Thus the supervisor asks the graduate student to perform a regression analysis assuming the following model form: o = Ee Note: Please keep at least 2 significant digits/figures after the decimal point for E and B (e.g., input 0.33 for 1/3, 1.13 for 1.127551). E = B =
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- 2. Three tensile tests were carried out on an aluminum bar. In each test, the strain was measured at the same values of stress. The results were Stress (MPa) 34.5 69.0 103.5 138.0 Strain (Test 1) 0.46 0.95 1.48 1.93 Strain (Test 2) 0.34 1.02 1.51 2.09 Strain (Test 3) 0.73 1.10 1.62 2.12 Where the units of strain are mm/m. Use linear regression to estimate the modulus of elasticity of the bar (modulus of elasticity = stress/strain). %3DThe quality control engineer at Palmer Industries is interested in estimating the tensile strength of steel wire based on its outside diameter and the amount of molybdenum in the steel. As an experiment, she selected 25 pieces of wire, measured the outside diameters, and determined the molybdenum content. Then she measured the tensile strength of each piece. The results of the first four are recorded in the table. Tensile Strength Outside Diameter Amount of Molybdenum Place (PSI ) Y (mm) X1 (Units) X2 A 11 0.3 6 B 9 0.2 5 C 16 0.4 8 D 12 0.3 7 Using a statistical software package, the QC engineer determined the multiple regression equation to be Y’=-0.5+20X1+1X2. a) Based on the equation, what is the estimated tensile strength of a steel wire having an outside diameter of .35 mm and 6.4 units of molybdenum? b) Interpret the value of b1 in the equation.Wrinkle recovery angle and tensile strength are the two most important characteristics for evaluating the performance of crosslinked cotton fabric. An increase in the degree of crosslinking, as determined by ester carboxyl band absorbance, improves the wrinkle resistance of the fabric (at the expense of reducing mechanical strength). The accompanying data on x = absorbance and y = wrinkle resistance angle was read from a graph in the paper "Predicting the Performance of Durable Press Finished Cotton Fabric with Infrared Spectroscopy".† x 0.115 0.126 0.183 0.246 0.282 0.344 0.355 0.452 0.491 0.554 0.651 y 334 342 355 363 365 372 381 392 400 412 420 Here is regression output from Minitab: Predictor Constant absorb S = 3.60498 Coef 321.878 156.711 SOURCE Regression Residual Error Total SE Coef 2.483 6.464 R-Sq = 98.5% DF 1 9 10 SS 7639.0 117.0 7756.0 T 129.64 24.24 0.000 0.000 R-Sq (adj) = 98.3% MS 7639.0 13.0 F P 587.81 (a) Does the simple linear regression model appear to be…
- Wrinkle recovery angle and tensile strength are the two most important characteristics for evaluating the performance of crosslinked cotton fabric. An increase in the degree of crosslinking, as determined by ester carboxyl band absorbance, improves the wrinkle resistance of the fabric (at the expense of reducing mechanical strength). The accompanying data on x = absorbance and y = wrinkle resistance angle was read from a graph in the paper "Predicting the Performance of Durable Press Finished Cotton Fabric with Infrared Spectroscopy".t 半 0.115 0.126 0.183 0.246 0.282 0.344 0.355 0.452 0.491 0.554 0.651 334 342 355 363 365 372 381 392 400 412 420 Here is regression output from Minitab: Predictor Coef SE Coef P Constant 321.878 2.483 129.64 0.000 absorb 156.711 6.464 24.24 0.000 S = 3.60498 R-Sq = 98.5% R-Są (adj) - 98.3% SOURCE DF MS F P Regression 1 7639.0 7639.0 587.81 0.000 Residual Error 9 117.0 13.0 Total 10 7756.0 (a) Does the simple linear regression model appear to be…Wrinkle recovery angle and tensile strength are the two most important characteristics for evaluating the performance of crosslinked cotton fabric. An increase in the degree of crosslinking, as determined by ester carboxyl band absorbance, improves the wrinkle resistance of the fabric (at the expense of reducing mechanical strength). The accompanying data on x = absorbance and y = wrinkle resistance angle was read from a graph in the paper "Predicting the Performance of Durable Press Finished Cotton Fabric with Infrared Spectroscopy".t x 0.115 0.126 0.183 0.246 0.282 0.344 0.355 0.452 0.491 0.554 0.651 y 334 342 355 363 365 372 381 400 392 412 420 Here is regression output from Minitab: Predictor Constant absorb S = 3.60498 Coef 321.878 156.711 SOURCE Regression Residual Error Total R-Sq= 98.5% DF SE Coef 2.483 6.464 1 9 10 SS 7639.0 117.0 7756..0 T 129.64 24.24 P 0.000 0.000. R-Sq (adj) 98.3% MS 7639.0 13.0 F 587.81 (a) Does the simple linear regression model appear to be appropriate?