A civil engineer is analyzing the compressive strength of concrete. Compressive

Question

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with sigma^2 = 1000(psi)^2. A random sample of 12 specimens has a mean compressive strength of x = 3250 psi. (a) Construct a 95% two-sided confidence interval on mean compressive strength. (b) Construct a 99% two-sided confidence interval on mean compressive strength. Compare the width of this confidence interval with the width of the one found in part (a). (a) Given that the compressive strength is normal distribution with sigma^2 = 1000 (psi)^2. A random sample of size 12 specimens has a mean compressive strength of 3250 psi i.e., x = 3250 n= 12 The formulae for 95% two-sided confidence interval on the mean compressive strength is given by x - z_alpha/2 sigma/Squareroot n lessthanorequalto mu lessthanorequalto x + z_alpha/2 sigma/Squareroot n Here z_alpha/2 = 1.96 Substitute the values we have x - z_alpha/2 sigma/Squareroot n lessthanorequalto x + z_alpha/2 sigma/Squareroot n = 3250 - 1.96 x Squareroot 1000/Squareroot 12 lessthanorequalto mu lessthanorequalto 3250 + 1.96 xSquareroot 1000/Squareroot 12 = 3250 - 17.8923 lessthanorequalto mu lessthanorequalto 3250 + 17.8923 Therefore 3232.1077 lessthanorequalto mu lessthanorequalto 3267.8923

Explanation / Answer

variance = s.d^2
given that the value for variance is given as 1000
and then by the defination we have s.d = sqrt(1000)

variance = 1000 (psi)^2 , its the value for measurments (psi)^2 but does n't it mean variance is 1000^2

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