a)
sample mean, xbar = 20.95
sample standard deviation, s = 0.3512
sample size, n = 4
degrees of freedom, df = n - 1 = 3
Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 3.182
ME = tc * s/sqrt(n)
ME = 3.182 * 0.3512/sqrt(4)
ME = 0.56
CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (20.95 - 3.182 * 0.3512/sqrt(4) , 20.95 + 3.182 *
0.3512/sqrt(4))
CI = (20.39 , 21.51)
b)
Option B)
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