Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 27 students, she finds 3 who eat cauliflower. Obtain and interpret a 90% confidence interval for the proportion of students who eat cauliflower on Jane’s campus using Agresti and Coull’s method.


  1. ANSWER- BETWEEN 9.38% and 12.9%
    Step-by-step explanation:
    Let X be the number of students who eat cauliflower.
    Therefore, X=3
    Let N be the total number of students surveyed.
    Therefore, N=27
    Thus, p’=N/X=3/27=1/9=0.1111
    Now, for 90% confidence level, Z=1.645
    The formula for the interval range of proportion of students is :
    Putting the values we get:
    p=0.1111+-[1 645*0.0109]
    Thus, Jane is 90% confident that the population proportion p, for students who eat cauliflower in her campus is between 9.38%and 12.9% (after converting the answer we got to percentage).


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