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applied statistics and probability for engineers

Questions and Answers of Applied Statistics And Probability For Engineers

=+a 1-degree increase in curing temperature when concentration, catalyst ratio, and curing time all remain fixed.
=+b. Estimate, in a way that conveys information about precision and reliability, the average change in durability press rating associated with
=+temp 0.011226 0.004973 2.26 0.033 time 0.10197 0.05874 1.74 0.095 S = 0.8365 R-Sq = 69.2% R-Sq(adj) = 64.3%Analysis of Variance Source DF SS MS F P Regression 4 39.3769 9.8442 14.07 0.000 Error 25
=+a. Fitting the model with the four independent variables as predictors resulted in the following Minitab output. Does the fitted model appear to be useful?The regression equation is durpr =
=+66. The accompanying data was taken from the article“Applying Stepwise Multiple Regression Analysis to the Reaction of Formaldehyde with Cotton Cellulose” (Textile Research J., 1984:
=+C. Total 14 75551276 Fitting the complete second-order model resulted in SSResid 5 805,534. Carry out a test at significance level .01 to decide whether at least one of the second-order predictors
=+gluconic acid production (mg/L). The accompanying SAS output resulted from a request to fit the model with predictors x1, x2, and x3 only.Source DF Sum of Squares Mean Square FValue Pr > F Model 3
=+65. The article cited in Exercise 64 also examined the effect of x1 5 pH, x2 5 sucrose concentration (g/L), and x3 5 spore population (106 cells/ml) on y 5
=+C. Total 14 8717252 Fitting the complete second-order model resulted in SSResid 5 541,632. Carry out a test at significance level .01 to decide whether at least one of the second-order predictors
=+production (mg/L). The accompanying SAS output resulted from a request to fit the model with predictors x1, x2, and x3 only.Source DF Sum of Squares Mean Square FValue Pr > F Model 3 5861301
=+Uranium Ore Residue of the Leaching Stage using Statistical Experimental Design” (Annals of Nuclear Energy, 2013: 48–52) reported on a twostage bioleaching process of vanadium by using the
=+dissolution of metals by this method can be done in a two-stage bioleaching process: (1) microorganisms are grown in culture to produce metabolites(e.g. organic acids) and (2) ore is added to the
=+64. The use of microorganisms to dissolve metals from ores has offered an ecologically friendly and less expensive alternative to traditional methods. The
=+c. The standardized residuals from fitting the simple linear regression model are (in increasing order of x values) 2.04, 2.05, .14, .13, .22, .21,.11, 2.37, 21.04, .36, .63, 21.08, 2.43, 2.34,
=+b. Carry out a test of appropriate hypotheses to see whether there is in fact a linear relationship between the two variables.
=+a. The article included the statement “the linear correlation coefficient, r 2 5 .89.” Is this entire statement correct? If not, why, and what part is correct?
=+1998: 87–93) included a scatterplot of y 5 final settled height fraction versus x 5 initial solids concentration (g/L), from which the following data was read:x: .5 .9 1.1 1.7 2.0 2.2 2.7 3.0
=+the performance of suspended-growth activatedsludge processes. The article “Sludge Volume Index Settleability Measures” (Water Environ. Research,
=+c. Does the cubic predictor appear to provide useful information about y over and above that provided by the linear and quadratic predictors?State and test the appropriate hypotheses.d. When x 5
=+b. Fitting a quadratic model to the data results in R2 5 .780. Calculate adjusted R2 for this model and compare to adjusted R2 for the cubic model.
=+a. What proportion of observed variation in energy output can be attributed to the model relationship?
=+Predictor Coef SE Coef T P Constant -133.787 8.048 -16.62 0.000 x 12.7423 0.7750 16.44 0.000 x**2 -0.37652 0.02444 -15.41 0.000 x**3 0.0035861 0.0002529 14.18 0.000 s = 0.168354 R-Sq = 98.0%
=+and Exergy Efficiency for Solar Box and Parabolic Cookers” (J. of Energy Engr., 2007: 53–62).x: 23.20 23.50 23.52 24.30 25.10 26.20 y: 3.78 4.12 4.24 5.35 5.87 6.02 x: 27.40 28.10 29.30 30.60
=+62. The accompanying data on y 5 energy output (W)and x 5 temperature difference (K) was provided by the authors of the article “Comparison of Energy
=+c. Even if the y values had been much closer together, so that the model could be judged useful, would there be any way to check model adequacy to decide whether a quadratic regression model
=+b. Fit the simple linear regression model, and state whether you agree with the discussant’s assessment.
