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- 1. Measures and Strengths of Association Remember that while we may find two variables to be involved in a relationship, we also want to know the strength of the association. Each type of variable has its own measure to determine this though. Three measures will be discussed in this paper, Lambda, Gamma, and Pearson’s r. Lambda Lambda is a measure of association which should be used when both variables are nominal. Essentially this means that knowing a person’s attribute on one variable will help you guess their attribute on the other (Babbie et al., 2014). Gamma Gamma is used to explore the relationship between two ordinal variables. It can also be used to measure association between one dichotomous nominal and one ordinal variable (Babbie et al., 2014, p. 227). Unlike lambda, gamma indicates a strength of an association and a direction. The closer to -1.00 or +1.00, the stronger the relationship, whereas the closer to 0 the weaker the relationship. You can determine the direction of a relationship the following way: A negative association is indicated by a negative sign. This means that as one variable increases the other decreases- the variables are moving away from each other. For example, as social class increases, prejudice decreases. On the other hand, a positive association, indicated by a plus or positive sign, means that both
- 2. variables change in the same direction, either increase or decrease. For example, as social class increases, so too does prejudice or as social class decreases, so too does prejudice. Correlation Coefficient- Pearson’s r Pearson’s r, also known as the correlation coefficient, is the test measure used to determine the association between interval/ratio variables. This measure is similar to Gamma in how it can be understood and establish direction of association. Value of Measures of Association 0.00 + or - .01 to .09 + or - .10 to .29 + or - .30 to .99 Strength of Association None- no assocation at all Weak- uninteresting association Moderate- worth making note of Evidence of a strong association- extremely interesting Perfect- strongest association possible 1.00 Measures of Association in SPSS Analyze – Descriptive Statistics – Crosstabs Place your dependent variable in the Row and your independent variable in the Column. Click "Statistics" to choose which test you will run for the measure/strength of association. You will select Lambda for
- 3. nominal variables, Gamma for ordinal variables (or one ordinal and one dichotomous nominal), or Pearson’s r for interval/ratio variables. Measures of Association in SPSS- Understanding Output Lambda The test above is looking at the relationship between one’s political affiliation and their race. We look at the value .036, which is the measure when political party is the DV (see in table). This means that we can improve our guessing of political affiliation by 4% if we know that person’s race. Based on our notes above, this is a pretty weak relationship and an uninteresting association overall. Would we still continue? Yes, maybe. It is important to note that this is one element of the larger picture we are searching for. Gamma The relationship between one’s subjective class identification and opinion on government welfare spending is shown above. Gamma is used because both variables are ordinal. The value of .071 means that knowing a person’s subjective class status improves our estimate of his/her opinion on government welfare spending by 7%. Based on the chart above, this relationship is weak. It is also not statistically significant since .144>.05. Notice both IV and DV's codings are consistent (all from low to high), thus the
- 4. positive sign of .071 indicates IV and DV are positively associated, though not statistically significant. Pearson’s r For this test we are looking at the correlation between a person’s age and the number of hours they spend watching television. Keep in mind that the survey only questioned adults over the age of 18. So, if you are interested in seeing number of hours for children, this is not a good data set for that. We want to square the correlation here (.130 x .130) to determine the coefficient of determination. This new measure = .0169. We interpret this as approximately 2% of the variation in time watching television is determined by a person’s age. Fast Forward Thinking This data, coupled with the statistic from the test of significance (ex. Chi-square, t Test, or ANOVA), will help you to interpret your findings:
- 5. Mechanical Engineering Department ME 406- Manufacturing and Design Project A (Engineering Statistics and Manufacturing) Title of the Project Quality Control, Regression Analysis, and Models Fitting Term 171 Class Section- xx Project Team – Group X_X xxxxxxxxx Name 1 (Group Coordinator) xxxxxxxxx Name 2 xxxxxxxxx Name 3 xxxxxxxxx Name 4 Assigned on 13th November 2017 Due on 4th January 2018 (Before Exam starts) Note Please Attempt each question as asked using the software where ever mentioned. Full Report generated by STATGRAPHICS Must be included in Problems 1 .All Excel
