In each of the 15 consecutive years, 1000 high school completers were randomly selected and the number who enrolled in college was determined. The results are listed below. Find the values you would use for the centerline, the upper control limit, and the lower control limit.
601, 625, 619, 626, 619, 619, 650, 670, 656, 629, 633, 618, 652, 639, 667.
A) p=9.52, UCL=9.72, LCL=9.32
B) p=0.63, UCL=0.64, LCL=0.62
C) p=0.63, UCL=0.68, LCL=0.58
D) p=0.63, UCL=0.68, LCL=0
To find the centerline, we need to calculate the average enrollment rate over the 15 years:
Centerline = (601 + 625 + 619 + 626 + 619 + 619 + 650 + 670 + 656 + 629 + 633 + 618 + 652 + 639 + 667) / 15 = 636.2
To determine the control limits, we can use the following formulas:
UCL = p̂ + 3√(p̂(1-p̂)/n)
LCL = p̂ - 3√(p̂(1-p̂)/n)
where p̂ is the average enrollment rate and n is the sample size (1000).
p̂ = 636.2/1000 = 0.6362
UCL = 0.6362 + 3√(0.6362*(1-0.6362)/1000) = 0.6807
LCL = 0.6362 - 3√(0.6362*(1-0.6362)/1000) = 0.5918
Therefore, the values we would use for the centerline, upper control limit, and lower control limit are:
Centerline = 636.2
UCL = 0.6807
LCL = 0.5918
The answer is C) p=0.63, UCL=0.68, LCL=0.58.
Consider a process that includes careful testing of each manufactured defibrillator. Listed below are the number of defective defibrillators in 20 successive batches of 10,000. Find the values you would use for the centerline, the upper control limit, and the lower control limit.
20, 14, 22, 27, 12, 12, 18, 23, 25, 19, 24, 28, 21, 25, 17, 19, 17, 22, 15, 20
A) p=0.040000, UCL=0.041039, LCL=0.038961
B) p=0.002000, UCL=0.003340, LCL=0.000660
C) p=0.002000, UCL=0.021200, LCL=0
D) p=0.040000, UCL=0.041470, LCL=0.038530
To find the centerline, we need to calculate the average proportion of defective defibrillators over the 20 batches:
Centerline = (20 + 14 + 22 + 27 + 12 + 12 + 18 + 23 + 25 + 19 + 24 + 28 + 21 + 25 + 17 + 19 + 17 + 22 + 15 + 20) / (20 * 10000) = 0.0204
To determine the control limits, we can use the following formulas:
UCL = p̂ + 3√(p̂(1-p̂)/n)
LCL = p̂ - 3√(p̂(1-p̂)/n)
where p̂ is the average proportion of defective defibrillators and n is the sample size (10,000).
p̂ = 0.0204
UCL = 0.0204 + 3√(0.0204*(1-0.0204)/10000) = 0.021132
LCL = 0.0204 - 3√(0.0204*(1-0.0204)/10000) = 0.019668
Therefore, the values we would use for the centerline, upper control limit, and lower control limit are:
Centerline = 0.0204
UCL = 0.021132
LCL = 0.019668
The answer is A) p=0.040000, UCL=0.041039, LCL=0.038961.
A drugstore considers a wait for more than 5 minutes to be a defect. Each week 100 customers are randomly selected and timed at the checkout line. The number of defects for 20 consecutive weeks are given below.
4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 12, 6, 6, 6, 7, 6, 7, 8, 7
what is the upper control limit?
To determine the upper control limit for a p-chart, we need the following information:
- p̂: the average proportion of defects over the 20 weeks.
- UCL: the upper control limit for the proportion of defects.
To find p̂, we add up the number of defects and divide by the total number of observations (20 * 100 = 2000):
p̂ = (4 + 4 + 5 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 6 + 12 + 6 + 6 + 6 + 7 + 6 + 7 + 8 + 7) / 2000 = 0.0605
To find UCL, we use the following formula:
UCL = p̂ + 3√(p̂(1 - p̂)/n)
where n is the sample size (100).
UCL = 0.0605 + 3√(0.0605(1 - 0.0605)/100) = 0.1063
Therefore, the upper control limit for the proportion of defects is 0.1063, or approximately 10.6%.
A drugstore considers a wait for more than 5 minutes to be a defect. Each week 100 customers are randomly selected and timed at the checkout line. The number of defects for 20 consecutive weeks are given below.
4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 12, 6, 6, 6, 7, 6, 7, 8, 7
what is the correct upper control limit?
A) UCL=0.047
B) UCL=0.133
C) UCL=0.120
D) UCL=0.061
To determine the upper control limit for a p-chart, we need the following information:
- p̂: the average proportion of defects over the 20 weeks.
- UCL: the upper control limit for the proportion of defects.
To find p̂, we add up the number of defects and divide by the total number of observations (20 * 100 = 2000):
p̂ = (4 + 4 + 5 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 6 + 12 + 6 + 6 + 6 + 7 + 6 + 7 + 8 + 7) / 2000 = 0.0605
To find UCL, we use the following formula:
UCL = p̂ + 3√((p̂(1-p̂))/n)
where n is the sample size (100).
UCL = 0.0605 + 3√((0.0605(1-0.0605))/100) = 0.1203
Therefore, the correct upper control limit for the proportion of defects is 0.120, or approximately 12.0%.
The answer is C) UCL=0.120.