Historically,the average time to service a customer complaint has been 3 days and the standard deviation has been 0.50 day.Management would like to specify the control limits for an $\bar { X }$ -chart with a sample size of 10 and 3 sigma limits.Suppose the next sample of 10 yielded the following result: 3.2,2.1,3.6,2.8,3.9,3.5,2.7,4.1,2.6,and 3.3.What conclusion can be drawn?
A) Assuming the process variability is in control, the process average is also in statistical control.
B) Assuming the process average is in control, the process average is out of statistical control.
C) The sample size should be increased.
D) No conclusion can be drawn because there is insufficient data.

Question

Quizplus · Accepted Answer

We can calculate the sample mean ($\bar { X }$) by adding the 10 values and dividing by 10: 
$$\bar { X }=\frac{3.2+2.1+3.6+2.8+3.9+3.5+2.7+4.1+2.6+3.3}{10}=3.15$$
The upper control limit (UCL) can be calculated using the formula: 
$$UCL=\bar { X }+3\frac{\sigma}{\sqrt{n}}$$
where $\sigma$ is the population standard deviation, and n is the sample size. Plugging in the given values, we get: 
$$UCL=3.15+3\frac{0.50}{\sqrt{10}}=4.05$$
The lower control limit (LCL) can be calculated using the formula: 
$$LCL=\bar { X }-3\frac{\sigma}{\sqrt{n}}$$
Plugging in the given values, we get: 
$$LCL=3.15-3\frac{0.50}{\sqrt{10}}=2.25$$
Since all 10 sample values are within the control limits, we can conclude that the process is in statistical control. Therefore, choice (A) is correct. Choice (B) is incorrect because we concluded that the process average is in control, not out of control. Choice (C) is not necessary because we already have enough data with a sample size of 10. Choice (D) is also incorrect because we were able to draw a conclusion based on the given data.

Historically,the Average Time to Service a Customer Complaint Has Been Xˉ\bar { X }Xˉ

Historically,the Average Time to Service a Customer Complaint Has Been $\bar { X }$