- 6. sheets be included in PROBLEMS where Excel has been used such as in Q Each problem should be started on separate page after pasting the the problem statement at the top of the page. Your Solution will be typewritten. A hard copy as (WORD +EXCELL SHEETS) be provided along with a Hard copy in PDF format. Plus Software generated reports A soft copy as a CD (WORD +EXCELL SHEETS) be provided along with a soft copy copy in PDF format. Plus Software generated reports. CD Must have mentioned that it is STSTISTICS PROJECT and it shoulf have All Project students Name and ID mentioned on CD and a on a text file inside the cD. Project copies both word and PDF will be uploaded on WebCT as will be explained later. PROBLEM # 1 (1.5 point) For alpha distribution Find
- 7. a) Plot F(t) versus t/C for =0.05 , 0.1.0.15,0.2 ,0.25 and 0.3 (all curves should be in one Figure for range of t/C varying from (0+=0.05) to 4). Hint use Normal Distribution Function(select True) of Excel subroutine for Z=- b) Find pdf for t varying from -∞ to +∞ ,(note at t=0 function is undefined). Note where Z=- c) Plot f(t) versus t/C for =0.05 , 0.1.0.15,0.2 ,0.25 and 0.3 .all curves in one Figure for range of t/C varying from (0+=0.05) to 4 . Hint use Normal Distribution Function (select False) of Excel subroutine for Z=- d) Median value of T, t0.5 e) Quantile, tp which is the solution of F(tp)=p . f) Mode of T, (value of t where PROBLEM # 2 (1.0 point) Use EXCEL and attach all spreadsheet analysis with solution) Fifty measurements of the ultimate tensile strength of wire are
- 8. given in the accompanying table. a) Group the data and make an appropriate normalized histogram (with total area of histogram be 1 ) to approximate the PDF. b) Calculate and for the distribution from the ungrouped data. c) Using and from part b, draw a normal distribution through the normalized histogram .histogram. Ultimate Tensile Strength 103,779 102,325 102,325 103,799 102,906 104,651 105,377 100,145 104,796 105,087 104,796 103,799 103,197 106,395 106,831
- 10. 103,633 105,232 106,540 106,104 102,616 106,831 101,744 100,726 103,924 101,598 Source: Data from E. B. Haugen, Probabilistic Mechanical Design Wiley, New York, 1980 (c) Determine the mean, median, and mode from the ungrouped data. (d) Determine the range and standard deviation from the ungrouped data (e) Plot the cumulative frequency distribution on normal- probability paper, and determine the mean and standard deviation. (f), for the data given in Table .what are the 95 percent confidence limits on the mean of the population?
- 11. PROBLEM # 3 (1.0 point) (Use EXCEL and attach all spreadsheet analysis with solution) Three sets of identical twenty five fatigue specimens were tested at the three different level of stresses.. The number of cycles to failure. The results are expressed as log, were as follows. TABLE 1: FATIGUE LIFE DATA NUMBER OF CYCLES TO FAILURES No. S1 S2 S3 380 MPa 340 MPa 300 MPa 1 34200 125500 954000 2
- 15. 21 94000 420700 2921700 22 97200 428500 3046500 23 99600 664800 3105500 24 116700 776100 3523200 25 122500 793900 4311700 Assume that the data at each stress level (S1, S2 and S3) is lognormally distributed. Using directly the data in table determine · What is the mean fatigue life ( μ) and its standard deviation (σ
- 16. )? · What is the mean Ln of fatigue life ( μlnN) and its standard deviation Ln of fatigue life (σlnN)? · What are the Parameters of Lognormal distributions at S1,S2and S3. · Fit lognormal distribution to the data using linear regression model using Excel or Use Statgraphics to fit the lognormal models to data for each stress level and comment on how good the fit is. PROBLEM # 4 (1.5 point) Q4. (a) The lifetime of a mechanical switch produced by a company has been determined to have a population mean of μ = 2000 h and σ = 200 h. The temper of a phosphor bronze leaf spring in the switch is changed slightly by the supplier. To determine whether this has changed the product, a sample of 100 switches is tested to give the sample values and . Has there been a change in the product? (0.75 point) Q4. (b) A vendor of steel wire advertises a mean breaking load of 10,000 lb. A sample of eight tests shows a mean breaking load of 9250 lb and a standard deviation of 110 lb. Do our tests support the vendor’s claim? (0.75 point)
