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?

7.2.3.2.

Sample sizes required

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, , 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 , a discrepancy from the hypothesized variance. For example:

1. Is the true variance larger than its hypothesized value by ?
2. Is the true variance smaller than its hypothesized value by ?

That is, the tests of interest are:

1. H₀:

2. H₀:

Interpretation

The experimenter wants to assure that the probability of erroneously accepting the null hypothesis of unchanged variance is at most . The sample size, N, required for this type of detection depends on the factor, ; the significance level, ; and the risk, .

First choose the level of significance and beta risk

The sample size is determined by first choosing appropriate values of and and then following the directions below to find the degrees of freedom, , from the chi-square distribution.

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 and the corresponding value of for each value of degrees of freedom in the table where

1. 

   

2. 

   

The value of where is closest to is the correct degrees of freedom and

N = + 1

Hints on using software packages to do the calculations

The quantity is the critical value from the chi-square distribution with degrees of freedom which is exceeded with probability . It is sometimes referred to as the percent point function (PPF) or the inverse chi-square function. The probability that is evaluated to get 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. A buyer is unwilling to accept a shipment if is greater than 55 ohm^.cm for a particular lot. This problem falls under case (1) above. The question is how many samples are needed to assure risks of = 0.05 and = .01.

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 = 1 200

Dataplot 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

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 which is closest to 0.01 is 0.010099; this has degrees of freedom = 169. Therefore, the minimum sample size needed to guarantee the risk level is N = 170.

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:

1. 1 in A1
2. CHIINV{(1-), A1}/R in B1
3. CHIDIST(B1,A1) in C1

   In EXCEL, CHIINV{(1- ), A1} is the critical value of the chi-square distribution that is exceeded with probabililty . This example requires CHIINV(.95,A1). CHIDIST(B1,A1) is the cumulative density function up to B1 which, for this example, needs to reach 1 - = 1 - 0.01 = 0.99. The EXCEL screen is shown below.

   

Iteration step

Then:

1. From TOOLS, click on "GOAL SEEK"
2. Fill in the blanks with "Set Cell C1", "To Value 1 - " and "By Changing Cell A1".
3. Click "OK"

Clicking on "OK" iterates the calculations until C1 reaches 0.99 with the corresponding degrees of freedom shown in A1: