Answer to Question #288659 in Statistics and Probability for Müg

Question

A rivet is to be inserted into a hole. If the Standard deviation of hole diameter exceeds

0,01mm, there is an unacceptably high probability that the rivet will not fit. A random sample

of n=15 part is selected and the hole diameter is measured. The sample Standard deviation of

the hole diameter measurements is s=0.008mm. Is there strong evidence to indicate that the

standard deviation of hole diameter exceeds 0.01mm? Use α=0.01

Answer 1

$$\sigma_0=0.01\ mm, \sigma_0^2=0.0001\ mm^2$$

$$s=0.008\ mm, s^2=0.000064\ mm^2$$

The following null and alternative hypotheses need to be tested:

$H_0:\sigma^2\leq0.0001$

$H_1:\sigma^2>0.0001$

This corresponds to a right-tailed test test, for which a Chi-Square test for one population variance will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=15-1=14$ degrees of freedom, and the the rejection region for this right-tailed test is $R = \{\chi^2: \chi^2 >29.1412\}.$

The Chi-Squared statistic is computed as follows:

$$\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{(15-1)(0.000064)}{0.0001}=8.96$$

Since it is observed that $\chi^2 = 8.96 \le 29.1412= \chi_c^2,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population variance $\sigma^2$ is greater than $0.0001,$ at the $0.01$ significance level.

Therefore, there is not enough evidence to claim that the standard deviation of hole diameter exceeds $0.01\ mm$ at the $0.01$ significance level.