For option E it’s either smaller or larger Please solve ASAP I need it in 5 min
ID: 423745 • Letter: F
Question
For option E it’s either smaller or largerPlease solve ASAP I need it in 5 min The defect rate for data entry of insurance claims has historically been about 2.00%. This exercise contains only parts a, b, c, d, and e. a) If you wish to use a sample size of 100, the 3-sigma control limits are: (enter your response as a number between 0 and 1, rounded to three decimal places). (enter your response as a number between 0 and 1, rounded to three decimal places). UCL LCL b) If the sample size is 50, the 3? control limits are: UCLp enter your response as a number between 0 and 1, rounded to three decimal places). LCLp (enter your response as a number between O and 1, rounded to three decimal places). c) If the sample size is 100, the 2? control limits are: UCLp enter your response as a number between O and 1, rounded to three decimal places). LCLp(enter your response as a number between O and 1, rounded to three decimal places). d) If the sample size is 50, the 2? control limits are: UCLp(enter your response as a number between 0 and 1, rounded to three decimal places). LCLp (enter your response as a number between 0 and 1, rounded to three decimal places). Enter your answer in each of the answer boxes.
Explanation / Answer
In this case we will use the p-charts.
p-bar = 2% = 2/100 = 0.02
(a) UCL = pbar + 3*(p bar*(1-pbar)/n)^0.5 = 0.02+3*(0.02*(1-0.02)/100)^0.5 = 0.0620
LCL = pbar - 3*(p bar*(1-pbar)/n)^0.5 = 0.02-3*(0.02*(1-0.02)/100)^0.5 = -0.022. As it cannot be negative it will be 0.
(b) UCL = pbar + 3*(p bar*(1-pbar)/n)^0.5 = 0.02+3*(0.02*(1-0.02)/50)^0.5 = 0.0794
LCL = pbar - 3*(p bar*(1-pbar)/n)^0.5 = 0.02-3*(0.02*(1-0.02)/50)^0.5 = -0.0394. As it cannot be negative it will be 0.
( c) UCL = pbar + 2*(p bar*(1-pbar)/n)^0.5 = 0.02+2*(0.02*(1-0.02)/100)^0.5 = 0.0480
LCL = pbar - 2*(p bar*(1-pbar)/n)^0.5 = 0.02-2*(0.02*(1-0.02)/100)^0.5 = -0.008. As it cannot be negative it will be 0.
(d) UCL = pbar + 2*(p bar*(1-pbar)/n)^0.5 = 0.02+2*(0.02*(1-0.02)/50)^0.5 = 0.0596
LCL = pbar - 2*(p bar*(1-pbar)/n)^0.5 = 0.02-2*(0.02*(1-0.02)/50)^0.5 = -0.0196. As it cannot be negative it will be 0.
(e) When the sample size is larger then sigma is smaller.
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