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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?

The testing of H0 for a single population mean Given a random sample of measurements, Y1, ..., YN, there are three types of questions regarding the true standard deviation of the population that can be addressed with the sample data. They are:
  1. Does the true standard deviation agree with a nominal value?
  2. Is the true standard deviation of the population less than or equal to a nominal value?
  3. Is the true stanard deviation of the population at least as large as a nominal value?
Corresponding null hypotheses The corresponding null hypotheses that test the true standard deviation, sigma, against the nominal value, sigma(0) are:
  1. H0: sigma = sigma(0)
  2. H0: sigma <= sigma(0)
  3. H0: sigma >= sigma(0)
Test statistic The basic test statistic is the chi-square statistic

Chi-Square = (N-1)*s**2/(sigma(0)**2)

with N - 1 degrees of freedom where s is the sample standard deviation; i.e.,

s = SQRT{(1/(N-1))*SUM[i=1 to N][(Y(i) - Ybar)**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:

Chi-Square >= Chi-Square(1-alpha/2,N-1)  or
             Chi-Square <= Chi-Square(alpha/2,N-1);  
             Chi-Square >= Chi-Square(1-alpha,N-1);
             Chi-Square <= Chi-Square(alpha,N-1)

where Χ 2α/2 is the alpha/2 critical value from the chi-square distribution with N - 1 degrees of freedom and similarly for cases (2) and (3). Critical values can be found in the chi-square table in Chapter 1.

Warning Because the chi-square distribution is a non-negative, asymmetrical distribution, care must be taken in looking up critical values from tables. For two-sided 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-Square = (N-1)*s**2/(sigma(0)**2) = 9*(13.97)**2/100 = 17.56

is compared with the critical value, Χ 20.95, 9 = 16.92.

Since the test statistic (17.56) exceeds the critical value (16.92) of the chi-square distribution with 9 degrees of freedom, the manufacturer's claim is rejected.

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