Answer to Question #279121 in Statistics and Probability for Syaa

Question

Suppose that the thickness of a part used in a semiconductor is its critical dimension and that measurements of the thickness of a random sample of 18 such parts have the variance S2 = 0.68, where the measurements are in thousandths of an inch. The process is considered to be under control if the variation of the thickness is given by a variance not greater than 0.36. Assuming that the measurements constitute a random sample from a normal population, test the claim at the α =0.05 significance level.

Answer 1

The hypotheses to be tested are,

$H_0: \sigma^2 =0.36\space vs\space H_A: \sigma^2 > 0.36$

$n=18,\space s^2=0.68$

The test statistic is,

$\chi^2 =\frac{(n-1) \times s^2}{\sigma^2}$

$\chi^2 = \frac{17 \times 0.68}{0.36} = 32.111$

The table value at $\alpha=0.05$ with $n-1=18-1=17$ degrees of freedom is, $\chi^2_{0.05,17}=27.8571$ and the null hypothesis is rejected if $\chi^2\gt\chi^2_{0.05,17}$.

Since $\chi^2=32.111\gt\chi^2_{0.05,17}=27.8571,$ we reject the null hypothesis and we conclude that there is sufficient evidence to show that the population variance is greater than 0.36 level at 5% level of significance and the process is out of control .