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7/2/2015 IENG 486 Statistical Quality & Process Control 1 IENG 486 - Lecture 05 Interpreting Variation Using Distributions
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7/2/2015 IENG 486 Statistical Quality & Process Control2 Assignment: Reading: Chapter 1: (1.1, 1.3 – 1.4.5) Cursory – get Fig. 1.12., p.34; Deming Management,1.4.4 Liability Chapter 2: (2.2 – 2.7) Cursory – Define, Measure, Analyze, Improve, Control Chapter 3: (3.1, 3.3.1, 3.4.1) HW 1: Chapter 3 Exercises: 1, 3, 4 – using exam calculator 10 (use Normal Plots spreadsheet from Materials page) 43, 46, 47 (use Exam Tables from Materials page – Normal Dist.)
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7/2/2015 IENG 486 Statistical Quality & Process Control3 Distributions Distributions quantify the probability of an event Events near the mean are most likely to occur, events further away are less likely to be observed 35.0 2.5 37 ( ) 41.4 ( +2 ) 32.6 ( -2 ) 43.6 ( +3 ) 30.4 ( -3 ) 39.2 ( + ) 34.8 ( - )
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7/2/2015 IENG 486 Statistical Quality & Process Control4 Normal Distribution Notation: r.v. This is read: “x is normally distributed with mean and standard deviation .” Standard Normal Distribution r.v. (z represents a Standard Normal r.v.)
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7/2/2015 IENG 486 Statistical Quality & Process Control5 Simple Interpretation of Standard Deviation of Normal Distribution
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7/2/2015 IENG 486 Statistical Quality & Process Control6 Standard Normal Distribution The Standard Normal Distribution has a mean ( ) of 0 and a variance ( 2 ) of 1 (thus, standard deviation is also 1) Total area under the curve, (z), from z = – to z = is exactly 1 The curve is symmetric about the mean Half of the total area lays on either side, so: (– z) = 1 – (z) z (z)
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7/2/2015 IENG 486 Statistical Quality & Process Control7 Standard Normal Distribution How likely is it that we would observe a data point more than 2.57 standard deviations beyond the mean? Area under the curve from – to z = 2.57 is found by using the table on pp. 693-694, looking up the cumulative area for z = 2.57, and then subtracting the cumulative area from 1. z (z)
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7/2/2015 IENG 486 Statistical Quality & Process Control9 Standard Normal Distribution How likely is it that we would observe a data point more than 2.57 standard deviations beyond the mean? Area under the curve from – to z = 2.57 is found by using the table on pp. 693-694, looking up the cumulative area for z = 2.57, and then subtracting the cumulative area from 1. Answer: 1 –.99492 =.00508, or about 5 times in 1000 z (z)
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7/2/2015 IENG 486 Statistical Quality & Process Control10 What if the distribution isn’t a Standard Normal Distribution? If it is from any Normal Distribution, we can express the difference from a sample mean to the population mean in units of the standard deviation, and this converts it to a Standard Normal Distribution. Conversion formula is: where: x is the sample location point, is the population mean, and is the population standard deviation.
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7/2/2015 IENG 486 Statistical Quality & Process Control11 What if the distribution isn’t even a Normal Distribution? The Central Limit Theorem allows us to take the sum of several means, regardless of their distribution, and approximate this sum using the Normal Distribution if the number of observations is large enough. Most assemblies are the result of adding together components, so if we take the sum of the means for each component as an estimate for the entire assembly, we meet the CLT criteria. If we take the mean of a sample from a distribution, we meet the CLT criteria (think of how the mean is computed).
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7/2/2015 IENG 486 Statistical Quality & Process Control12 Example: Process Yield Specifications are often set irrespective of process distribution, but if we understand our process we can estimate yield / defects. Assume a specification calls for a value of 35.0 2.5. Assume the process has a distribution that is Normally distributed, with a mean of 37.0 and a standard deviation of 2.20. Estimate the proportion of the process output that will meet specifications.
