Question

The results of inspection of DNA samples taken over the past 10 days are given below. Sample size is 100. 10 Day Defectives 2 3 4 6 6 0 6 a) The upper and lower 3-sigma control chart limits are: UCLp(enter your response as a number between 0 and 1, rounded to three decimal places). LCL(enter your response as a number between 0 and 1, rounded to three decimal places). b) Given the limits in part a, is the process in control for the following days as shown in the table below? Day Number of Defects In-Control 12 13 13

Answer 1

a)

Total number of defectives = 37

p-bar = Total number of defectives/(no of samples * sample size) = 37/(10*100) = 0.037

UCL = p-bar + 3*sqrt((p-bar*(1-p-bar))/sample size)

= 0.037 + 3*sqrt((0.037*(1-0.037))/100)

= 0.093628526 = 0.094 (Rounded to 3 decimal places)

LCL = p-bar - 3*sqrt((p-bar*(1-p-bar))/sample size)

= 0.037 - 3*sqrt((0.037*(1-0.037))/100)

= -0.019628526 = 0

b)

Defect proportion for day 11 = no of defects/sample size = 6/100 = 0.06....In Control

Defect proportion for day 12 = no of defects/sample size = 13/100 = 0.13 .....Out of control (breaches UCL)

Defect proportion for day 13 = no of defects/sample size = 11/100 = 0.11 .....Out of control (breaches UCL)