Question

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.

Answer 1

Let $x$ be the number of students who eat cauliflower.

Therefore, $x=4$.

Let $n$ be the total number of students surveyed.

Therefore, $n=39$

Thus, $\hat p$$=\frac{4}{39} =0.10256$

Now, for 90% confidence level, from the table we know that $Z=1.645$.

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

$p= \hat p\pm Z\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Plugging in the values we get:

$p=0.10256\pm 1.645\sqrt{\frac{0.10256(1-0.10256)}{39}}=0.10256\pm 0.04858=0.15114, 0.05398$

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).