Question 1

If $\bar x$ = 23 ounces, ? = 0.4 ounces, and n = 16, the ±3? control limits will be
A) 21.8 to 24.2 ounces.
B) 23 ounces.
C) 22.70 to 23.30 ounces.
D) 22.25 to 23.75 ounces.
E) 23.3 to 25.7 ounces.

Accepted Answer

The control limits are calculated as $\bar x \pm 3(\frac{\sigma}{\sqrt{n}})$. Here, $\sigma = 0.4$, $n = 16$, so the standard error is $\frac{0.4}{\sqrt{16}} = 0.1$. Thus, the control limits are $23 \pm 3(0.1) = 22.70$ to $23.30$ ounces.

Question 2

Consumer's risk is the probability of
A) accepting a good lot.
B) rejecting a good lot.
C) rejecting a bad lot.
D) accepting a bad lot.
E) shipping a bad lot.

Accepted Answer

Consumer's risk is the probability of accepting a bad lot, which means the risk of consumers accepting a product that is actually of low quality or defective.

Question 3

To set $\bar x$ -chart upper and lower control limits, one must know the process central line, which is the
A) average of the sample means.
B) total number of defects in the population.
C) percent defects in the population.
D) size of the population.
E) average range.

Accepted Answer

The process central line on an $\bar x$ -chart, which is used for monitoring the mean of a process, is determined by the average of the sample means. This provides a baseline from which the control limits are calculated.

Question 4

Up to three standard deviations above or below the centerline is the amount of variation that statistical process control allows for
A) Type I errors.
B) about 95.5% variation.
C) natural variation.
D) all types of variation.
E) assignable variation.

Accepted Answer

The amount of variation allowed for in statistical process control is known as natural variation, which is measured by the spread around the centerline in terms of standard deviations. Typically, the range of acceptable variation is considered to be up to three standard deviations above or below the centerline. This range accounts for about 99.7% of all data points, leaving only about 0.3% outside this range as outliers. Therefore, the correct answer is C, natural variation.

Question 5

Statistical process control charts
A) display the measurements on every item being produced.
B) display upper and lower limits for process variables or attributes, and signal when a process is no longer in control.
C) indicate to the process operator the average outgoing quality of each lot.
D) indicate to the operator the true quality of material leaving the process.
E) display the measurements on every item being purchased.

Accepted Answer

Statistical process control charts display upper and lower limits for process variables or attributes, and signal when a process is no longer in control. They are used to monitor and control a process to ensure that it is producing products within specification limits. They do not display measurements of every item being produced or purchased, indicate the average outgoing quality of each lot, or display the true quality of material leaving the process.

Question 6

The usual purpose of an R-chart is to signal whether there has been a
A) gain or loss in dispersion.
B) change in the percent defective in a sample.
C) change in the central tendency of the process output.
D) change in the number of defects in a sample.
E) change in sample size.

Accepted Answer

An R-chart is used to monitor the variability or dispersion of a process over time. It shows whether there has been a gain or loss in dispersion by plotting the range of each subgroup. This helps in identifying if there is any change in the spread of the process output.

Question 7

Ten samples of a process measuring the number of returns per 100 receipts were taken for a local retail store. The number of returns were 10, 9, 11, 7, 3, 12, 8, 4, 6, 11. Find the standard deviation of the sampling distribution. (Hint- Use p-bar formula)
A) There is not enough information.
B) .081
C) 8.1
D) .0273
E) .0863

Accepted Answer

The formula to calculate the standard deviation of the sampling distribution using p-bar is:
SD = sqrt [p-bar(1-p-bar)/n], where p-bar is the sample proportion and n is the sample size.

To find p-bar (sample proportion), we need to add up the number of returns in all the samples and divide by the total number of receipts in all the samples:
p-bar = (10+9+11+7+3+12+8+4+6+11) / (10*100) = 0.0805

Now, we can plug in the values in the formula:
SD = sqrt [0.0805*(1-0.0805)/10*100] = 0.0273

Therefore, the standard deviation of the sampling distribution is 0.0273, which is answer choice D.

Question 8

Which of the following is true about cutting costs at Canada Bankers Life Assurance Company?
A) They outsourced their call centre.
B) Customer satisfaction increased.
C) Customer service improved.
D) Quality metrics showed improvements.
E) Wastage decreased.

Accepted Answer

Outsourcing the call centre is a common cost-cutting measure that companies take to reduce expenses.

Question 9

A manager wishes to build a 3-sigma range chart for a process. The sample size is five, the mean of sample means is 16.01, and the average range is 5.3. From Table S6.1 in the text, the appropriate value of D₃ is 0, and D₄ is 2.115. The UCL and LCL for this range chart are A) 33.9 and 11.2. B) 33.9 and 0. C) 11.2 and 0. D) 6.3 and 0. E) 31.91 and 0.11.

