Question

In: Operations Management

Q6. A quality analyst took the following samples as a pilot study to construct the mean...

Q6. A quality analyst took the following samples as a pilot study to construct the mean and range charts. The data below shows the weight in ounces. Construct the mean and range charts showing all control (action) –upper and lower limits, and warning upper and lower limits.

Sample Number

Item1

Item2

Item3

Item4

Sample 1

11.9

12.5

12.4

12.7

Sample 2

12.5

12.5

12.4

12.8

Sample 3

12.2

12.8

12.7

12

Sample 4

12.3

12.4

12.8

12.4

Sample 5

12.1

12.8

12.4

11.9

Sample 6

12.6

12.4

12.1

12.3

The following day, the analyst took another 6 samples to check if the process under control or not. Theses samples are as follows:

Sample Number

Item1

Item2

Item3

Item4

Sample 1

11.9

11.5

12.4

12.3

Sample 2

12.5

11.5

13.4

11.8

Sample 3

12.2

11.8

11.7

12

Sample 4

14.3

11.4

11.8

13.4

Sample 5

15.1

11.8

12.4

10.9

Sample 6

11.6

10.4

10.1

15.3

Is the process under control? Why? What actions should you take?

Factors for calculating control limits for control charts

For MEANS

Sample size (n)

Constant (dn)

Factors for warning limits 1.96/dn√n

Factors for action limits 3.09√n

2

1.128

1.23

1.94

3

1.693

0.67

1.05

4

2.059

0.48

0.75

5

2.326

0.38

0.59

6

2.334

0.32

0.50

7

2.704

0.27

0.43

8

2.847

0.24

0.38

9

2.970

0.22

0.35

10

3.078

0.20

0.32

Factors for calculating control limits for control charts

For RANGES

Sample size

Upper warning limit

Lower warning limit

Upper action limit

Lower action limit

2

2.81

0.04

4.12

0.00

3

2.17

0.18

2.98

0.04

4

1.93

0.29

2.57

0.10

5

1.81

0.37

2.34

0.16

6

1.72

0.42

2.21

0.21

7

1.66

0.46

2.11

0.26

8

1.62

0.50

2.04

0.29

9

1.58

0.52

1.99

0.32

10

1.56

0.54

1.94

0.35

Solutions

Expert Solution

Sample No. Item1 Item2 Item3 Item4 Sample Mean Sample Range
Sample 1 11.90 12.50 12.40 12.70 12.38 0.80
Sample 2 12.50 12.50 12.40 12.80 12.55 0.40
Sample 3 12.20 12.80 12.70 12.00 12.43 0.80
Sample 4 12.30 12.40 12.80 12.40 12.48 0.50
Sample 5 12.10 12.80 12.40 11.90 12.30 0.90
Sample 6 12.60 12.40 12.10 12.30 12.35 0.50
Avg. 12.4125 0.6500
Mean chart Range chart
Sample size 4 4
Action limits factor 0.75
UCL 12.900 1.671
LCL 11.925 0.065
CL 12.413 0.650
Warning limits factor 0.48
UAL 12.725 1.255
LAL 12.101 0.189

Using the above [LCL, UCL] and [LAL, UAL] ranges, plot the second set of samples as follows:

Sample Number Item1 Item2 Item3 Item4 Sample Mean Sample Range
Sample 1 11.90 11.50 12.40 12.30 12.03 0.90
Sample 2 12.50 11.50 13.40 11.80 12.30 1.90
Sample 3 12.20 11.80 11.70 12.00 11.93 0.50
Sample 4 14.30 11.40 11.80 13.40 12.73 2.90
Sample 5 15.10 11.80 12.40 10.90 12.55 4.20
Sample 6 11.60 10.40 10.10 15.30 11.85 5.20

The process is out of control both in mean and range charts.


