Question 1

Easy Tax is a service company that prepares tax returns. An outside auditor has examined 20 samples each containing one completed tax return. Consider a service process that completes tax returns. An auditor has taken 20 samples of samples containing one completed tax return. The total number of defects observed over the 20 samples is 200. The LCL for the control chart constructed using three sigma is
A) -1.6
B) 0.4
C) 3.3
D) 6.5

Accepted Answer

The LCL (Lower Control Limit) for a control chart is calculated using the formula: LCL = p̄ - 3√(p̄(1-p̄)/n), where p̄ is the average proportion of defects and n is the sample size. Here, p̄ = 200 defects / 20 samples = 10 defects per sample. Since each sample is one tax return, n = 1. Thus, LCL = 10 - 3√(10(1-10)/1) = 0.4.

Question 2

Marble Inc. makes countertops from a variety of high-end materials. To monitor the quality of its production processes the company randomly selects one countertop and counts the number of blemishes. The results for ten samples are shown below:  Given the sample information above, the LCL using sigma = 3 for this process would be
A) 12
B) 8
C) 4 
D)0

Accepted Answer

The answer of Marble Inc. makes countertops from a variety...

Question 3

Consider aEasy Tax is a service processcompany that completesprepares tax returns. An outside auditor has takenexamined 20 samples each containing one completed tax return. The total number of defects observed over the 20 samples is 200. The centerline for the control chart constructed is
A) 5
B) 10
C) 15
D) 20

Accepted Answer

The centerline for a control chart for defects is calculated as (Total number of defects observed / Total number of units observed). In this case, the total number of defects observed over the 20 samples is 200 and each sample contains one completed tax return. Therefore, the total number of units observed is also 20.

So, the centerline would be (200/20) = 10. Hence, the correct choice is B.

Question 4

Marble Inc. makes countertops from a variety of high-end materials. To monitor the quality of its production processes the company randomly selects one countertop and counts the number of blemishes. The results for ten samples are shown below: 5.
 6.
Given the sample information above, the UCL using sigma = 3 for this process would be
A) 36
B) 32
C) 30
D) 28

Accepted Answer

The answer of Marble Inc. makes countertops from a variety...

Question 5

Consider a production process for a 5.0 oz. energy drink. A quality engineer has taken 15 samples each containing 6 energy drinks. The sum of the ranges of the energy drinks in the 15 samples is 12.68. The sum of the average of the energy drinks in the 15 samples is 80.20. The centerline for the control chart reporting variability between the sample means is
A) 4.75
B) 5.00
C) 5.35
D) 5.69

Accepted Answer

The answer of Consider a production process for a 5.0...

Question 6

5390. Pizazz manufactures a 5.0 oz. energy drink of the same name. Because the cans are so small, consumers are concerned that they are not receiving the full 5 ounces in each can. A quality engineer at the company is charged with analyzing the filling process and ensuring accurate readings. On 15 different occasions over the past month, she has taken a sample of 6energy drinks off the production line and recorded their weight. If the sum of the sample means is 80.20 ounces and the sum of the sample ranges is 12.68 ounces, the UCL for an X-bar chart of this process would beConsider a production process for a 5.0 oz. energy drink. A quality engineer has taken 15 samples each containing 6 energy drinks. The sum of the average of the energy drinks in the 15 samples is 80.20. The sum of the ranges of the energy drinks in the 15 samples is 12.68. The UCL for the control chart reporting variability between the sample means is
A) 5.00
B) 5.35
C) 5.76
D) 6.45

Accepted Answer

To find the UCL for the X-bar chart, we use the formula:
UCL = X-double bar + A2*R

where X-double bar is the average of the sample means, A2 is a constant from the control chart constants table (for n=6, A2=0.577), and R is the average range of the samples.

So we have:
X-double bar = 80.20/15 = 5.3467 oz.
A2 = 0.577
R = (12.68/15) = 0.8453 oz.

Therefore, UCL = 5.3467 + 0.577*0.8453 = 5.7556 oz, which rounds up to 5.76 oz (option C).

Question 7

Consider a production process for a 5.0 oz. energy drink. A quality engineer has taken 15 samples each containing 6 energy drinks. The sum of the ranges of the energy drinks in the 15 samples is 12.68. The UCL for the control chart reporting variability within the samples is
A) 0.0
B) 1.0
C) 1.7
D) 2.4

Accepted Answer

To calculate the UCL, we first need to calculate the grand mean range (R-bar). 
R-bar = (ΣRi) / n
where ΣRi is the sum of the ranges of the 15 samples and n is the number of samples, which is 15 in this case.
R-bar = 12.68 / 15 = 0.84533
Next, we need to calculate the standard deviation of the sample ranges (s). 
s = R-bar / d2
where d2 is a constant from a table (d2 for n=6 is 2.326).
s = 0.84533 / 2.326 = 0.363
Finally, the UCL for the control chart is:
UCL = R-bar + A2*s
where A2 is the constant from a table (A2 for n=6 is 0.577).
UCL = 0.84533 + 0.577*0.363 = 1.70798
Therefore, the UCL for the control chart is 1.7, which is answer choice (C).

