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, β, 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 δ, 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 β. The sample size, N, required for this type of detection depends on the factor, δ; the significance level, α; and the risk, β.|
|First choose the level of significance and beta risk||The sample size is determined by first choosing appropriate values of α and β and then following the directions below to find the degrees of freedom, ν, from the chi-square distribution.|
|The calculations should be done by creating a table or spreadsheet||
Then generate a table of degrees of freedom, ν, say between 1 and 200. For case (1) or (2) above, calculate βν and the corresponding value of Cν for each value of degrees of freedom in the table where
The value of ν where Cν is closest to β is the correct degrees of freedom and
|Hints on using software packages to do the calculations||The quantity Χ 21-α, ν is the critical value from the chi-square distribution with ν degrees of freedom which is exceeded with probability α. It is sometimes referred to as the percent point function (PPF) or the inverse chi-square function. The probability that is evaluated to get Cν 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 δ 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 α = 0.05 and β = 0.01?|
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 Cν
for a given value of ν. The procedure is:
1. Define constants. α = 0.05 β = 0.01 δ = 55 σ02 = 100 R = 1 + δ/σ02 2. Create a function, Cnu. Cnu = F( F -1(α, ν)/R, ν ) - β F(x, ν) returns the probability of a chi-square random variable with ν degrees of freedom that is less than or equal to x and F -1(α, ν) returns x such that F(x, ν) = α. 3. Find the value of ν for which the function, Cnu, is zero.Using this procedure, Cnu is zero when ν is 169.3. Therefore, the minimum sample size needed to guarantee the risk level is N = 170.
Alternatively, we can determine the sample size by simply printing computed values of Cnu for various values of ν.
1. Define constants. α = 0.05 δ = 55 σ02 = 100 R = 1 + δ/σ02 2. Generate Cnu for values of ν from 1 to 200. Bnu = F -1(α, ν) / R Cnu = F(Bnu, ν)
The values of Cnu generated for ν between 165 and 175 degrees of freedom are shown below.
ν Bnu Cnu 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 Cnu closest to 0.01 is 0.0101, which is associated with ν = 169 degrees of freedom. Therefore, the minimum sample size needed to guarantee the risk level is N = 170.