AnswersProbability and Statistics with Applications Honors Sem-2-FL-1210300 2024-2025Unit Test — Unit testQuestion 7
7
QuizMultiple Choice
Unit Test — Unit test
Question 7 • Probability and Statistics with Applications Honors Sem-2-FL-1210300 (2024-2025)
A local school board wants to estimate the difference in the proportion of households with school-aged children that would support starting the school year a week earlier, and the proportion of households without school-aged children that would support starting the school year a week earlier. They survey a random sample of 40 households with school-aged children about whether they would support starting the school year a week earlier, and 30 households respond yes. They survey a random sample of 45 households that do not have school-aged children, and 25 respond yes. Assuming the conditions for inference have been met, what is the 90% confidence interval for the difference in proportions of households that would support starting the school year a week earlier?Find the z-table here.
Answer
A
(0.75 minus 0.56) plus-or-minus 1.65 StartRoot StartFraction 0.75 (1 minus 0.75) Over 45 EndFraction + StartFraction 0.56 (1 minus 0.56) Over 40 EndFraction EndRoot
B
(0.75 minus 0.56) plus-or-minus 1.96 StartRoot StartFraction 0.75 (1 minus 0.75) Over 85 EndFraction + StartFraction 0.56 (1 minus 0.56) Over 85 EndFraction EndRoot
C
(0.75 minus 0.56) plus-or-minus 1.65 StartRoot StartFraction 0.75 (1 minus 0.75) Over 40 EndFraction + StartFraction 0.56 (1 minus 0.56) Over 45 EndFraction EndRoot
D
(0.75 minus 0.56) plus-or-minus 1.96 StartRoot StartFraction 0.75 (1 minus 0.75) Over 40 EndFraction + StartFraction 0.56 (1 minus 0.56) Over 45 EndFraction EndRoot