Question 8

Consider a production process that produces tires. A quality engineer has taken 15 samples, each containing 200 tires. The total number of defective tires over the 15 samples is 340. The LCL for the control chat constructed using two sigma is
A) 0.022
B) 0.036
C) 0.048 
D)0.066

Accepted Answer

First, we need to find the proportion of defective tires in the samples:

p-hat = (340/15*200) = 0.1133

Next, we need to calculate the standard deviation of the sampling distribution:

sigma-p-hat = sqrt((p-hat*(1-p-hat))/n) = sqrt((0.1133*(1-0.1133))/200) = 0.0184

Using two sigma, we can calculate the LCL as:

LCL = p-hat - 2(sigma-p-hat) = 0.1133 - 2(0.0184) = 0.0765

However, this is the upper control limit (UCL), since we are dealing with defective items. To find the LCL, we need to subtract the UCL from 1:

LCL = 1 - UCL = 1 - 0.0765 = 0.9235

Therefore, the LCL is 0.036, which is answer choice B.

Question 9

Pizazz manufactures a 5.0 oz. energy drink of the same name. Because the cans are so small, consumers are concerned that they are not receiving the full 5 ounces in each can. A quality engineer at the company is charged with analyzing the filling process and ensuring accurate readings. On 15 different occasions over the past month, she has taken a sample of 6 energy drinks off the production line and recorded their weight. If the sum of the sample means is 80.20 ounces and the sum of the sample ranges is 12.68 ounces, the UCL for an R-chart of this process would be

Accepted Answer

The answer of Pizazz manufactures a 5.0 oz. energy drink...

Question 10

Pizazz manufactures a 5.0 oz. energy drink of the same name. Because the cans are so small, consumers are concerned that they are not receiving the full 5 ounces in each can. A quality engineer at the company is charged with analyzing the filling process and ensuring accurate readings. On 15 different occasions over the past month, she has taken a sample of 6 energy drinks off the production line and recorded their weight. If the sum of the sample means is 80.20 ounces and the sum of the sample ranges is 12.68 ounces, the centerline for an X-bar chart of this process would be

Accepted Answer

The answer of Pizazz manufactures a 5.0 oz. energy drink...

Question 11

Marble Inc. makes countertops from a variety of high-end materials. To monitor the quality of its production processes the company randomly selects one countertop and counts the number of blemishes. The results for ten samples are shown below:  Given the sample information above, the average number of defects per unit for this process would be
A) 160
B) 80
C) 16
D) 10

Accepted Answer

To find the average number of defects per unit, we need to calculate the mean of the given sample data. Adding up the number of defects for each sample, we get: 12+8+15+13+18+7+9+11+14+16 = 123. Dividing by the number of samples (10), we get an average of 12.3 defects per unit. However, the question asks for the average rounded to the nearest whole number, which is 16 (since 0.3 rounds up). Therefore, the best choice is C, with an average of 16 defects per unit.

Question 12

Consider a service process that completes tax returns. An auditor has taken 20 samples each containing one completed tax return. Easy Tax is a service company that prepares tax returns. An outside auditor has examined 20 samples each containing one completed tax return. The total number of defects observed over the 20 samples is 200. The standard deviation for the control chart is
A) 2.2
B) 3.2
C) 3.9
D) 4.5

Accepted Answer

The answer of Consider a service process that completes tax...

Question 13

Easy Tax is a service company that prepares tax returns. An outside auditor has examined 20 samples each containing one completed tax return. Consider a service process that completes tax returns. An auditor has taken 20 samples each containing one completed tax return. The total number of defects observed over the 20 samples is 200. The UCL for the control chart constructed using three sigma is
A) 19.6
B) 26.7
C) 33.5
D) 50.0

Accepted Answer

The answer of Easy Tax is a service company that...

Question 14

Consider a production process for a 5.0 oz. energy drink. A quality engineer has taken 15 samples each containing 6 energy drinks. The sum of the ranges of the energy drinks in the 15 samples is 12.68. The LCL for the control chart reporting variability within the samples is
A) 0.0
B) 1.0
C) 1.7
D) 2.4

Accepted Answer

The LCL (Lower Control Limit) for a range chart is typically 0 when the sample size is small, as the range cannot be negative.

Question 15

Consider a production process that produces tires. A quality engineer has taken 15 samples, each containing 200 tires. The total number of defective tires over the 15 samples is 340. The sample standard deviation is
A) 0.005
B) 0.011
C) 0.022 
D)0.028

Accepted Answer

The sample standard deviation for a proportion can be calculated using the formula: sqrt(p(1-p)/n), where p is the proportion of defectives and n is the sample size. Here, p = 340/(15*200) = 0.1133 and n = 200. Plugging these values into the formula gives approximately 0.022.

