Answer to Question #115387 in Statistics and Probability for Jennie campos

Question

The standard deviation of the breaking strengths of certain cables produced by a company is given as 240 lbs. After a change was introduced in the process of manufacture of these cables, the breaking strengths of a sample of 8 cables showed a standard deviation of 300 lbs. investigate the significance of the apparent increase in variability, using a significance level of (a) 0.05 and (b)0.01.

Answer 1

$\sigma_0=240\\ N=8\\ s=300\\ \text{We assume that the breaking strengths has normal distribution}.\\ a)\alpha=0.05\\ H_0:\sigma^2=\sigma_0^2=240^2, H_1: \sigma^2\neq\sigma_0^2=240^2\\ \text{We will use the following random variable:}\\ \chi^2=\frac{(N-1)s^2}{\sigma_0^2}\text{ where } S^2\text{ is a random value}\\ \text{of corrected variance}.\\ \text{This random variable has }\chi^2\text{-distribution with } \\ k=N-1=8-1=7 \text{ degree of freedom}.\\ \text{Observed value:}\\ \chi^2\approx 10.9375.\\ \text{Critical values:}\\ \chi^2_{critical_{left}}=\chi^2_{critical}(1-\alpha/2;k)\approx 1.6899.\\ \chi^2_{critical_{right}}=\chi^2_{critical}(\alpha/2;k)\approx 16.013.\\ \text{Critical region: }(-\infty;1.6899)\cup (16.013;\infty)\\ \text{Our observed value does not fall into the critical region}.\\ \text{So we accept } H_0.\\ \text{Variability did not change after a change}\\ \text{in the process of manufacture}.\\ b)\alpha=0.01\\ \text{Critical values:}\\ \chi^2_{critical_{left}}\approx 0.9893\\ \chi^2_{critical_{right}}\approx 20.278\\ \text{Critical region: }(-\infty;0.9893)\cup (20.278;\infty)\\ \text{Our critical value does not fall into the critical region.}\\ \text{So we accept } H_0.\\ \text{Variability did not change after a change}\\ \text{in the process of manufacture}.$