
Type I and Type II Errors and Their Application
[Editor's Note: This article has been updated since its original publication to reflect a more recent version of the software interface.]
Type I and Type II errors are two wellknown concepts in quality engineering, which are related to hypothesis testing. Often engineers are confused by these two concepts simply because they have many different names. We list a few of them here.
Type I errors are also called:
 Producer’s risk
 False alarm
 error
Type II errors are also called:
 Consumer’s risk
 Misdetection
 error
Type I and Type II errors can be defined in terms of hypothesis testing.
 A Type I error () is the probability of rejecting a true null hypothesis.
 A Type II error () is the probability of failing to reject a false null hypothesis.
Or simply:
 A Type I error () is the probability of telling you things are wrong, given that things are correct.
 A Type II error () is the probability of telling you things are correct, given that things are wrong.
The above statements are summarized in Table 1.
One concept related to Type II errors is "power." Power is the probability of rejecting H0 when H1 is true. The value of power is equal to 1. It is the power to detect the change.
In this article, we will use two examples to clarify what Type I and Type II errors are and how they can be applied.
Example 1  Application in Manufacturing
Assume an engineer is interested in controlling the diameter of a shaft. Under the normal (in control) manufacturing process, the diameter is normally distributed with mean of 10mm and standard deviation of 1mm. She records the difference between the measured value and the nominal value for each shaft. If the absolute value of the difference, D = M  10 (M is the measurement), is beyond a critical value, she will check to see if the manufacturing process is out of control.
What is the probability that she will check the machine but the manufacturing process is, in fact, in control? Or, in other words, what is the probability that she will check the machine even though the process is in the normal state and the check is actually unnecessary?
Assume that there is no measurement error. Under normal manufacturing conditions, D is normally distributed with mean of 0 and standard deviation of 1. If the critical value is 1.649, the probability that the difference is beyond this value (that she will check the machine), given that the process is in control, is:
So, the probability that she is going to check the machine but find the process is, in fact, in control is 0.1. This probability is the Type I error, which may also be called false alarm rate, α error, producer’s risk, etc.
The engineer realizes that the probability of 10% is too high because checking the manufacturing process is not an easy task and is costly. She wants to reduce this number to 1% by adjusting the critical value. The new critical value is calculated as:
Using the inverse normal distribution, the new critical value is 2.576. From the above equation, it can be seen that the larger the critical value, the smaller the Type I error.
By adjusting the critical line to a higher value, the Type I error is reduced. However, the engineer is now facing a new issue after the adjustment. Some customers complain that the diameters of their shafts are too big. Instead of having a mean value of 10, they have a mean value of 12, which means that the engineer didn’t detect the mean shift and she needs to adjust the manufacturing process to its normal state.
What is the probability of failing to detect the mean shift under the current critical value, given that the process is indeed out of control? In this case, the mean of the diameter has shifted. The answer for this question is found by examining the Type II error.
The engineer asks a statistician for help. The statistician uses the following equation to calculate the Type II error:
Here, is the mean of the difference between the measured and nominal shaft diameters and is the standard deviation. The mean value of the diameter shifting to 12 is the same as the mean of the difference changing to 2. From the above equation, we can see that the larger the critical value, the larger the Type II error.
The result tells us that there is a 71.76% probability that the engineer cannot detect the shift if the mean of the diameter has shifted to 12. This is the reason why oversized shafts have been sent to the customers, causing them to complain.
It seems that the engineer must find a balance point to reduce both Type I and Type II errors. If she increases the critical value to reduce the Type I error, the Type II error will increase. If she reduces the critical value to reduce the Type II error, the Type I error will increase. The engineer asks the statistician for additional help.
The statistician notices that the engineer makes her decision on whether the process needs to be checked after each measurement. This means the sample size for decision making is 1. The statistician suggests grouping a certain number of measurements together and making the decision based on the mean value of each group. By increasing the sample size of each group, both Type I and Type II errors will be reduced. However, a large sample size will delay the detection of a mean shift. The smallest sample size that can meet both Type I and Type II error requirements should be determined. The engineer provides her requirements to the statistician.
The engineer wants:
 The Type I error to be 0.01.
