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7. Product and Process Comparisons
7.2. Comparisons based on data from one process

7.2.2.

Are the data consistent with the assumed process mean?

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 mean of the population that can be addressed with the sample data. They are:
  1. Does the true mean agree with a known standard or assumed mean?
  2. Is the true mean of the population less than a given standard?
  3. Is the true mean of the population at least as large as a given standard?
Typical null hypotheses The corresponding null hypotheses that test the true mean, mu, against the standard or assumed mean, mu(0) are:
  1. H0: mu = mu(0)

  2. H0: mu <= mu(0)

  3. H0: mu >= mu(0)
Test statistic where the standard deviation is not known The basic statistics for the test are the sample mean and the standard deviation. The form of the test statistic depends on whether the poulation standard deviation, sigma, is known or is estimated from the data at hand. The more typical case is where the standard deviation must be estimated from the data, and the test statistic is
t = (Ybar - mu(0)/(s/SQRT(N))

where the sample mean is

Ybar = (1/N)*SUM[i=1 to N]Y(i)

and the sample standard deviation is

s = SQRT{(1/(N-1))*SUM[i=1 to N][Y(i) - Ybar)**2}

with N - 1 degrees of freedom.

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:
  1. t |   ≥   t1-α/2, N-1
  2. t   ≥   t1-α, N-1
  3. t   ≤   tα, N-1
where t1-α/2, N-1 is the 1-α/2 critical value from the t distribution with N - 1 degrees of freedom and similarly for cases (2) and (3). Critical values can be found in the t-table in Chapter 1.
Test statistic where the standard deviation is known If the standard deviation is known, the form of the test statistic is
z = (Ybar - mu(0))/(sigma/SQRT(N))

For case (1), the test statistic is compared with z1-α/2, which is the 1-α/2 critical value from the standard normal distribution, and similarly for cases (2) and (3).

Caution If the standard deviation is assumed known for the purpose of this test, this assumption should be checked by a test of hypothesis for the standard deviation.
An illustrative example of the t-test The following numbers are particle (contamination) counts for a sample of 10 semiconductor silicon wafers:
50  48  44  56  61  52  53  55  67  51

The mean = 53.7 counts and the standard deviation = 6.567 counts.

The test is two-sided Over a long run the process average for wafer particle counts has been 50 counts per wafer, and on the basis of the sample, we want to test whether a change has occurred. The null hypothesis that the process mean is 50 counts is tested against the alternative hypothesis that the process mean is not equal to 50 counts. The purpose of the two-sided alternative is to rule out a possible process change in either direction.
Critical values For a significance level of alpha = 0.05, the chances of erroneously rejecting the null hypothesis when it is true are 5 % or less. (For a review of hypothesis testing basics, see Chapter 1).

Even though there is a history on this process, it has not been stable enough to justify the assumption that the standard deviation is known. Therefore, the appropriate test statistic is the t-statistic. Substituting the sample mean, sample standard deviation, and sample size into the formula for the test statistic gives a value of

t = 1.782

with degrees of freedom N - 1 = 9. This value is tested against the critical value

t1-0.025;9 = 2.262

from the t-table where the critical value is found under the column labeled 0.975 for the probability of exceeding the critical value and in the row for 9 degrees of freedom. The critical value is based on alpha/2 instead of alpha because of the two-sided alternative (two-tailed test) which requires equal probabilities in each tail of the distribution that add to alpha.

Conclusion Because the value of the test statistic falls in the interval (-2.262, 2.262), we cannot reject the null hypothesis and, therefore, we may continue to assume the process mean is 50 counts.
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