- 17. PROBLEM # 5 Control Charts (1 point) Problem 5-Refer your Text Book above ( See Ebook provided as text book)- -Solve with appropriate calculations, tables and charts 5.1-Problem 8.1 Parts (a) and (b) –P454 5.2 -Problem 8.28 Parts (a) and (b) –P470 PROBLEM # 6 Regression Analysis (MUST USE STATGRAPHICS_ ATTACH COMPLETE REPORT PLUS ONE PAGE SUMMARY OF EACH FITTED MODEL) 2 points Developing Cutting Forces Empirical Models of a Counter Boring Process in Aluminum. Counter boringis an operation to enlarge the hole made using drilling. Counter boring or finish boring is a deep hole drilling process that requires a work piece with a pre-existing bore. Counter boring is used to enlarge the drilled hole to the proper depth and machine a square shoulder on the bottom to secure maximum clamping action from the faster. The drilling used to produce a circular hole by removing solid metal. The counter bore tool has a guide, called a pilot, which keep it positioned correctly in the hole. Counter boring tools are often used on low power machines were a small diameter solid boring tool is used for the pre-bore and then a counter boring
- 18. tool is used to finish the job. Counter boring is also used when there is a heat treat process required after the initial hole is drilled or if a stepped hole is required. Pilot of diameter d , which is the predrilled hole size in the workpiece diameter of already drilled hole.. D Dia of the enlarged hole Visit the link and download STATGRAPHICS FREE FOR 30 DAYS AND USE MULTIPLE REGRESSION MODULE (SEE EXAMPLES ON WEBSITE) TO DEVELOP FOLLOWING MODELS> http://www.statgraphics.com/centurion-xvii Regression Analysis http://www.statgraphics.com/regression- analysis And Quality Control Module (PROICESS CAPABILITY BANALYSIS) from the link http://www.statgraphics.com/process-capability-analysis Following are the results of Cutting Forces measurements experiments performed at KFUPM Workshop by Professor Anwar K Sheikh. The results are being shared for regression analysis learning objectives. Cutting Force Fz as a Function of V,D,d and f Fz
- 19. Newton Speed , V mm/minutes Feed , f Mm/revolution d mm D mm Ln (FZ) Ln(Speed) Ln(feed) Ln(D) 52 2463.007 0.03 3.5 6.5 3.951244 7.809138 -3.50656 1.252763 78 2463.007 0.05
- 48. Moment Mz as a Function of V,D,d and f Mz Newton-meter Speed , V mm/minutes Feed , f Mm/ revolution d mm D mm Ln (Mz) Ln(Speed) Ln(feed) Ln(D) 39 2463.007 0.03 3.5 6.5 3.663562 7.809138 -3.50656
- 80. 9.5 18 6.476972 8.962757 -2.52573 2.251292 Using STATGRAPHICS -MULTIPLE LINEAR REGRESSION MODULE develop the empirical model for cutting force Fz and Torque (Moment) Mz can be writing as following Proposed Model 1 Force (Model 1 F) Proposed Model 1 Moments (Model 1 M) Proposed Model 2 Force (Model 1 F) .In spread sheet create a new column of val;ues. Proposed Model 2 Moments (Model 1 M). In spread sheet create a new column of values.
- 81. Find A,f ,b and c d e etc in Fz model using first data table ,and Find B,d,e,f d using second Data table for each of the odel.. Write one page summary based upon completer regression report generated by STATGRAPHICS for each fitted model .Its goodness of fit as measured by R2 values and other important coefficients tabulated and plotted in the report. (Attach PDF copy of each full report of STATGRAPHICS output and your EXCEL File used as input data. c b a V D f A Fz ´ ´ ´ = f e d V
- 84. 2 ) ( 2 2 d D - PLEASE LET ME KNOW IF YOU NEED ANY ADDITIONAL INFO FROM ME. Please help I need a good grade. In week 3, we used epsilons and 10-percent-point rule to determine if a potential relationship between two variables is worth examining further. This week, we'll use tests of "measures of association" to figure out the exact strength of a relationship between two variables. In addition, we'll learn how to interpret SPSS outputs for measures of association tests such as lambda, gamma, and Pearson's r， along with other possible tests. Remember that these tests are specific to the level of measurement that your variables are. In other words, one test may not work in a different relationship test. Here are the guidelines: 1. Both DV and IV are nominal variables: Lambda (when it is
- 85. not a 2X2 table) 1. If it is a 2X2 table: Phi 2. Both DV and IV are ordinal variables: Gamma 3. One variable ordinal AND the other variable dichotomous nominal (like Yes/No, male/female, etc.): Gamma 1. One variable ordinal AND the other variable nominal (not dichotomous, has more than 2 categories): Cramer’s V. 4. Both DV and IV are I/R variables: Pearson's r To interpret the output: Keep in mind measures of association is a statistical procedure based on Proportional Reduction of Error (PRE). Thus the format of interpretation will be: ......knowing the IV will reduce error in predicting the DV by *%. Please note: Don't just say "IV" and "DV" in your explanation. You need to enter your variables names for IV and DV, and replace * for the exact test value from the output. If the value of Lambda is .34, then it will be interpreted as 34%. Ok, now it is time for you to try! Be sure to test the strength of association of your final project for this week's forum discussion. You can download the class handout attached at the bottom of the page, or Click here for details.