A95% confidence interval estimate for the difference between the mean of all treated rats and the mean of all untreated rats was computed to be (1.08, 15.06). does this interval give us evidence to say that the means differs significantly at alpha = .05? why or why not?

Step-by-step explanation:

Given that a 95% confidence interval estimate for the difference between the mean of all treated rats and the mean of all untreated rats was computed to be (1.08, 15.06).

This gives mean = 8.07

This confidence interval says that we can be 95% confidence that mean difference would be atleast 1.08 and atmost 15.06

Since 0 is not contained in this interval we can say that there is evidence that the means differ significantly at alpha = 5%

nothjing

step-by-step explanation:

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)