2. Three tensile tests were carried out on ar 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). %3D
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- An engineer wants to compare the tensile strengths of steel bars that are produced using a conventional method and an experimental method. (The tensile strength of a metal is a measure of its ability to resist tearing when pulled lengthwise.) To do so, the engineer randomly selectoThe table below shows the results from the specific gravity (S.G.) test performed in a soil laboratory including twenty samples of sand. Determine the Coefficient of Quartile Variation.The tensile adhesion of each of 20 randomly selected specimens of a metal tested. The resulting measurements (in megapascals). Use them to calculate a 90% condence interval for the true mean tensile adhesion strength of the metal. tensile adhesion of 20 specimens: 11.4 11.9 13.6 8.8 19.5 15.8 16.7 17.6 14.1 11.4 11.9 12.7 7.9 18.5 15.4 15.4 7.5 14.9 10.1 19.8 Using Excel, Total sum of 20 specimen measurements= 274.9 Average = 13.745 Use the given amounts to find the 90% confidence interval of the true mean.
- Spider silk is the strongest known material, natural or man-made, on a weight basis. A study examined the mechanical O Assignment Score: 25% O Resources Give Up? Check Answer Question 13 of 20 properties of spider silk using 21 female golden orb weavers, Nephila clavipes. The data on silk yield stress rn the amount of force per unit area needed to reach permanent deformation of the silk strand. The data are exprece megapascals (MPa). 164.0 478.7 251.3 351.7 173.0 448.9 300.6 362.0 272.4 740.2 329.0 327.2 270.5 332.1 288.8 176.1 282.2 236.1 358.2 270.5 290.7 (a) Use the software of your choice to make a dotplot of these data. Select the correct description of the shape, center, and spread of the distribution. O The distribution is unimodal and essentially symmetric except for a high outlier. The center is approximately 291 MPa. The spread is from 164.0 to 478.7 MPa. 0% O The distribution is unimodal and extremely left-skewed except for a high outlier. The center is approximately 240…To assess the strength of the recycled concrete aggregated products, the measurements of the resiliency modulus (MPa) were collected on 20 selected specimens. Identify the unit, population of units, variable of interest, statistical population, and sample.The 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.
- In the 2015 AFC Championship game, there was a charge that the New England Patriots underinflated their footballs for an advantage. The Patriots' opponents during the championship game were the Indianapolis Colts. Measurements of the Colts' footballs were taken. The balls should be inflated to between 12.5 and 13.5 pounds per square inch (psi). The accompanying data consists of six of the eight measurements on the Colts' footballs. Complete parts (a) and (b) below. 12.75, 12.50, 12.15, 12.35, 12.55, 12.30 The test statistic is = (Round to two decimal places as needed.) The p-value is = (Round to three decimal places as needed.) b. Use the data to construct a 95% confidence interval for the mean psi for the Colts' footballs. How does this confidence interval support your conclusion in part (a)? The 95% confidence interval is ___ psi to ___ psi. (Round to two decimal places as needed. Use ascending order.)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 "spring-like effect" in a golf club could be determined by measuring the coefficient of restitution (the ratio of the outbound velocity to the inbound velocity of a golf ball fired at the clubhead). Twelve randomly selected drivers produced by two clubmakers are tested and the coefficient of restitution measured. The data follow: Club 1: 0.8406, 0.8104, 0.8234, 0.8198, 0.8235, 0.8562, 0.8123, 0.7976, 0.8184, 0.8265, 0.7773, 0.7871 Club 2: 0.8305, 0.7905, 0.8352, 0.8380, 0.8145, 0.8465, 0.8244, 0.8014, 0.8309, 0.8405, 0.8256, 0.8476 Test the hypothesis that both brands of ball have equal mean overall distance. Use α = 0.05 and assume equal variances. Question: Reject H0 if t0 < ___ or if t0 > ___.
- How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? Students doing a laboratory exercise sample 20 pieces of wood and find an average load capacity of 30,841 pounds. We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume the characteristics of materials vary normally. Suppose that the strength of pieces of wood like these follow a normal distribution with a population standard deviation of 3,000 pounds (As you can see all three necessary assumptions are met). a. Assuming that all evergreen wood has a known "load" capacity average of 30,000 pounds. Make a two-sided hypothesis (null and alternative statement) about Douglas Fir "load" capacity compared to the overall average. b. Apply the formula for finding our test statistic (show your work or describe the process) Specifically, use the formula for the z-statistic (pasted below and on…Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05. Q) If a participant has the same corneal thickness in their normal eye as the eye with Glaucoma, what would be the value for difference: measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma.Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05. Hypothesis: H0: μd=0Ha: μd≠0 Using output Test statistics : t=0.134P value=0.897 Degrees of freedom (df): df=7 Level of significance: α=0.05 Decision: P value > 0.05 thus we fails to reject null hypothesis. Question a)Write a report summarizing your findings. When writing the report consider that medical staff estimate that a difference of 4.5 microns or more could impact on their ability to interpret eye pressure correctly. b) Define for the hypothesis stated in part b) what a type 1 error and type 2 error would mean. Is it possible…