Question

An article in Technometrics (1999, Vol. 41, pp. 202–211) studied the capability of a gauge by measuring the weight of paper. The data for repeated measurements of one sheet of paper are in the following table. Construct a 90% one-sided, upper confidence interval for the standard deviation of these measurements. Assume population is approximately normally distributed.

Round your answer to 3 decimal places.

3.481

3.448

3.485

3.475

3.472

3.477

3.472

3.464

3.472

3.470

3.470

3.470

3.477

3.473

3.474

Answer 1

sample mean is x=(1/n)$\sum X_i$ ​ =52.08/15=3.472

The sample variance is given as,

s²=(1/(n-1))$\sum X_i(x_i-x)^2$ =0.000966/14=0.000069

confidence level, α =90%=0.90

degrees of freedom

df=14

χ²= 7.79

A 90% upper confidence bound for the variance is found as follows:

$\sigma^2<(n-1)s^2/\chi^2$

$\sigma^2<(14)0.000069/7.79;\sigma^2<0.000124005135$

σ<0.0111357593