One hundred insurance claims are inspected daily at a local insurance company. The following data was collected over the last 30 days. Construct a p chart. Assume any points outside of the control limits were due to assignable causes and corrected and reduced to 0. Construct a revised chart. Day Number Defective 1 3 2 3 3 3 4 2 5 0 6 3 7 0 8 1 9 7 10 3 11 2 12 0 13 0 14 4 15 1 16 2 17 4 18 0 19 1 20 1 21 0 22 2 23 8 24 2 25 1 26 3 27 0 28 1 29 2 30 4

Total number of defectives = 63

Sample size = n = 100

Number of samples= 30

Proportion of defectives = Pbar = 63/ ( 30 x 100 ) = 0.021

Upper control limit = UCL

= pbar + 3 x Square root ( Pbar x ( 1 – Pbar)/ n)

= 0.021 + 3 x Square root ( 0.021 x 0.979/ 100 )

= 0.021 + 3 x 0.0143

= 0.021 + 0.0429

= 0.0639

Lower Control Limit , LCL

= Maximum ( 0, p bar -3 x Square root ( Pbar x ( 1 – pbar)/n)

= Maximum ( 0, 0.021 – 0.0429)

= Maximum ( o, – 0.0219)

= 0

UCLp = 0.0639

LCLp = 0

With sample size being 100 , the maximum permissible number of defects = 100 x UCLp = 6.39

Following data however go beyond upper control limit of 6.39:

7 on day 9

8 on day 23

Above 2 sets of data are outside control limits and hence they are made to ZERO

The revised proportion of defective = pbar = ( 63 – 7 -8) / ( 100 x 30 ) = 48 /3000 = 0.016

The revised control limits as follows :

Upper control limit = UCL

= pbar + 3 x Square root ( Pbar x ( 1 – Pbar)/ n)

= 0.016 + 3 x Square root ( 0.016 x 0.984/100)

= 0.016 + 3 x 0.0125

= 0.016 + 0.0375

= 0.0535

Lower Control Limit , LCL

= Maximum ( 0, p bar -3 x Square root ( Pbar x ( 1 – pbar)/n)

= Maximum ( 0, 0.016 – 0.0375 )

= Maximum ( 0, – 0.0215)

= 0

UCLp = 0.0535

LCLp = 0