Question

From past experience, it has been found that the strengths of concrete produced is normally distributed with a variance of 12.25N/mm2. If the Contract Specifications require that a material engineer should be 90% confident that his estimate does not deviate from the true mean strength of concrete by 3 N/mm2.

(1) How many concrete cubes should Materials Engineer take in each batch?

(2) Assuming if the Materials Engineer managed to take only 2 specimens in each batch, because one of his field technicians called in sick. What impact does this development have on the level of confidence in his estimate?

Answer 1

(1) We have Variance ơ2 12.25 Confidence level 0.90 Margin of error d 3 The required sample size would be, To calculate critical z value for 0.90 confidence level, using NORM.S.INV excel unction we can get the z value, but it gives values for left tailed probability, area to the left of z is 5% + 90%-95%-0.95 NORM.S.INV(0.95) gives 1.644853627 So, z -1.6449 n-(1.6349). 12.25-3682545257 N 4 So, Materials Engineer should take 4 concrete cubes in each batch (2) If instead of 4 material engineer takes only n 2 concrete cubes, then 12.25 = 12.25 1.469387755 z- 1.212183053

Now, to find the confidence level for z 1.2122 i.e. P(-1.2122 < Z < 1.2122) = P(Z < 1.2122)-P(Z <-1.2122) P(Z < 1.2122) - [1- P(Z < 1.2122)] Since, ф(2)-1-ф(z) = 2 *P(Z < 1.2122)-1 we can -NORM.SDISTtz, cumulative) excel function to calculate P(Z -NORM.SDIST(1.2122,TRUE) gives 0.887278842, hence PZ < 1.2122) = 0.8872 く 1.2122. So, P(-1.2122 < Z< 1.2122) 2 0.8872-1-0.7746 this development will reduce the level of confidence to 77.46% in his estimate.