Related Solutions

A quality analyst wants to construct a sample mean chart for controlling a packaging process. Each...
A quality analyst wants to construct a sample mean chart for controlling a packaging process. Each day last week, he randomly selected six packages and weighed each. The data from that activity appears below.                   Weight (in grams) Day Package 1 Package 2 Package 3 Package 4 Package 5 Package 6 Monday 23 22 23 24 23 21 Tuesday 23 21 19 21 24 20 Wednesday 20 19 20 21 19 20 Thursday 18 19 20 19 21 22 Friday...
A quality analyst wants to construct a sample mean chart for controlling a packaging process. He...
A quality analyst wants to construct a sample mean chart for controlling a packaging process. He knows from past experience that whenever this process is in control, package weight is normally distributed with a mean of 20 ounces and a standard deviation of two ounces. Each day last week, he randomly selected four packages and weighed each: Day Weight (ounces) Monday 23 22 23 24 Tuesday 23 21 19 21 Wednesday 20 19 20 21 Thursday 18 19 20 19...
A study took random samples from ocean beaches. A total of 40 samples of size 250...
A study took random samples from ocean beaches. A total of 40 samples of size 250 mL were taken, and the mean number of plastic microparticles of sediment was 18.3 with a standard deviation of 8.2. Test for evidence that the mean number of microparticles is different from acceptable levels 15. What are the null and alternative hypotheses for this test?       Calculate the hypothesis test standard error. Calculate the t test statistic. Show all work. Draw/label/shade a curve. Be sure...
To investigate water quality, in early September 2016, the Ohio Department of Health took water samples...
To investigate water quality, in early September 2016, the Ohio Department of Health took water samples at 24 beaches on Lake Erie in Erie County. Those samples were tested for fecal coliform, which is the E.coli bacteria found in human and animal feces. An unsafe level of fecal coliform means there is a higher chance that disease‑causing bacteria are present and more risk that a swimmer will become ill if she or he should accidentally ingest some of the water....
To investigate water quality, in early September 2016, the Ohio Department of Health took water samples...
To investigate water quality, in early September 2016, the Ohio Department of Health took water samples at 2424 beaches on Lake Erie in Erie County. Those samples were tested for fecal coliform, which is the E.coli bacteria found in human and animal feces. An unsafe level of fecal coliform means there is a higher chance that disease‑causing bacteria are present and more risk that a swimmer will become ill if she or he should accidentally ingest some of the water....
1) A financial analyst would like to construct a 95% confidence interval for the mean earnings...
1) A financial analyst would like to construct a 95% confidence interval for the mean earnings of a company. The company's earnings have a standard deviation of $12 million. What is the minimum sample size required by the analyst if he wants to restrict the margin of error to $2 million? 2) A budget airline wants to estimate what proportion of customers would pay $10 for in-flight wireless access. Given that the airline has no prior knowledge of the proportion,...
Q1: A study of environmental air quality measured suspend particular matter in air samples at two...
Q1: A study of environmental air quality measured suspend particular matter in air samples at two sites. DATA is listed in the table. Site 1 22 68 36 32 42 24 28 38 40 Site 2 38 34 36 40 39 34 33 32 37 (a) Calculate the mean and standard deviation for each group. (5) (b) Test the hypotheses that air quality for two sites are different. (5) Level of Significance= 0.05
Eleven students and 14 professors took part in a study to find mean commuting distances. The...
Eleven students and 14 professors took part in a study to find mean commuting distances. The mean number of miles traveled by students was 5.6 and the standard deviation was 2.8. The mean number of miles traveled by professors was 14.3 and the standard deviation was 9.1. Perform a test of the null hypothesis that the population variances are equal.
construct an AOV table including expected mean squares in R studio A study was designed to...
construct an AOV table including expected mean squares in R studio A study was designed to evaluate the effectiveness of new treatments to reduce the systolic blood pressure of patients determined to have high blood pressure. Three drugs were selected for evaluation. There are numerous nondrug treatments for reducing blood pressure including various combinations of a controlled diet, exercise programs, biofeedback and so on. the researchers randomly selected three non drug treatments for examination in the study. The age of...
The following data were recorded as part of a study on sustainable farming techniques that took...
The following data were recorded as part of a study on sustainable farming techniques that took place in Boone County, IA. (Davis, Adam S. et al. Increasing Cropping System Diversity Balances Productivity, Profitability and Environmental Health. PLOS ONE. October 10, 2012. DOI:10.1371/journal.pone.0047149) Means are for the April-November growing seasons. (1): Construct a two-way scatter plot for “air temperature” against the “total precipitation” and on a separate graph construct a two-way scatter plot for “air temperature” against “log of total precipitation”....
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT