I need help. You are interested in finding a 95% confidence interval for the average commute that non-residential students have to their college. The data below show the number of commute miles for 12 randomly selected non-residential college students. Round answers to 3 decimal places where possible.

(11.847 ; 15.813)

Step-by-step explanation:

We are given 12 samples which are :

8, 20, 20, 11, 18, 12, 6, 5, 7, 22, 12, 25

We use a T-distribution to find the confidence interval since the sample size. is small, n < 30

Using a calculator :

The sample mean, xbar = 13.83

Sample standard deviation, s = 6.87

The confidence interval, C.I

C.I = xbar ± Tcritical * s/√n)

Tcritical at 95%, df = n - 1, 12 - 1 = 11

Tcritical(0.05, 11) = 2.20

Hence,

C.I = 13.83 ± 2.20(6.87/√12)

C.I = 13.83 ± 1.9831981

C. I = (13.83 - 1.983 ; 13.83 + 1.983)

C. I = (11.847 ; 15.813)

alright so i think for this one it would to draw it out. this may be hard for me to try and draw, but i will do my best on here:

        r

  /       |     \

notice how it forms 2 triangles inside one bigger triangle.

so if we know the angle spr is the same as psr that is saying the measure of angle p is the same as angle s. this means the big triangle psr is either isosceles or equilateral. in both cases, the triangle is perfectly symmetrical when split down the middle, which is in this case rq. because we know it is perfectly symmetrical on both sides, we know that the two triangles inside the bigger one are equal.