7.
Product and Process Comparisons
7.2. Comparisons based on data from one process


The testing of \(H_0\) for a single population mean 
Given a random sample of measurements, \(Y_1, \, \ldots, \, Y_N\),
there are three types of questions regarding the true standard
deviation of the population that can be addressed with the sample
data. They are:


Corresponding null hypotheses 
The corresponding null hypotheses that test the true standard
deviation, \(\sigma\),
against the nominal value, \(\sigma_0\),
are:


Test statistic  The basic test statistic is the chisquare statistic $$ \chi^2 = \frac{(N1)s^2}{\sigma_0^2} \, , $$ with \(N1\) degrees of freedom where \(s\) is the sample standard deviation; i.e., $$ s = \sqrt{\frac{1}{N1} \sum_{i=1}^N \left( Y_i  \bar{Y} \right)^2} \, . $$  
Comparison with critical values  For a test at significance level \(\alpha\), where \(\alpha\) is chosen to be small, typically 0.01, 0.05 or 0.10, the hypothesis associated with each case enumerated above is rejected if: $$ \begin{eqnarray} \mbox{1. } & \chi^2 \ge \chi^2_{1\alpha/2} \,\, \mbox{ or } \,\, \chi^2 \le \chi^2_{\alpha/2} \\ & \\ \mbox{2. } & \chi^2 \ge \chi^2_{1\alpha} \\ & \\ \mbox{3. } & \chi^2 \le \chi^2_{\alpha} \, , \\ \end{eqnarray} $$ where \(\chi^2_{\alpha/2}\) is the \(\alpha/2\) critical value from the chisquare distribution with \(N1\) degrees of freedom and similarly for cases (2) and (3). Critical values can be found in the chisquare table in Chapter 1.  
Warning  Because the chisquare distribution is a nonnegative, asymmetrical distribution, care must be taken in looking up critical values from tables. For twosided tests, critical values are required for both tails of the distribution.  
Example 
A supplier of 100 ohm^{.}cm silicon wafers claims that his
fabrication process can produce wafers with sufficient consistency
so that the standard deviation of resistivity for the lot does not
exceed 10 ohm^{.}cm. A sample of \(N\)
= 10 wafers taken
from the lot has a standard deviation of 13.97 ohm.cm. Is the
suppliers claim reasonable? This question falls under
null hypothesis (2) above. For a
test at significance level, \(\alpha\)
= 0.05, the test statistic,
$$ \chi^2 = \frac{(N1)s^2}{\sigma_0^2} = \frac{9(13.97)^2}{100} = 17.56\, , $$
is compared with the critical value, \(\chi_{0.95, \, 9}^2\)
= 16.92.
Since the test statistic (17.56) exceeds the critical value (16.92) of the chisquare distribution with 9 degrees of freedom, the manufacturer's claim is rejected. 