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Control charts will be covered in detail in a future class. They are a very important tool for assessing stability of a process over time and have application in all phases of a project.
One of several issues identified by the project team is customer complaints due to shipping. The CTQ is on‐time shipment and the KPOV being measured is the number of days late.
The customers have requested that all shipments be on time and nothing be shipped earlier that 5 days.
Using the above file, calculate the days late from the SHIP DATE – DUE DATE using the Minitab calculator function Calc >> Calculator
Using the tools you learned in this module, analyze the data to determine if the variation in Cycle Time is due to shape, center, spread, stability, or some combination.Be prepared to share your results with the class.
The data represents shipping performance compared to due date.. Ship Delta is the difference between the ship date and due date measure in number of days. A negative value represents an early shipment, positive is late.
The data represents shipping performance compared to due date.. Ship Delta is the difference between the ship date and due date measure in number of days. A negative value represents an early shipment, positive is late.
File: SHIPPING PERFORMANCE .xlsGraphical Tools >> Histogram and Descriptive Statistics
The data represents shipping performance compared to due date.. Ship Delta is the difference between the ship date and due date measure in number of days. A negative value represents an early shipment, positive is late.
File: SHIPPING PERFORMANCE .xlsAnalyze >> Distribution
Launch JMP and create a new data table.Compute and display the normal probability.
1. Add one row to the data table2. Double-click the column header, "Column 1."3. In the dialog box that follows, select "New
Property --> Formula." Screenshot:4. Click "Edit Formula."5. In the dialog box that follows, double-click on
the words, "no formula." Enter "Normal Distribution(1.0)" Note: JMP requires that the user enter a z-value rather than x, mean, and standard deviation.
6. Note: When you finish editing the function, JMP will display it in a graphically-editable form. Functions can also be entered graphically using the "Functions" list and the arithmetic buttons. This approach is more intuitive, though more time-consuming. The "Normal Distribution" function is found under "Probability" when the functions are shown grouped.
7. Click "OK" to close the function dialog box. 8. Click "OK" in the column properties dialog box. 9. Repeat Step 5 and enter (-2.0)
JMP computes the probability and displays it in the first
row of Column 1.Above upper spec = 1 - .8413 = .1587 or 15.87%
Below Lower spec = .0228 or 2.28%
Amount out of spec = .1587 + .0228 = .1815 or 18.15%
A project is looking at the number of discrepancies in company travel expense reports with the intent of reducing the cycle time to process the expenses.
A project is looking at the number of discrepancies in company travel expense reports with the intent of reducing the cycle time to process the expenses.
File: Pareto.mtwAnalyze >> Quality and Process >> Pareto Plot
A project is looking at the number of discrepancies in company travel expense reports with the intent of reducing the cycle time to process the expenses.
Selling price for used cars of a particular model and option package were captured. Construct a scatter plot to see the relationship between the 2 variables.
This is typical of data collected in Excel and it will need to be stacked by rows.Create columns Age (yrs) and Car Price
Selling price for used cars of a particular model and option package were captured. Construct a scatter plot to see the relationship between the 2 variables.
Data was collected for 4 cars at each age. This is typical of data collected in Excel and it will need to be stacked by rows.
Selling price for used cars of a particular model and option package were captured. Construct a scatter plot to see the relationship between the 2 variables.
This is typical of data collected in Excel and it will need to be transposed by rows.Create columns Age (yrs) and Car PriceTables >> Transpose
The histogram shows a distribution which appears “normal”’ or bell=shaped. However, the P-Value for the normality test (P less than .05 means non-normal) shows a possible bi-modal distribution. Since the sample size is large (>30) there is not much tolerance for minor departures from normality. This passes the “eyeball” test so assume normality and go on, but look for reasons for this distribution
Percent above upper spec = 1 - .802 = .198
Percent below lower spec = .250
Total out of spec = .198 + .250 = .448or 44.8% defective
Minitab 17
NOTE – the non-statistical thinker would look at the data and note that the mean is less than zero, and might “observe” that the average shipping time is actually early – so what’s the problem?
A box plot using ship day of the week as a grouping variable shows a possible source of the clustering. Shipments on Thursday and Friday tend to be later than the rest of the week.
Minitab 17
The run chart shows a cyclic variation. The P-values (less than .05 is significant) show that the data is clustered and shows a lack of randomness. This could account for the abnormality with the distribution.
Looking at the Order Entry data, a Pareto chart shows that “No Problem” has the largest number –nothing interesting here.
A Box Plot using status as the grouping variable shows that pricing is the biggest contributor to the data entry time. Credit and new part also contribute.
The histogram shows adistribution which appears“normal”’ or bell=shaped.However, the P-Value forthe normality test (P lessthan .05 means non-normal) shows a possiblebi-modal distribution. Sincethe sample size is large(>30) there is not muchtolerance for minordepartures from normality.This passes the “eyeball”test so assume normalityand go on, but look forreasons for this distribution
The histogramshows a distributionwhich appears“normal”’ orbell=shaped.However, the P-Value for thenormality test (P lessthan .05 means non-normal) shows apossible bi-modaldistribution. Sincethe sample size islarge (>30) there isnot much tolerancefor minor departuresfrom normality.
The run chart shows a cyclic variation. The P-values (less than .05 is significant) show that the data is clustered and shows a lack of randomness.
A box plot using ship day of the week as a grouping variable shows a possible source of the clustering. Shipments on Thursday and Friday tend to be later than the rest of the week.