Question

Problem 3) Jane Hochkiss, director of production, has decided to record the number of defective labels in random daily samples on control charts. Jane estimates that 1.5 percent loose labels is typical when the labeling process is in control. Twelve daily samples, each consisting of 200 pairs of jeans, were selected and examined. The number of defective labels found in each sample is shown below.

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sample number
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
11.00
12.00

number of defectives
2.00
3.00
5.00
2.00
7.00
1.00
3.00
-   
5.00
3.00
9.00
2.00

​

Required:

1.)Prepare a p chart using 3-sigma control limits for the UCL and LCL for the chart.
2.)Conclude if the labeling operation is in control.

Answer 1

1)

total number of defective sample=   42
total samples =    2400
p-bar = Total number of defects/Total sample =   0.0175
  

  
Sp=√(p(1-p)/n)=√(0.0175*(1-0.0175)/200)=   0.0093
UCL=p bar+z*Sp=0.0175+3*0.0093=   0.0453
LCL=p bar-z*Sp=0.0175-3*0.0093=   -0.0103
since, LCL cannot be negative, so, LCL=0  

b)

since, all data points are within LCL and UCL, so,

the labeling operation is in control