Jane wants to estimate the proportion of students on her campus who eat cauliflower. after surveying 39 students, she finds 4 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.

Let be the number of students who eat cauliflower.

Therefore, .

Let be the total number of students surveyed.

Therefore,

Thus,

Now, for 90% confidence level, from the table we know that .

The formula for the interval range of proportion of students is :

Plugging in the values we get:

Thus, Jane is 90% confident that the population proportion p, for students who eat cauliflower in her campus is between 5.398% and 15.114% (after converting the answer we got to percentage).

welll its a very hard question but ill

step-by-step explanation:

first you take the 6 dollars and add 10 to it then add the 50 to that then multiply by 8 with xy and your answer will be

hope this

11

step-by-step explanation:

32.50-7.50=25

225/25=11