A 90% confidence interval for the true percentage of college students who like Brussels sprouts is (1.8%, 4.6%). What is the point estimator of the true percentage of college students who like Brussels sprouts?

And replacing the info from the problem we have:

So then the best estimator for the true proportion p is given by or equivalent to 3.2 %

Step-by-step explanation:

We want to find a confidence interval for a proportion p who represent the parameter of interest.

The confidence interval would be given by this formula:

For this case the 90% confidence interval is given by (1.8%=0.018, 4.6%=0.046) after apply the last formula

Since the confidence interval is symmetrical we can estimate the point estimator of the true percentage with this formula:

And replacing the info from the problem we have:

So then the best estimator for the true proportion p is given by or equivalent to 3.2 %

answer: idk

step-by-step explanation:

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the 14th row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100