7.
Product and Process Comparisons
7.2. Comparisons based on data from one process 7.2.3. Are the data consistent with a nominal standard deviation?


Sample sizes to minimize risk of false acceptance  The following procedure for computing sample sizes for tests involving standard deviations follows W. Diamond (1989). The idea is to find a sample size that is large enough to guarantee that the risk, \(\beta\), of accepting a false hypothesis is small.  
Alternatives are specific departures from the null hypothesis 
This procedure is stated in terms of changes in the variance, not
the standard deviation, which makes it somewhat difficult to
interpret. Tests that are generally of interest are stated in terms
of \(\delta\),
a discrepancy from the hypothesized variance. For example:


Interpretation  The experimenter wants to assure that the probability of erroneously accepting the null hypothesis of unchanged variance is at most \(\beta\). The sample size, \(N\), required for this type of detection depends on the factor, \(\delta\); the significance level, \(\alpha\); and the risk, \(\beta\).  
First choose the level of significance and beta risk  The sample size is determined by first choosing appropriate values of \(\alpha\) and \(\beta\) and then following the directions below to find the degrees of freedom, \(\nu\), from the chisquare distribution.  
The calculations should be done by creating a table or spreadsheet  First compute $$ R = 1 + \frac{\delta}{\sigma^2_0} \, . $$ Then generate a table of degrees of freedom, \(\nu\), say between 1 and 200. For case (1) or (2) above, calculate \(\beta_\nu\) and the corresponding value of \(C_\nu\) for each value of degrees of freedom in the table where $$ \begin{eqnarray} \mbox{ 1. } & \beta_\nu & = \chi^2_{1\alpha, \nu} / R \\ & & \\ & C_\nu & = \mbox{Pr }(\chi^2_\nu < \beta_\nu) \\ & & \\ \mbox{ 2. } & \beta_\nu & = \chi^2_{\alpha, \nu} / R \\ & & \\ & C_\nu & = \mbox{Pr }(\chi^2_\nu > \beta_\nu) \end{eqnarray} $$ The value of \(\nu\) where \(C_\nu\) is closest to \(\beta\) is the correct degrees of freedom and $$ N = \nu + 1 \, . $$  
Hints on using software packages to do the calculations  The quantity \(\chi_{1\alpha, \, \nu}^2\) is the critical value from the chisquare distribution with \(\nu\) degrees of freedom which is exceeded with probability \(\alpha\). It is sometimes referred to as the percent point function (PPF) or the inverse chisquare function. The probability that is evaluated to get \(C_\nu\) is called the cumulative density function (CDF).  
Example  Consider the case where the variance for resistivity measurements on a lot of silicon wafers is claimed to be 100 (ohm^{.}cm)^{2}. A buyer is unwilling to accept a shipment if \(\delta\) is greater than 55 ohm^{.}cm for a particular lot. This problem falls under case (1) above. How many samples are needed to assure risks of \(\alpha\) = 0.05 and \(\beta\) = 0.01?  
Calculations 
If software is available to compute the roots (or zero values) of a
univariate function, then we can determine the sample size by finding
the roots of a function that calculates \(C_\nu\)
for a given value of \(\nu\).
The procedure is:
Alternatively, we can determine the sample size by simply printing computed values of \(C_\nu\) for various values of \(\nu\).
The values of \(C_\nu\) generated for \(\nu\) between 165 and 175 degrees of freedom are shown below. \(\nu\) \(\beta_\nu\) \(C_\nu\)
165 126.4344 0.0114 166 127.1380 0.0110 167 127.8414 0.0107 168 128.5446 0.0104 169 129.2477 0.0101 170 129.9506 0.0098 171 130.6533 0.0095 172 131.3558 0.0092 173 132.0582 0.0090 174 132.7604 0.0087 175 133.4625 0.0085The value of \(C_\nu\) closest to 0.01 is 0.0101, which is associated with \(\nu\) = 169 degrees of freedom. Therefore, the minimum sample size needed to guarantee the risk level is \(N\) = 170. The calculations used in this section can be performed using both Dataplot code and R code. 