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7/2/2015 IENG 486 Statistical Quality & Process Control13 Continuous & Discrete Distributions Continuous Probability of a range of outcomes is the area under the PDF (integration) Discrete Probability of a range of outcomes is the area under the PDF (sum discrete outcomes) 35.0 2.5 37 ( ) 41.4 ( +2 ) 32.6 ( -2 ) 43.6 ( +3 ) 30.4 ( -3 ) 39.2 ( + ) 34.8 ( - ) 35.0 2.5 364032 42303834 ()()
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7/2/2015 IENG 486 Statistical Quality & Process Control14 Discrete Distribution Example Sum of two six-sided dice: Outcomes range from 2 to 12. Count the possible ways to obtain each individual sum - forms a histogram What is the most frequently occurring sum that you could roll? Most likely outcome is a sum of 7 (there are 6 ways to obtain it) What is the probability of obtaining the most likely sum in a single roll of the dice? 6 36 =.167 What is the probability of obtaining a sum greater than 2 and less than 11? 32 36 =.889
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7/2/2015 IENG 486 Statistical Quality & Process Control15 How do we know what the distribution is when all we have is a sample? Theory – “CLT applies to measurements taken consisting of many assemblies…” Experience – “past use of a distribution has generated very good results…” “Testing” – combination of the above … in this case, anyway! If we know the generating function for a distribution, we can construct a grid (probability paper) that will allow us to observe a straight line when sufficient data from that distribution are plotted on the grid Easiest grid to create is the Standard Normal Distribution … because it is an easy transformation to “standard“ parameters
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7/2/2015 IENG 486 Statistical Quality & Process Control16 Normal Probability Plots Take raw data and count observations (n) Set up a column of j values (1 to j) Compute (z j ) for each j value (z j ) = (j - 0.5)/n Get z j value for each (z j ) in Standard Normal Table Find table entry( (z j )), then read index value (z j ) Set up a column of sorted, observed data Sorted in increasing value Plot z j values versus sorted data values Approximate with sketched line at 25% and 75% points
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7/2/2015 IENG 486 Statistical Quality & Process Control17 Interpreting Normal Plots Assess Equal-Variance and Normality assumptions Data from a Normal sample should tend to fall along the line, so if a “fat pencil” covers almost all of the points, then a normality assumption is supported The slope of the line reflects the variance of the sample, so equal slopes support the equal variance assumption Theoretically: Sketched line should intercept the z j = 0 axis at the mean value Practically: Close is good enough for comparing means Closer is better for comparing variances If the slopes differ much for two samples, use a test that assumes the variances are not the same
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Normal Probability Plots Tools for constructing Normal Probability Plots: (Normal) Probability Paper In-class handout Normal Plots Template Materials Page on course website Interpretation Fat Pencil Test 7/2/2015 IENG 486 Statistical Quality & Process Control18
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7/2/2015 IENG 486 Statistical Quality & Process Control19 Bonus Problem A four-sided die and an eight-sided die are rolled once, and the outcomes summed: Construct a histogram for the distribution. What is the most likely resulting sum? What is the standard deviation of the distribution? The min and max of this distribution and the pair of six-sided dice are the same. Which, if either, of the two processes would be better for obtaining their mean value(s)? – Why?
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7/2/2015 IENG 486 Statistical Quality & Process Control20 Exam I Bonus* Problems We collected the Grip Strength Data, and it is posted on the Materials Page - download the spreadsheet, and in groups of two, answer the following (assume =.05, when needed): Is there a significant difference in dominant and weaker hand strength? Do male grip strengths vary more than females? Does the right side of the room have a stronger grip than the left side of the room? Is there a difference in grip strength between gray/white shirted people and all others? Does foot size make a difference in grip strength? Does foot size correlate with grip strength – if, so, how? Are SDSM&T students stronger than the average public? Assume: mean Male strength is 104.3.0 lbs, std dev is 28.3 lbs (right hand) mean Female strength is 62.8 lbs, std dev is 17.0 lbs (right hand)
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