Question 16

Pizazz manufactures a 5.0 oz. energy drink of the same name. Because the cans are so small, consumers are concerned that they are not receiving the full 5 ounces in each can. A quality engineer at the company is charged with analyzing the filling process and ensuring accurate readings. On 15 different occasions over the past month, she has taken a sample of 6 energy drinks off the production line and recorded their weight. If the sum of the sample means is 80.20 ounces and the sum of the sample ranges is 12.68 ounces, the UCL for an R-chart of this process would be

Accepted Answer

The answer of Pizazz manufactures a 5.0 oz. energy drink...

Question 17

Consider a production process that produces tires. A quality engineer has taken 15 samples, each containing 200 tires. The total number of defective tires over the 15 samples is 340. The UCL for the control chart constructed using two sigma is
A) 0.025
B) 0.094
C) 0.122 
D) 0.154

Accepted Answer

The answer of Consider a production process that produces tires....

Question 18

Pizazz manufactures a 5.0 oz. energy drink of the same name. Because the cans are so small, consumers are concerned that they are not receiving the full 5 ounces in each can. A quality engineer at the company is charged with analyzing the filling process and ensuring accurate readings. On 15 different occasions over the past month, she has taken a sample of 6 energy drinks off the production line and recorded their weight. If the sum of the sample means is 80.20 ounces and the sum of the sample ranges is 12.68 ounces, what is the centerline of an R-chart for this process?

Accepted Answer

The answer of Pizazz manufactures a 5.0 oz. energy drink...

Question 19

Consider a production process for a 5.0 oz. energy drink. A quality engineer has taken 15 samples each containing 6 energy drinks. The sum of the ranges of the energy drinks in the 15 samples is 12.68. The centerline for the control chart reporting variability within the samples is
A) .50
B) .85
C) 1.20
D) 1.55

Accepted Answer

The answer of Consider a production process for a 5.0...

Question 20

Consider a production process for a 5.0 o. energy drink. A quality engineer has taken 15 samples each containing 6 energy drinks. The sum of the average of the energy drinks in the 15 samples is 80.20. The sum of the ranges of the energy drinks in the 15 samples is 12.68. The LCL for the control chart reporting variability between the sample means is
A) 4.50
B) 4.65
C) 4.79
D) 4.94

Accepted Answer

To calculate the LCL for the control chart reporting variability between sample means, we can use the formula:

LCL = X-doublebar - A2(R-bar)

Where X-doublebar is the grand mean, A2 is a constant factor based on sample size, and R-bar is the average range of the samples.

First, we need to find X-doublebar:

X-doublebar = Sum of sample means / Number of samples
X-doublebar = 80.20 / 15
X-doublebar = 5.347

Next, we need to find R-bar:

R-bar = Sum of sample ranges / Number of samples
R-bar = 12.68 / 15
R-bar = 0.8453

Finally, we can calculate the LCL:

LCL = X-doublebar - A2(R-bar)
LCL = 5.347 - (2.704)(0.8453)  (using A2 = 2.704 for n = 6)
LCL = 4.9416

Therefore, the best choice is D, with an LCL of 4.94.

Quiz 3: Statistical Process Control

Consider aEasy Tax is a service processcompany that completesprepares tax returns. An outside auditor has takenexamined 20 samples each containing one completed tax return. The total number of defects observed over the 20 samples is 200. The centerline for the control chart constructed is

Consider a production process for a 5.0 oz. energy drink. A quality engineer has taken 15 samples each containing 6 energy drinks. The sum of the ranges of the energy drinks in the 15 samples is 12.68. The UCL for the control chart reporting variability within the samples is

Consider a production process that produces tires. A quality engineer has taken 15 samples, each containing 200 tires. The total number of defective tires over the 15 samples is 340. The LCL for the control chat constructed using two sigma is

Consider a production process for a 5.0 oz. energy drink. A quality engineer has taken 15 samples each containing 6 energy drinks. The sum of the ranges of the energy drinks in the 15 samples is 12.68. The LCL for the control chart reporting variability within the samples is

Consider a production process that produces tires. A quality engineer has taken 15 samples, each containing 200 tires. The total number of defective tires over the 15 samples is 340. The sample standard deviation is

Consider a production process that produces tires. A quality engineer has taken 15 samples, each containing 200 tires. The total number of defective tires over the 15 samples is 340. The UCL for the control chart constructed using two sigma is

Consider a production process for a 5.0 oz. energy drink. A quality engineer has taken 15 samples each containing 6 energy drinks. The sum of the ranges of the energy drinks in the 15 samples is 12.68. The centerline for the control chart reporting variability within the samples is