“…The sample mean and variance can be used to evaluate the confidence intervals of the true values of the distributions, assuming normality, as discussed in Section . Using the Student’s t distribution, the “true” mean is expected to lie in the following range y̅ − t a , N normalR − 1 N R S y̅ < μ y < y̅ + t a , N normalR − 1 N R S y̅ For a confidence interval a = 95% with N R – 1 = 2 degrees of freedom, it can be calculated that t 0.95,2 = 4.30 and, consequently, y̅ − 3.04 S y̅ < μ y < y̅ + 3.04 S y̅ According to the chi-squared distribution (χ 2 ), the “true” variance is in the following range N R − 1 χ false( 1 − a false) / 2 , N normalR − 1 2 S y̅ 2 < σ y 2 < N R − 1 χ false( 1 + a false) / 2 , N normalR − 1 2 S y̅ 2 In the analyzed case, using the same previously defined confidence level and degree of freedom, χ 0.025,2 2 = 0.0506 and χ 0.975,2 2 = 7.377, achieving<...…”