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

7.2.7.

How can we determine whether the proportion of defectives produced by a process has changed from the "nominal" value?

Testing proportions is based on the binomial distribution

The proportion defective produced by a process is usually monitored using statistics based on the binomial distribution. 

The Binomial Distribution

The observed number of defective, x, when a random sample of size n is selected from a continuous manufacturing process (or a large population or lot), is a random variable that follows the binomial distribution. 

Binomial example

Let us do an example: 

Consider tossing a coin and betting on "tails". Equating getting a "head" with "defective", we might ask what is the probability of getting exactly 2 heads in six tosses of a coin? 

With n = 6, x = 2, p = 1/2, we get 

What is the probability of get 2 heads or less? 

This is the probability of 0 heads, or 1 head, or 2 heads, which is p(0) + p(1) + p(2) = 1/64 + 6/64 + 15/64 = 22/64 or 11/32. 

The formulas for the mean and variance of the binomial distribution are np and np(1-p) respectively. Usually one writes q instead of 1-p. So for n = 6 and p = .5 the mean = 3 and the variance = 1.5 

For large samples (n > 30) the normal distribution is used to approximate the binomial distribution.

How to test proportions

Testing the Proportion Defective

To test if the fraction defectives has changed from a nominal or long established value, we test the null hypothesis:

H_o: p = p_o, the nominal value vs
H_a: p > p_o or p < p_o (or p is different from p_o) 

The test statistic is

and we compare this test statistic to normal distribution tail values for the appropriate one or two sided hypothesis test with total error a. If the magnitude of z is too large, we reject H_o and conclude p has changed.

Example of testing proportions

Example

A new way of processing wafers was introduced. Of 200 tested wafers, 26 show some type of defect. That is a proportion of 0.13. 

Since the old method had demonstrated a 0.10 proportion of defectives over a long period of time, management questioned the reliability of the new method. To validate the ineffectiveness of the new way, the following hypothesis was tested: 

H_o: p = p_o, the new process has the same defect level as the old process vs.
H_a: p > 0.10, the new process has a greater proportion of defectives.

As stated above, this is a one-tail test since we are most concerned about protecting against a switch to a less reliable process. Choosing a = 0.05, we will reject the null, when the test statistic z is greater than the critical value 1.645. 

Substituting the numerical values into the test statistic, we get

Since the test statistic (1.41) is less than the critical value (1.645) we cannot reject the null hypothesis of equality and conclude that the new way of processing produces more defectives. The new process may, indeed, be worse - but we need more evidence before we can make that conclusion at the 95% confidence level.