4. A hospital is using X and R charts to record the time it takes to process pat
ID: 331403 • Letter: 4
Question
4. A hospital is using X and R charts to record the time it takes to process patient account information. A sample of five applications is taken each day. The first four weeks' (20 days') data give the following values: (15 Points) Xbar - 16 Minutes and Rbar 7 Minutes If the upper and lower specifications are 21 minutes and 13 minutes, respectively, calculate and interpret the indices: Estimate of the Population Standard Deviation (Sigma), 6 Sigma, Cp.and Cpk. Interpret the results. 5. Explain what is meant by variation and what occurs when a process is in control or out of control. The text refers to this as chance or assignable cause. Give examples of an assignable cause and a chance cause. (10 points) 6. Using the data in Question 1 prepare a XmR Control Chart (20 points) a. What is the Mean and the Moving range for the data. b. Interpret the control chart 7. What do control limits represent? What do Specification limits represent? (15 points) a. Describe the three cases that compare specification limits to control limits. b. What is the impact of each on quality?Explanation / Answer
Question 4-
X bar = 16 minutes, Rbar= 7 minutes
Upper specification (USL) = 21 minutes
Lower specification (LSL) = 13 minutes
No. of applications taken each day (n) = 5
Standard deviation (?) = Rbar / d2
For sample size n=5, value of d2= 2.326
Hence Standard deviation (?) = 7/2.326
= 3.009
Hence 6? = 6*3.009
= 18.054
Lets calculate Cp (Potential capability index)
Cp = (USL- LSL)/process capability (6?)
Putting the vallues as given above-
Cp= (21-13)/ 18.054
= 8/18.054
= 0.44
Cpk = 1/3 Min (Zu, Zi } , where Zu = (USL-u)/?
Zu = (21-16)/ 3.009
= 5/3.009
= 1.66
Lets calculate Zi = (u-LSL)/ ?
Zi = (16-13)/3.009
= 3/3.009
= 0.997
Considering Zi value (0.997) is lower than Zu (1.66), Zi would be considered for calculation
1/3 (0.997)
= 0.33
Cp is the potential capability index which proves that if properly centred, how a process or observation can work relative to the set specification. Cp value of 0.44 proves that process is not stable as desired Cp should always be >1.
Likely Cpk =0.33 is also bound to produce non confirming parts.
Question 5-
Variation refers to the deviation from any set process. Variation is caused by two factors i.e. Assignable cause variation and non assignable cause variation. Depenind upon the process and final goal variation can be desirable, but most of the times nondesirable. Variations which is caused by nonassignable cause are due to some unexplainable causes and doesn’t imact the process much. But the variations, which arise due to the assignable casues, can impact the process and in result can give rise to higher varation such as higher standard deviation and lower process capabilty.
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