1 / 75

750 likes | 930 Views

Regression Analysis. Defense Resources Management Institute. Unscheduled Maintenance Issue:. 36 flight squadrons Each experiences unscheduled maintenance actions (UMAs) UMAs costs $1000 to repair, on average. You’ve got the Data… Now What?. Unscheduled Maintenance Actions (UMAs).

Download Presentation
## Regression Analysis

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**Regression Analysis**Defense Resources Management Institute**Unscheduled Maintenance Issue:**• 36 flight squadrons • Each experiences unscheduled maintenance actions (UMAs) • UMAs costs $1000 to repair, on average.**You’ve got the Data… Now What?**Unscheduled Maintenance Actions (UMAs)**What do you want to know?**• How many UMAs will there be next month? • What is the average number of UMAs ?**UMAs Next Month**95% Confidence Interval**Average UMAs**95% Confidence Interval**Model: Cost of UMAs for one squadron**If the cost per UMA = $1000, the Expected cost for one squadron = $60,000**Model: Total Cost of UMAs**Expected Cost for all squadrons = 60 * $1000 * 36 = $2,160,000**Model: Total Cost of UMAs**Expected Cost for all squadrons = 60 * $1000 * 36 = $2,160,000 How confident are we about this estimate?**~ 95%**mean (=60) standard error =12/36 = 2**~56 ~58 60 ~62 ~64**(1 standard unit = 2) ~ 95%**95% Confidence Interval on our estimate of UMAs and costs**• 60 + 2(2) = [56, 64] • low cost: 56 * $1000 * 36 = $2,016,000 • high cost: 64 * $1000 * 36 = $2,304,000**What do you want to know?**• How many UMAs will there be next month? • What is the average number of UMAs ? • Is there a relationship between UMAs and and some other variable that may be used to predict UMAs? • What is that relationship?**Relationships**• What might be related to UMAs? • Pilot Experience ? • Flight hours ? • Sorties flown ? • Mean time to failure (for specific parts) ? • Number of landings / takeoffs ?**Regression:**• To estimate the expected or mean value of UMAs for next month: • look for a linear relationship between UMAs and a “predictive” variable • If a linear relationship exists, use regression analysis**Regression analysis:**describes and evaluates relationships between one variable (dependent or explained variable), and one or more other variables (called the independent or explanatory variables).**What is a good estimating variable for UMAs?**• quantifiable • predictable • logical relationship with dependent variable • must be a linear relationship: Y = a + bX**Describing the Relationship**• Is there a relationship? • Do the two variables (UMAs and sorties or experience) move together? • Do they move in the same direction or in opposite directions? • How strong is the relationship? • How closely do they move together?**Correlation Coefficient**• Statistical measure of how closely two variables are moving together in a coordinated fashion • Measures strength and direction • Value ranges from -1.0 to +1.0 • +1.0 indicates “perfect” positive linear relation • -1.0 indicates “perfect” negative linear relation • 0 indicates no relation between the two variables**Sorties vs. UMAs**r = .9788**Experience vs. UMAs**r = .1896**A Word of Caution...**• Correlation does NOT imply causation • It simply measures the coordinated movement of two variables • Variation in two variables may be due to a third common variable • The observed relationship may be due to chance alone**What is the Relationship?**• In order to use the correlation information to help describe the relationship between two variables we need a model • The simplest one is a linear model:**One Possibility**Sum of errors = 0**Another Possibility**Sum of errors = 0**Which is Better?**• Both have sum of errors = 0 • Compare sum of absolute errors:**One Possibility**Sum of absolute errors = 6**Another Possibility**Sum of absolute errors = 6**Which is Better?**• Sum of the absolute errors are equal • Compare sum of errors squared:**The Correct Relationship: Y = a + bX + U**Y systematic random 100 90 80 70 60 50 X 100 110 120 130**The correct relationship:**• Y = a + bX + U Y systematic random 100 90 80 70 60 50 X 100 110 120 130**Least-Squares Method**• Penalizes large absolute errors • Y- intercept: • Slope:**Assumptions**• Linear relationship: • Errors are random and normally distributed with mean = 0 and variance = • Supported by Central Limit Theorem

More Related