=+a. Why is the relationship between these two variables clearly not deterministic?
=+61. In a discussion of the article “Tensile Behavior of Slurry Infiltrated Mat Concrete (SIMCON)” (ACI Materials J., 1998: 77–79), the discussant presented data on y 5 toughness (psi) and x 5
=+d. Predict astringency for a single wine sample whose tannin concentration is .6, and do so in a way that conveys information about reliability and precision.
=+c. Estimate true average astringency when tannin concentration is .6, and do so in a way that conveys information about reliability and precision.
=+b. Calculate and interpret a confidence interval for the slope of the true regression line.
=+a. Fit the simple linear regression model to this data. Then determine the proportion of observed variation in astringency that can be attributed to the model relationship between astringency
=+ Relevant summary qualities are as follows:^xi 5 19.404, ^yi 5 2.549, ^x 2i 5 13.248032^y 2i 5 11.835795, ^xi yi 5 3.497811 Sxx 5 13.248032 2 (19.404)2 y32 5 1.48193150, Syy 5 11.82637622 Sxy 5
=+reported on an investigation to assess the relationship between perceived astringency and tannin concentration using various analytic methods. Here is data provided by the authors on x 5 tannin
=+60. Astringency is the quality in a wine that makes a wine drinker’s mouth feel slightly rough, dry, and puckery.The paper “Analysis of Tannins in Red Wine Using Multiple Methods: Correlation
=+c. The largest value of DCO is much greater than the other values. Does this observation appear to have had a substantial impact on the fitted equation?
=+b. Predict the value of DNOy that would result from making one more observation when DCO is 400, and do so in a way that conveys information about precision and reliability. Does it appear that
=+a. Fit an appropriate model to the data and judge the utility of the model.
=+59. The accompanying data was read from a scatterplot in the article “Urban Emissions Measured with Aircraft” (J. of the Air and Waste Mgmt. Assoc., 1998: 16–25). The response variable is
=+ Noticing the relatively small P-value for the moisture predictor, a fellow student concludes that, based on the model utility test, there is a useful linear relationship between the two
=+ Here is the Mintab output from a request to fit a simple linear regression model of y on x:The regression equation is density = 545 - 5.46 moisture Predictor Coef SE Coef T p Constant 545.23 28.19
=+58. Suppose data was collected on y 5 bulk density(kg/m3) and x 5 moisture content (%) for a sample of six seeds of a particular type resulting in the accompanying scatterplot.5.0 500 Scatterplot
=+29 pillars using the parameter estimates given in the output. Then label each pillar as “stable” if the estimated probability is at least .75 and “unstable”otherwise. How many of the
=+c. Determine an estimate (as the authors did) for the probability of pillar stability for each of the
=+b. Provide interpretations for e 2.774 and e 5.668.
=+a. Use the output to determine whether the two predictor variables appear to have a significant impact on pillar stability. Use 5 .1.
=+ The corresponding logistic regression output from R is given here:Coefficients:Estimate Std.Error z value Pr(>|z|)(Intercept) -13.146 5.184 -2.536 0.0112 x1 2.774 1.477 1.878 0.0604 x2 5.668 2.642
=+Mining Sci., 2013: 55–60) used a logistic regression model to predict pillar stability. The article reported the following data on x1 5 pillar height to width ratio, x2 5 pillar strength to
=+57. Pillar stability is a most important factor to ensure safe conditions in underground mines. The authors of “Developing Coal Pillar Stability Chart Using Logistic Regression” (Intl. J. of
=+and the likelihood of the consultant being able to perform a certain complex task.Success: 8 13 14 18 20 21 21 22 25 26 28 29 30 32 Failure: 4 5 6 6 7 9 10 11 11 13 15 18 19 20 23 27 Interpret the
=+Predictor Coef StDev Z P Ratio Lower Upper Constant –0.5727 0.6024 –0.95 0.342 age 0.004296 0.005849 0.73 0.463 1.00 0.99 1.02 Logistic Regression Table for Exercise 56 Odds 95% CI Predictor
=+56. The following data resulted from a study commissioned by a large management consulting company to investigate the relationship between amount of job experience (months) for a junior consultant
=+55. Kyphosis refers to severe forward flexion of the spine following corrective spinal surgery. A study carried out to determine risk factors for kyphosis reported the accompanying ages (months)
=+b. Explain what happens to the odds when x is increased by 1. Your explanation should involve the .5 that appears in the formula for (x).c. For what value of x are the odds 1? 5? 10?