Accepted Answer

The UCL and LCL for the range chart can be calculated using the formula: UCL = D4 * R_bar LCL = D3 * R_bar where D4 and D3 are constants from Table S6.1, and R_bar is the average range. From the problem statement, the average range is given as 5.3, and the appropriate value of D₃ is 0, and D₄ is 2.115. Therefore: UCL = 0 * 5.3 = 0 LCL = 2.115 * 5.3 = 11.2 Thus, the best choice is C. The UCL for the range chart is 0, and the LCL is 11.2.

Question 10

A manager wants to build 3-sigma control limits for a process. The target value for the mean of the process is 10 units, and the standard deviation of the process is 6. If samples of size 9 are to be taken, the UCL and LCL will be
A) -8 and 28.
B) 16 and 4.
C) 12 and 8.
D) 4 and 16.
E) 8 and 12.

Accepted Answer

The 3-sigma control limits are calculated using the formula: mean ± 3*(standard deviation/sqrt(sample size)). Here, mean = 10, standard deviation = 6, and sample size = 9. Thus, UCL and LCL are 10 ± 3*(6/sqrt(9)) = 10 ± 6 = 16 and 4, respectively.

Question 11

The mean and standard deviation for a process for which we have a substantial history are μ = 120 and σ = 2. For the x-bar chart, a sample size of 16 will be used. What is the mean of the sampling distribution?
A) 1/8 (0.125)
B) 7.5
C) 2
D) 40
E) 120

Accepted Answer

The mean of the sampling distribution is equal to the mean of the process, which is given as 120.

Question 12

According to the text, the most common choice of limits for control charts is usually
A) ± 1 standard deviation.
B) ± 2 standard deviations.
C) ± 3 standard deviations.
D) ± 3 standard deviations for means and ± 2 standard deviations for ranges.
E) ± 1 standard deviations for means and ± 2 standard deviations for ranges.

Accepted Answer

The most common choice for control chart limits is ± 3 standard deviations, as this captures 99.73% of the data if the underlying distribution is normal. This provides a practical balance between detecting meaningful changes in process performance while minimizing false alarms. Other choices (such as ± 2 standard deviations) are less sensitive to changes in process performance, while choices that vary between mean and range limits can be more difficult to use effectively.

Question 13

For an x-bar chart where the standard deviation is known, the Upper Control Limit
A) is 3 . ? below the mean of sample means for a 3? control chart.
B) is 3 . ? above the mean of sample means for a 3? control chart.
C) is 3 . ?/ $\sqrt { n }$ below the mean of sample means for a 3? control chart.
D) is 3 . ?/ $\sqrt { n }$ above the mean of sample means for a 3? control chart.
E) Cannot be calculated unless the average range is known.

Accepted Answer

The Upper Control Limit (UCL) for an x-bar chart, when the population standard deviation (σ) is known, is calculated as the mean of sample means plus 3 times the standard deviation of the sampling distribution of the mean. The standard deviation of the sampling distribution of the mean is given by σ/√n, where n is the sample size. Therefore, the UCL is 3σ/√n above the mean of sample means.

Question 14

Which of the following is true of a p-chart? A) The lower control limit is found by subtracting a fraction from the average number of defects. B) The lower control limit indicates the minimum acceptable number of defects. C) The lower control limit equals D₃ times p-bar. D) The lower control limit may be at zero. E) The lower control limit is the same as the lot tolerance percent defective.

Accepted Answer

The lower control limit on a p-chart may be at zero because it is based on the variability in proportion of defects in the sample, not a fixed minimum acceptable number. The formula for calculating the lower control limit is LCL = p-bar - 3sqrt(p-bar(1-p-bar)/n), where p-bar is the average proportion of defects and n is the sample size.

Question 15

The p-chart tells us whether there has been a
A) gain or loss in dispersion.
B) change in the percent defective in a sample.
C) change in the central tendency of the process output.
D) change in the number of defects in a sample.
E) none of the above.

Accepted Answer

The p-chart is used to monitor the proportion of nonconforming units in a sample, so it detects changes in the percent defective in a sample. The other options are not relevant to the p-chart.

Question 16

Jars of pickles are sampled and weighed. Sample measures are plotted on control charts. The ideal weight should be precisely 11 oz. Which type of chart(s) would you recommend?
A) p-charts
B) c-charts
C) $\bar x$ and R-charts
D) $\bar x$ -, but not R-charts
E) both p- and c-charts

Accepted Answer

$\bar x$ and R-charts are used for monitoring the mean and variability of a process where the measurements are continuous, such as the weight of pickle jars.

