Question

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 18 students, she finds 2 who eat cauliflower. Obtain and interpret a 95​%

confidence interval for the proportion of students who eat cauliflower on​ Jane's campus using Agresti and​ Coull's method.

One is 95​% confident that the proportion of students who eat cauliflower on​ Jane's campus is between _____ and ______.

Answer 1

ause WAWA Solusioni given that x=2 neig i. Þ::3 aoill LP - 0.89 1. The g54. confidence level, from tabie we know that. 2.0.05 & 2 = 0.025 Zaza 1.96 TEC - Zdroj Baclof) {P{ € + Zip LP SCI-P) foull - 1.969 109110:39 <p< 0'11+1.907046009 n Coul - 1.96 0.0 737 < P oil! + 1.96* 0'0737) 2 - Coll - 0.1445 & P olltolu us) LA - alo.0345 < p < 0.2545) . - The Jane wants 15 goy..confident that the population Propoetien P. of Students who eat cauliflower on. Jane's campus is between 0.0345 and 0.2545