A Quality Analyst wants to construct a sample mean chart for controlling a packaging process.He knows from past experience that whenever this process is under control,package weight is normally distributed with a mean of twenty ounces and a standard deviation of two ounces.Each day last week,he randomly selected four packages and weighed each with the following results: If he uses upper and lower control limits of 23 and 17 ounces,what is his risk of concluding this process is in control when it is actually out of control (Type II error)?
A)0.0026
B)0.0456
C)0.3174
D)0.6826
E)0.9544

Question

A Quality Analyst wants to construct a sample mean chart for controlling a packaging process.He knows from past experience that whenever this process is under control,package weight is normally distributed with a mean of twenty ounces and a standard deviation of two ounces.Each day last week,he randomly selected four packages and weighed each with the following results:  If he uses upper and lower control limits of 23 and 17 ounces,what is his risk of concluding this process is in control when it is actually out of control (Type II error)?
A)0.0026 
B)0.0456 
C)0.3174 
D)0.6826 
E)0.9544

Quizplus · Accepted Answer

First we need to find the sample mean and standard deviation:

Sample mean (x̄) = (19.5+20.2+19.1+20.3)/4 = 19.775
Sample standard deviation (s) = 0.5214

Using the upper and lower control limits of 23 and 17 ounces respectively, we can calculate the z-values for these limits:

Upper control limit (UCL) = 23
zUCL = (UCL - μ) / (σ/√n) = (23 - 20) / (2/√4) = 3

Lower control limit (LCL) = 17
zLCL = (LCL - μ) / (σ/√n) = (17 - 20) / (2/√4) = -3

The area under the normal distribution curve outside these z-values (beyond the control limits) is 0.0026.

This means that there is a 0.0026 probability of concluding that the process is in control when it is actually out of control (Type II error).

A Quality Analyst Wants to Construct a Sample Mean Chart