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raketka [301]
3 years ago
7

Process control is nothing more or less than monitoring quality as the work is being performed. Perhaps the most widely adopted

approach to process control is the use of control charts. Control charts can be developed and used to manage both characteristics (X-bar charts) and attributes (p-charts). Further, control charts can and should be used throughout the value-added process; that is, in purchasing, production, and logistics operations. Similarly, control charts can be used to improve quality in both manufacturing and service environments.
Three types of information must be available to set up a control chart. These are 1) a desired target, 2) a measure of variability around the target, and 3) a managerial assessment of required quality level. The target generally takes the form of a mean or average; the variability is usually measured via a variance (standard deviation); and the assessment of required quality level is expressed using a z-score. It is important to remember that the measure of variability be expressed in terms consistent with the "items" being plotted on the control chart--this usually requires that a standard deviation be transformed into a standard error. When using a control chart, the process operator looks for occurrences when the process is out of control (exceeds control limits) or when "bad" trends emerge. Over time, the causes of these occurrences can be identified and eliminated such that the process is improved continually. Finally, remember that the process must be in control when the control chart is set up (otherwise GIGO!).
For the following process, answer the following questions.
1. What is the grand mean or target?
2. What is the standard deviation?
3. What is the standard error?
4. With a Z-score of 3, what are the upper and lower limits?

Twenty samples of four items each were taken to set up the control chart. The design target = 2.80 cm.

Sample Measures within the Sample
1 2.81 2.79 2.78 2.80
2 2.81 2.79 2.80 2.83
3 2.79 2.77 2.80 2.81
4 2.81 2.78 2.79 2.81
5 2.82 2.79 2.80 2.78
6 2.78 2.79 2.82 2.80
7 2.80 2.80 2.81 2.79
8 2.79 2.80 2.80 2.82
9 2.78 2.78 2.81 2.80
10 2.81 2.79 2.80 2.81
11 2.78 2.82 2.79 2.80
12 2.80 2.82 2.81 2.79
13 2.78 2.79 2.78 2.81
14 2.81 2.82 2.80 2.79
15 2.78 2.82 2.81 2.79
16 2.79 2.80 2.79 2.82
17 2.80 2.81 2.80 2.78
18 2.80 2.80 2.79 2.82
19 2.78 2.82 2.79 2.79
20 2.77 2.81 2.82 2.80
Mathematics
1 answer:
ella [17]3 years ago
8 0

Answer:

a. Grand Mean = 2.799

b. Standard Deviation = 0.01387

c. Standard Error = 0.0016

d. The range with a Z score of 3 = (2.75739, 2.84061)

Step-by-step explanation:

a. Grand mean :

Given that

Sum of 80 values = 223.91

Total observation = 80

Grand mean = 223.91/80 = 2.799

b. With the excel command "STDEV[range of cells]", we get the standard deviation as = 0.01387

c. Standard error =  Standard deviation / root of n

= 0.01387/\sqrt{80}\\ = 0.0016

d. The range with a Z score of 3 :

lower limit = mean - (3 * SD) = 2.799 - (0.04161) = 2.75739

upper limit = mean + (3 * SD) = 2.799 + (0.04161) = 2.84061

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Complete each statement with a number that makes the statement true.
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Answer:

Step-by-step explanation:

1). (d). 5°C < 7°C

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2 years ago
How do you do the equation 3.25 divided by 5.2
valentina_108 [34]
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5 0
3 years ago
The Insure.com website reports that the mean annual premium for automobile insurance in the United States was $1,503 in March 20
stira [4]

Answer:

a) Option A)

H_{0}: \mu \geq 1503\\H_A: \mu < 1503

b) Point estimate of difference = -78

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $1,503

Sample mean, \bar{x} = $1,425

Sample size, n = 25

Sample standard deviation, s = $160

We have to carry a hypothesis test that the mean annual premium in Pennsylvania is lower than the national mean annual premium.

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5 0
3 years ago
The domain and range
larisa86 [58]
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hope it helps

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