A professor finds the mean score of a special project due every week for the 21 students in his class. The mean scores are recorded in the table. He wants to make a control chart for the process mean and already knows the mean of the ranges, R=27.3. Find the centerline X, the upper central limit (UCL), and the lower central limit (LCL).

Question

A professor finds the mean score of a special project due every week for the 21 students in his class. The mean scores are recorded in the table. He wants to make a control chart for the process mean and already knows the mean of the ranges, R=27.3. Find the centerline X, the upper central limit (UCL), and the lower central limit (LCL). 
Week: 1, 2, 3, 4, 5
Mean score: 86.2, 89.1, 76.8, 78.7, 83.5
A) x= 82.8600, UCL= 97.1948, LCL= 68.5252
B) x= 19.7286, UCL= 24.4515, LCL= 15.0057
C) x= 82.8600, UCL= 87.5829, LCL=78.1371
D) x= 82.8600, UCL= 98.6121, LCL= 67.1079

Bot · Answer

The first step is to find the overall mean score, which is the centerline X:

X = (86.2 + 89.1 + 76.8 + 78.7 + 83.5) / 5 = 82.86

The formula for the UCL and LCL are:

UCL = X + A*R
LCL = X - A*R

where A is the control chart constant, which depends on the sample size and the type of control chart being used. For this problem, since we have 21 samples, we can use the following table to find the value of A:

https://www.spcforexcel.com/knowledge/control-chart-basics/control-chart-constants

Using the table, we find that A = 1.73 for this situation.

Plugging in the values, we get:

UCL = 82.86 + 1.73*27.3 = 97.1948
LCL = 82.86 - 1.73*27.3 = 68.5252

Therefore, the answer is A) x= 82.8600, UCL= 97.1948, LCL= 68.5252.