Question

Consider a cylindrical engine part manufacturer that has designed a new machine to reduce the variance of the outside diameters of their product. A random sample of ??= 25 outside diameters produced by this machine has a mean of ??* = 3 and a variance of ??, = 0.00014. Assume that the population of all cylindrical engine part outside diameters that would be produced by the new machine is normally distributed, and let ??, denote the variance of this population. Find a 95% confidence interval for ??,.

Answer 1

Answer:  $0.000085\leq\sigma^2\leq0.000271$

Step-by-step explanation:

Given : Sample size : n= 25

Degree of freedom : n-1 = 24

Significance for 95% confidence = $\alpha=0.05$

Sample variance : $s^2=0.00014$

We assume that the population of all cylindrical engine part outside diameters that would be produced by the new machine is normally distributed

Using chi-square distribution table , Critical values are:

$\chi^2_{{n-1,\alpha/2}}=\chi^2_{24,\ 0.025}=39.364\\\\\chi^2_{{n-1,1-\alpha/2}}=\chi^2_{24,\ 0.975}=12.401$

Confidence interval for variance is given by:-

$\dfrac{(n-1)s^2}{\chi^2_{{n-1,\alpha/2}}}\leq \sigma^2\leq\dfrac{(n-1)s^2}{\chi^2_{{n-1,1-\alpha/2}}}\\\\\\=\dfrac{(24)(0.00014)}{39.364}\leq \sigma^2\leq\dfrac{(24)(0.00014)}{12.401}\\\\\\\approx0.000085\leq\sigma^2\leq0.000271$

∴ 95% confidence interval for variance :

$0.000085\leq\sigma^2\leq0.000271$