=+a. Tabulate values of x, (x), the odds (x)y[1 2(x)], and the log odds for x 5 0, 1, 2, . . . , 10.
=+2001: 3135–3138) investigated the spread of esophageal cancer to the lymph nodes. With x 5 size of a tumor (cm) and y 5 1 if the cancer does spread, consider the logistic regression model with a
=+54. It seems reasonable that the size of a cancerous tumor should be related to the likelihood that the cancer will spread (metastasize) to another site.The article “Molecular Detection of p16
=+Vars R-Sq R-Sq(adj) Cp S 1 2 3 4 d d d d 2 3 4 3 4 4 1 52.7 50.9 174.4 0.73030 X 2 67.9 65.4 112.5 0.61349 X X 3 77.5 75.0 73.1 0.52124 X X X 4 83.4 80.7 50.8 0.45835 X X X X 5 90.9 88.9 21.4
=+d. Here is output from Minitab’s best subsets option, with just the single best subset of each size identified. Which model(s) would you consider using (subject to checking model adequacy)?
=+c. From the output in part (b), we conjecture that none of the predictors involving x1 are providing useful information. When these predictors were eliminated, the value of SSResid for the reduced
=+S = 0.268703 R-Sq = 96.7% R-Sq(adj) = 93.4%Analysis of Variance Source DF SS MS F P Regression 14 29.4287 2.1020 29.11 0.000 Res. Error 14 1.0108 0.0722 Total 28 30.4395 Does at least one of the
=+b. Fitting the complete second-order model gave the following results:Predictor Coef SE Coef T P Constant -119.49 18.53 -6.45 0.000 x1 -0.1047 0.2839 -0.37 0.718 x2 28.678 3.625 7.91 0.000 x3
=+Regression 4 19.8882 4.9721 11.31 0.000 Res. Error 24 10.5513 0.4396 Total 28 30.4395 Calculate and interpret the values of R2 and adjusted R2. Does the model appear to be useful?
=+a. Fitting the model with the four xi’s as predictors yielded the following output:Predictor Coef SE Coef T P Constant -4.586 2.542 -1.80 0.084 x1 0.01317 0.02707 0.49 0.631 x2 1.6350 0.2707 6.04
=+53. The article “Response Surface Methodology for Protein Extraction Optimization of Red Pepper Seed” (Food Sci. and Tech., 2010: 226–231) gave data on the response variable y 5 protein yield
=+52. Refer to the wood specific gravity data presented in Exercise 49. The following R2 values resulted from regressing each predictor on the other four predictors (in the first regression, the
=+Data for Exercise 49 ONumber of predictors Variables included R2 Adjusted R2 1 9 .756 .748 2 2, 9 .833 .821 3 3, 9, 10 .852 .836 4 3, 6, 9, 10 .860 .839 5 3, 5, 6, 9, 10 .866 .840 6 3, 5, 6, 8, 9,
=+51. The article “The Analysis and Selection of Variables in Linear Regression” (Biometrics, 1976: 1–49) considered a data set of 32 observations on the following variables: y 5 fuel
=+T-Value 3.01 3.31 2.63 2.12 S 0.0180 0.0174 0.0185 0.0200 R-Sq 77.05 77.03 72.27 65.50 Response is spgrav on 5 predictors, with N = 20 Step 1 2 Constant 0.7585 0.5179%sprwood –0.00444 –0.00438
=+Step 1 2 3 4 Constant 0.4421 0.4384 0.4381 0.5179 sprngfib 0.00011 0.00011 0.00012 T-Value 1.17 1.95 1.98 sumrfib 0.00001 T-Value 0.12%sprwood –0.00531 –0.00526 –0.00498 –0.00438 T-Value
=+50. The accompanying Minitab output resulted from applying both the backward elimination method and the forward selection method to the wood specific gravity data given in Exercise 49. Explain for
=+which model(s) would you recommend investigating in more detail?s s s s s p u p u u f f p l l R-Sq i i e a a Vars R-Sq (adj) C-p s b b r b b 1 56.4 53.9 10.6 0.021832 X 1 10.6 5.7 38.5 0.031245 X
=+data (top of page 570) on y and the predictors x1 5 number of fibers/mm2 in springwood, x2 5 number of fibers/mm2 in summerwood, x3 5 springwood %, x4 5 % springwood light absorption, and x5 5
=+49. The article “Anatomical Factors Influencing Wood Specific Gravity of Slash Pines and the Implications for the Development of a High-Quality Pulpwood” (TAPPI, 1964: 401–404) reported the
=+Obs Velocity Viscosity Mesh size Response Standardized residual hii 1 2.14 10.00 .34 28.9 2.01721 .202242 2 4.14 10.00 .34 26.1 1.34706 .066929 3 8.15 10.00 .34 22.8 .96537 .274393 4 2.14 2.63 .34
=+ The standardized residuals and hii values resulted from the model that included the three independent variables as predictors. Are there any unusual observations?