Question 17

Plots of sample ranges indicate that the most recent value is below the lower control limit. What course of action would you recommend?
A) Since there is no obvious pattern in the measurements, variability is in control.
B) One value outside the control limits is insufficient to warrant any action.
C) Lower than expected dispersion is a desirable condition; there is no reason to investigate.
D) The process is out of control; reject the last units produced.
E) Variation is not in control; investigate what created this condition.

Accepted Answer

A value below the lower control limit indicates that the process is exhibiting excessive variability. Therefore, it is best to investigate what is causing this condition and take corrective action. Option A is incorrect because just because there is no obvious pattern does not mean that the variability is in control. Option B is incorrect because even one value outside the control limits indicates that the process is out of control. Option C is incorrect because lower-than-expected dispersion is not desirable if it indicates out-of-control variability. Option D is incorrect because one single data point is not enough to justify rejecting all units produced.

Question 18

An x-bar control chart was examined and no data points fell outside of the limits. Can this process be considered in control?
A) No, there could be a pattern to the points, the R-chart must be checked.
B) Yes, it is unlikely that there is a pattern in these points.
C) No, the number of samples must be known.
D) Yes.
E) Yes, there is not enough data to determine otherwise.

Accepted Answer

An x-bar control chart showing no data points outside the limits does not guarantee the process is in control. Patterns within the control limits or issues in the range (R-chart) could indicate a lack of control.

Question 19

The Central Limit Theorem
A) is the theoretical foundation of the c-chart.
B) states that the average of assignable variations is zero.
C) allows managers to use the normal distribution as the basis for building some control charts.
D) states that the average range can be used as a proxy for the standard deviation.
E) controls the steepness of an operating characteristic curve.

Accepted Answer

The Central Limit Theorem states that regardless of the distribution of the population, the distribution of the sample means will approach a normal distribution as sample size increases. This allows managers to use the normal distribution as the basis for building control charts, such as the X-bar and R charts.

Question 20

The type of inspection that classifies items as being either good or defective is
A) variable inspection.
B) attribute inspection.
C) fixed inspection.
D) all of the above.
E) none of the above.

Accepted Answer

Attribute inspection is the type of inspection that classifies items as being either good or defective based on specific requirements or characteristics. It is a pass-fail type of inspection and does not consider the degree of variation in the measured characteristic, which is the characteristic of variable inspection. Fixed inspection refers to a scheduled inspection that is performed at a predetermined time or event, and may be either attribute or variable inspection. Therefore, the best choice is B) attribute inspection.

Quiz 26: Statistical Process Control

If xˉ\bar xxˉ = 23 ounces, ? = 0.4 ounces, and n = 16, the ±3? control limits will be

Consumer's risk is the probability of

To set xˉ\bar xxˉ -chart upper and lower control limits, one must know the process central line, which is the

Up to three standard deviations above or below the centerline is the amount of variation that statistical process control allows for

Statistical process control charts

The usual purpose of an R-chart is to signal whether there has been a

Ten samples of a process measuring the number of returns per 100 receipts were taken for a local retail store. The number of returns were 10, 9, 11, 7, 3, 12, 8, 4, 6, 11. Find the standard deviation of the sampling distribution. (Hint- Use p-bar formula)

Which of the following is true about cutting costs at Canada Bankers Life Assurance Company?

A manager wishes to build a 3-sigma range chart for a process. The sample size is five, the mean of sample means is 16.01, and the average range is 5.3. From Table S6.1 in the text, the appropriate value of D3 is 0, and D4 is 2.115. The UCL and LCL for this range chart are

A manager wants to build 3-sigma control limits for a process. The target value for the mean of the process is 10 units, and the standard deviation of the process is 6. If samples of size 9 are to be taken, the UCL and LCL will be

The mean and standard deviation for a process for which we have a substantial history are μ = 120 and σ = 2. For the x-bar chart, a sample size of 16 will be used. What is the mean of the sampling distribution?

According to the text, the most common choice of limits for control charts is usually

For an x-bar chart where the standard deviation is known, the Upper Control Limit

Which of the following is true of a p-chart?

The p-chart tells us whether there has been a

Jars of pickles are sampled and weighed. Sample measures are plotted on control charts. The ideal weight should be precisely 11 oz. Which type of chart(s) would you recommend?

Plots of sample ranges indicate that the most recent value is below the lower control limit. What course of action would you recommend?

An x-bar control chart was examined and no data points fell outside of the limits. Can this process be considered in control?

The Central Limit Theorem

The type of inspection that classifies items as being either good or defective is

If $\bar x$ = 23 ounces, ? = 0.4 ounces, and n = 16, the ±3? control limits will be

To set $\bar x$ -chart upper and lower control limits, one must know the process central line, which is the

A manager wishes to build a 3-sigma range chart for a process. The sample size is five, the mean of sample means is 16.01, and the average range is 5.3. From Table S6.1 in the text, the appropriate value of D₃ is 0, and D₄ is 2.115. The UCL and LCL for this range chart are