Chaoyun L. answered 10/27/15
Tutor
1
(1)
Master of Math with concentration in Statistics
A)The degree of freedom here is 50-1=49
T1 = (x-μ)/(s/√n) = (2.39 -2.60)/ (0.07/√50) = -21.21 , with p-vlaue on t-table < 0.0005
T2= (x-μ)/(s/√n) = (2.81-2.60)/(0.07/√50) = 21.21, with p-value on t-table <0.0005
So 1-(< 0.0005)-(<0.0005 ) > 0.999.
Aproximately 100% between 2.39 and 2.81.
b) t1 = (x-μ)/(s/√n) = (2.53 - 2.60) / (0.07/√50) = -7.07
t2 = (x-μ)/(s/√n) = (2.67 - 2.60) / (0.07/√50) = 7.07
Still, aprosimately 100% between them.
This is what I am thinking of... Hope it helps.