=+48. A study carried out to investigate the relationship between a response variable relating to pressure drops in a screen-plate bubble column and the predictors x1 5 superficial fluid velocity,
=+b. Would the forward selection method of model selection have considered the best two-predictor model? Explain your reasoning.
=+a. Use the criteria discussed in the text to recommend the use of a particular model.
=+47. The accompanying table shows the smallest value of SSResid for each number of predictors k (k 5 1, 2, 3, 4) for a regression problem in which y 5 cumulative heat of hardening in cement, x1 5
=+b. Plot the standardized residuals against depth and against water content, and comment on the plots.
=+a. Construct a normal quantile plot of the standardized residuals to see whether it is plausible that the random deviations in the fitted model come from a normal distribution.
=+second-order model.Obs Shstren Depth Watcont Stresid NQuant 1 14.7 8.9 31.5 21.50075 21.20448 2 48.0 36.6 27.0 .53889 .89743 3 25.6 36.8 25.9 2.52893 2.65862 4 10.0 6.1 39.1 2.17350 2.26585 5 16.0
=+46. Exercise 41 of Section 11.5 gave data on y 5 shear strength of a soil specimen, x1 5 depth, and x2 5 water content. The data is presented again, along with the standardized residuals and
=+b. Construct a plot of the standardized residuals versus x and comment.
=+a. Does it appear plausible that the random deviations in the simple linear regression model equation are normally distributed?
=+45. Reconsider the data on x 5 inverse thickness and y 5 flux from Exercise 9 of Section 11.1. The values of the standardized residuals from a simple linear regression analysis and the
=+d. Fitting the complete second-order model (as did the article’s authors) resulted in SSResid 5.003579. Does it appear that at least one of the second-order predictors provides useful
=+2.0004968, and b25.102204 (the t-ratios for these two predictors are both highly significant).In addition, syn 5 .00286 when fiber length525 and hydraulic gradient51.2. Is there convincing
=+c. Fitting the model with just fiber length and hydraulic gradient as predictors gave the estimated regression coefficients a52.005315, b1 5
=+b. Does fiber content appear to provide useful information about velocity provided that fiber length and hydraulic gradient remain in the model? Carry out a test of hypotheses at 5 .05.
=+ How would you interpret the number –.0003020 in the Coef column on the output?
=+a. Here is output from fitting the model with the three xi’s as predictors:Predictor Coef SE Coef T p Constant -0.002997 0.007639 -0.39 0.697 fib cont -0.012125 0.007454 -1.63 0.111 fib lngth
=+44. Coir fiber, derived from coconut, is an eco-friendly material with great potential for use in construction. The article “Seepage Velocity and Piping Resistance of Coir Fiber Mixed Soils”
=+BHP -0.0033680 0.0003919 -8.59 0.000 Supp1_1 -1.7181 0.5977 -2.87 0.007 Supp1_2 -1.4840 0.6010 -2.47 0.019 Lub_1 -0.3036 0.5754 -0.53 0.602 Lub_2 0.8931 0.5779 1.55 0.133 s = 1.18413 R-sq = 77.5%
=+28.216 and R2 5 .849. Does this model appear to improve on the model with just BHP and predictors for supplier? Use = .05.Predictor Coef SE Coef T p Constant 21.5322 0.6782 31.75 0.000
=+d. From the output, it appears that lubrication regimen may not be providing useful information.A regression with the corresponding predictors removed resulted in SSResid 5 48.426. What is the
=+c. When BHP is 1000, material is from supplier 1, and no lubrication is used, syn 5 .524. Calculate a 95%PI for the springback that would result from making an additional observation under these
=+b. The accompanying Minitab output resulted from fitting the model of part (a) (the articles authors also used Minitab; amusingly, they employed a significance level of .06 in various tests of
=+a. What predictors would you use in a model to incorporate supplier and lubrication information in addition to BHP?

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