Roman C. answered 12/08/15
Tutor
5.0
(851)
Masters of Education Graduate with Mathematics Expertise
We first need the mean μ and standard deviation σ.
We also know that since we are testing at significance α=1-0.95 =0.05.
Thus we have:
n = 12
xbar = Σxi / n = (4045 + ... + 4324) / 12 = 4341.25
s = √ [Σ(xi - xbar)2 / (n-1)]
= √{ [(4045 - 4341.25)2 + ... + (4324 - 4341.25)2] / 11 }
= 395.87
SE = s/√n = 114.28
t*df=11 = 2.20 Using P(t ≤ t*) = 1-α/2 = 0.975
Thus the confidence interval is
xbar ± t* · SE = 4341.25 ± 2.20 · 114.28 = ( 4089.83 , 4592.67 )
We also know that since we are testing at significance α=1-0.95 =0.05.
Thus we have:
n = 12
xbar = Σxi / n = (4045 + ... + 4324) / 12 = 4341.25
s = √ [Σ(xi - xbar)2 / (n-1)]
= √{ [(4045 - 4341.25)2 + ... + (4324 - 4341.25)2] / 11 }
= 395.87
SE = s/√n = 114.28
t*df=11 = 2.20 Using P(t ≤ t*) = 1-α/2 = 0.975
Thus the confidence interval is
xbar ± t* · SE = 4341.25 ± 2.20 · 114.28 = ( 4089.83 , 4592.67 )