Quizplus · Accepted Answer

The correct 95% confidence interval for the difference in population proportions is calculated using the formula for the difference between two proportions. The formula involves the sample proportions (p1 and p2), the sample sizes (n1 and n2), and the standard error of the difference. The calculation results in a difference of 0.15 (53% - 38% = 15% or 0.15) with a margin of error that, based on the given options, matches the value provided in option C, which is ± 0.136. This margin of error accounts for the variability expected in the difference between two sample proportions.

A Study Determined That in 2009, 38% of Those in a Random