…17.7 Butterfly wings. Researchers studied the morphological attributes of monarch butterflies (Danaus plexippus), a species that undertakes large seasonal migrations over North America. They measured the forewing weight (in milligrams, mg) of a sample of 92 monarch butterflies, all of which had been reared in captivity in identical conditions.° Figure 17.4 shows the output from the statistical software JMP. (The data are also available in the Large.Butterfly the data file if you wish to practice working with your own software.) Estimate with 95% confidence the mean forewing weight of monarch butterflies reared in captivity. Follow the four- step process as illustrated in Example 17.2. 4 STEP そMP FWweight 30 25 20 15 10 11 12 13 14 15 8 9 10 Summary Statistics Mean 11.795652 Std Dev 1.1759413 Std Err Mean 0.1226004 Upper 95% Mean Lower 95% Mean 1 FIGURE 17.4 Software output (JMP) for the forewing weight of monarch 12.039183 11.552122 92 N. butterflies. Count
- The compressive strength of an alloy fastener used in aircraft construction is being studied. Ten loads were selected over the range 2500–4300 psi, and a number of fasteners were tested at those loads. The numbers of fasteners failing at each load were recorded. The complete test data are shown in Table E11-14.b) Civil engineers have developed a 3D model to predict the response of jointed concrete pavements to temperature variations. To check this model, they measured the change in traverse strain on six different occasions and compared this to the modelled values. The table below was obtained. Date Jul 25 Aug 4 Aug 16 - p-value, -58 69 35 Sep 3 -32 Oct 26 -40 Nov 3 -83 - conclusion, Percent Use the Summary output below to answer the following question Summary Output 4.4 (iii) Test whether the differences between the measurements are normally distributed. State your: - hypothesis, Ho Data are normally distributed HA Data are not normally distributed : - the test statistic, 99 - Decision (Reject, Retain), 95 90 80 70 60 50 40 30 20 Change in Strain Note for conclusion: -(if you think there is significant evidence data is normally distributed, enter 1) -(if you think there is significant evidence data is not normally distributed, enter O), Summary Output 4.4: 10 Field Measurement 5 1 -40…The compressive strength of concrete is being studied, and four different mixing techniques are being investigated. The föitowing data have been collected. Compressive Strength (psi) Mixing Technique Observations 1 3129 3000 2865 2890 2 3200 3300 2975 3150 3 2800 2900 2985 3050 4 2600 2700 2600 2765 (a) Test the hypothesis that mixing techniques affect the strength of concrete. Use a = 0.05. Calculate to 2 decimal places fo: i Does mixing technique affect concrete strength? (b) Find to 2 decimal places the P-value for the F-statistic computed in part (a). P-value = i (c) Analyze the following residual plots to determine model adequacy. Does the assumption of normality seem reasonable? Does the assumption of constant variance seem reasonable? >
- An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in x = 18.11 kgf/cm² for the modified mortar (m = 42) and y = 16.82 kgf/cm² for the unmodified mortar (n = 30). Let μ₁ and μ₂ be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. (a) Assuming that ₁ = 1.6 and ₂ = 1.3, test Ho: ₁ - ₂ = 0 versus H₂ : ₁ - ₂ > 0 at level 0.01. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. O Reject Ho. The data does not suggest that the difference in average tension bond strengths exceeds 0. O Fail to reject Ho. The data does not suggest that the difference in average tension bond strengths…An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in x = 18.11 kgf/cm² for the modified mortar (m = 42) and y = 16.82 kgf/cm² for the unmodified mortar (n = 32). Let μ₁ and μ₂ be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. (a) Assuming that 0₁ = 1.6 and ₂ = 1.3, test Ho: ₁ - ₂ = 0 versus H₂: M₁-M₂ > 0 at level 0.01. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. O Reject Ho. The data does not suggest that the difference in average tension bond strengths exceeds 0. O Fail to reject Ho. The data suggests that the difference in average tension bond strengths exceeds…An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsións have been added during mixing) to that of unmodified mortar resulted in x = 18.11 kgf/cm2 for the modified mortar (m = 42) and y = 16.88 kgf/cm2 for the unmodified mortar (n = 31). Let ₁ and ₂ be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. (a) Assuming that o₁ = 1.6 and ₂ = 1.3, test Ho: ₁ - ₂ = 0 versus H₂: H₁ - H₂> 0 at level 0.01. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) Z = P-value = State the conclusion in the problem context. O Fail to reject Ho. The data suggests that the difference in average tension bond strengths exceeds 0. Fail to reject Ho. The data does not suggest that the difference in average tension bond strengths…