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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?
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| 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,
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| 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
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| Interpretation |
The experimenter wants to assure that the probability of erroneously
accepting the null hypothesis of unchanged variance is at most
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| First choose the level of significance and beta risk |
The sample size is determined by first choosing appropriate values
of |
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| The calculations should be done by creating a table or spreadsheet |
First compute
Then generate a table of degrees of freedom, say between 1 and 200.
For case (1) or (2) above, calculate
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| Hints on using software packages to do the calculations |
The quantity
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| Example |
Consider the case where the variance for resistivity measurements
on a lot of silicon wafers is claimed to be 100 ohm.cm.
A buyer is unwilling to accept a shipment if
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| Calculations using Dataplot |
The procedure for performing these calculations using Dataplot
is as follows:
let d=55 let var = 100 let r = 1 + d/(var) let function cnu=chscdf(chsppf(.95,nu)/r,nu) - 0.01 let a = roots cnu wrt nu for nu = 1 200Dataplot returns a value of 169.5. Therefore, the minimum sample size needed to guarantee the risk level is N = 170. Alternatively, we could generate a table using the following Dataplot commands: let d=55 let var = 100 let r = 1 + d/(var) let nu = 1 1 200 let bnu = chsppf(.95,nu) let bnu=bnu/r let cnu=chscdf(bnu,nu) print nu bnu cnu for nu = 165 1 175 |
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| Dataplot output |
The Dataplot output, for calculations between 165 and
175 degrees of freedom, is shown below.
VARIABLES--
NU BNU CNU
0.1650000E+03 0.1264344E+03 0.1136620E-01
0.1660000E+03 0.1271380E+03 0.1103569E-01
0.1670000E+03 0.1278414E+03 0.1071452E-01
0.1680000E+03 0.1285446E+03 0.1040244E-01
0.1690000E+03 0.1292477E+03 0.1009921E-01
0.1700000E+03 0.1299506E+03 0.9804589E-02
0.1710000E+03 0.1306533E+03 0.9518339E-02
0.1720000E+03 0.1313558E+03 0.9240230E-02
0.1730000E+03 0.1320582E+03 0.8970034E-02
0.1740000E+03 0.1327604E+03 0.8707534E-02
0.1750000E+03 0.1334624E+03 0.8452513E-02
The value of |
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| Calculations using EXCEL | The procedure for doing the calculations using an EXCEL spreadsheet is shown below. The EXCEL calculations begin with 1 degree of freedom and iterate to the correct solution. | ||
| Definitions in EXCEL |
Start with:
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| Iteration step |
Then:
Clicking on "OK" iterates the calculations until C1 reaches 0.99 with the corresponding degrees of freedom shown in A1:
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