 The Type II error to be less than 0.1 if the mean value of the diameter shifts from 10 to 12 (i.e., if the difference shifts from 0 to 2).
Tables and curves for determining sample size are given in many books. These curves are called Operating Characteristic (OC) Curves. From the OC curves of Appendix A in reference [1], the statistician finds that the smallest sample size that meets the engineer’s requirement is 4.
This sample size also can be calculated numerically by hand. Assume the sample size is n in each group. The mean value and the standard deviation of the mean value of the deviation (difference between measurement and nominal value) of each group is 0 and under the normal manufacturing process. Based on the Type I error requirement, the critical value for the group mean can be calculated by the following equation:
Under the abnormal manufacturing condition (assume the mean of the deviation shifts to 2 and the standard deviation is unchanged), the Type II error is:
The smallest integer that satisfies the above two equations is the value of n.
Let’s set n = 3 first. The critical value is 1.4872 when the sample size is 3. Using this critical value, we get the Type II error of 0.1872, which is greater than the required 0.1. So we increase the sample size to 4. The critical value becomes 1.2879. The corresponding Type II error is 0.0772, which is less than the required 0.1. Therefore, the final sample size is 4.
By using the mean value of every 4 measurements, the engineer can control the Type II error at 0.0772 and keep the Type I error at 0.01.
Sometimes, engineers are interested only in onesided changes of their products or processes. For example, consider the case where the engineer in the previous example cares only whether the diameter is becoming larger. The hypothesis test becomes:
Assume the sample size is 1 and the Type I error is set to 0.05. The critical value will be 1.649. For detecting a shift of , the corresponding Type II error is . Readers can calculate these values in Excel or in Weibull++. The relation between the Type I and Type II errors is illustrated in Figure 1:
Figure 1: Illustration of Type I and Type II Errors
Example 2  Application in Reliability Engineering
Type I and Type II errors are also applied in reliability engineering. For example, in a reliability demonstration test, engineers usually choose sample size according to the Type II error.
A reliability engineer needs to demonstrate that the reliability of a product at a given time is higher than 0.9 at an 80% confidence level. She decides to perform a zero failure test. How many samples does she need to test in order to demonstrate the reliability with this test requirement?
The above problem can be expressed as a hypothesis test.
is the lower bound of the reliability to be demonstrated. The engineer must determine the minimum sample size such that the probability of observing zero failures given that the product has at least a 0.9 reliability is less than 20%. In other words, the sample size is determined by controlling the Type II error. The percentage of time that no more than f failures are expected during a passfail test is described by the cumulative binomial equation [2]:
The smallest integer that n can satisfy the above equation is the sample size.
Figure 2 shows Weibull++'s test design folio, which demonstrates that the reliability is at least as high as the number entered in the required inputs. From this analysis, we can see that the engineer needs to test 16 samples.
Figure 2: Determining Sample Size for Reliability Demonstration Testing
One might wonder what the Type I error would be if 16 samples were tested with a 0 failure requirement. In order to know this, the reliability value of this product should be known. Assume the engineer knows without doubt that the product reliability is 0.95. What is the Type I error if she uses the test plan given above? In other words, given a sample size of 16 units, each with a reliability of 95%, how often will one or more failures occur?
Using a sample size of 16 and the critical failure number of 0, the Type I error can be calculated as:
Therefore, if the true reliability is 0.95, the probability of not passing the demonstration testing is 0.5599. In this case, the test plan is too strict and the producer might want to adjust the number of units to test to reduce the Type I error.
Conclusion
In this article, we discussed Type I and Type II errors and their applications. It can be seen that a Type II error is very useful in sample size determination. In fact, power and sample size are important topics in statistics and are used widely in our daily lives. For example, these concepts can help a pharmaceutical company determine how many samples are necessary in order to prove that a medicine is useful at a given confidence level. So the next time a doctor tells you the effectiveness of a medicine is 99.9%, it is wise to ask how many patients were treated in the experiment and what the confidence level was.
References
[1] D. C. Montgomery and G.C. Runger, Applied Statistics and Probability for Engineers. 2nd Edition, John Wiley & Sons, New York, 1999.
[2] D. Kececioglu, Reliability & Life Testing Handbook, Volume 2. PrenticeHall, New Jersey, 1994.