2. The article "How to Optimize and Control the Wire Bonding Process" described an experiment carried out to assess the impact of the variables force (gm), x1 and temperature(degree Celsius), x2 on ball bond shear strength(gm),y. The following data were generated: Observation Force Temperature Strength 1 30 175 26.2 40 175 26.3 30 40 175 39.8 4 175 39.7 5 30 225 38.6 40 35.5 225 225 7 30 48.8 8 40 225 37.8 9 30 175 26.6 10 40 175 23.4 a. Find the regression equation representing the ball bond shear strength in terms force and temperature. b. Determine whether the data provide sufficient evidence to conclude that the force and temperature are useful for predicting the ball bond shear strength at 5% level of significance. 2 3
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- The following table shows the typical depth (rounded to the nearest foot) for nonfailed wells in geological formations in Baltimore County (The Journal of Data Science, 2009, Vol. 7, pp. 111-127). Geological Formation Group Number of Nonfailed Wells Nonfailed Well Depth Gneiss 1,515 255 Granite 26 218 Loch Raven Schist 3,290 317 Mafic 349 231 Marble 280 267 Prettyboy Schist 1,343 255 Other schists 887 267 Serpentine 36 217 Total 7,726 2,027 Let the random variable X denote the depth (rounded to the nearest foot) for nonfailed wells. Detemine the cumulative distribution function for X. Round your answers to four decimal places (e.g. 98.7654). x < 217 217Use the following equations to perform calculations necessary to complete the CFU. Use the table below to calculate Z, skewness, and kurtosis. Use the table below to calculate Z, skewness, and kurtosis. X Z Z^3 Z^4 4 4 5 6 7 8 9 9 10 10 12 Ez^3 Ez^4 N = 11 Mean = 7.63 SD = 2.66 Calculate the skewness for the data above and Round to two decimal points.Given the Z scores of: -1.5, 0.52, -1.0, 1.7 and 3.0: 1. Calculate the raw skewness and kurtosis scores for these data. 2. Calculate the standard error scores for skewness and kurtosis for these data. 3. Calculate the Z scores for skewness and kurtosis for these data.Wild irises are beautiful flowers found throughout the United States, Canada, and northern Europe. This problem concerns the length of the sepal (leaf-like part covering the flower) of different species of wild iris. Data are based on information taken from an article by R. A. Fisher in Annals of Eugenics (Vol. 7, part 2, pp. 179 -188). Measurements of sepal length in centimeters from random samples of Iris setosa (I), Iris versicolor (II), and Iris virginica (III) are as follows below. I II III 5.9 5.8 6.9 4.8 6.5 5.6 4.6 6.4 4.9 5.7 4.3 7.7 4.8 5.8 5.6 5.4 6.3 6.3 5.8 5.5 6.6 Shall we reject or not reject the claim that there are no differences among the population means of sepal length for the different species of iris? Use a 5% level of significance. (b) Find SSTOT, SSBET, and SSW and check that SSTOT = SSBET + SSW. (Use 3 decimal places.) SSTOT = SSBET = SSW = Find d.f.BET, d.f.W, MSBET, and MSW. (Use 4 decimal places for MSBET, and…Pepsi Cans. In Exercises 5-8, refer to the axial loads (pounds) of aluminum Pepsi cans that are 0.0109 in. thick, as listed in Data Set 30 “Aluminum Cans” in Appendix B. An axial load of a can is the maximum weight supported by the side, and it is important to have an axial load high enough so that the can isn’t crushed when the top lid is pressed onto the top. There are seven measurements from each of 25 days of production. If the 175 axial loads are in one column, the first 7 are from the first day, the next 7 are from the second day, and so on, so that the “subgroup size” is 7. Pepsi Cans: R Chart Treat the seven measurements from each day as a sample and construct an R chart. What does the result suggest?The authors of a paper investigated whether water temperature was related to how far a salamander would swim and whether it would swim upstream or downstream. Data for 14 streams with different mean water temperatures where salamander larvae were released are given (approximated from a graph that appeared in the paper). The two variables of interest are x = mean water temperature (°C) and y = net directionality, which was defined as the difference in the relative frequency of the released salamander larvae moving upstream and the relative frequency of released salamander larvae moving downstream. A positive value of net directionality means a higher proportion were moving upstream than downstream. A negative value of net directionality means a higher proportion were moving downstream than upstream. Mean Temperature (x) 6.22 8.01 8.67 10.51 12.4 12.04 12.55 17.93 18.34 19.84 20.3 19.12 17.68 19.57 USE SALT Net Directionality (y) -0.08 0.25 -0.14 0.00 0.08 0.03 -0.07 0.29 0.23 0.24 0.19…An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: 415 421 422 423 426 429 431 434 436 439 445 446 448 453 455 463 464 (a) Construct a boxplot of the data. O 420 420 430 430 440 440 450 The data appears to be centered near 438. The data is strongly skewed. There is one outlier. 450 Comment on any interesting features. (Select all that apply.) There are no outliers. The data appears to be centered near 428. There is little or no skew. 460 420 420 430 430 (b) Is it plausible that the given sample observations were selected from a normal distribution? Yes No 440 440 450 450 460 460 (c) Calculate a two-sided 95% confidence interval for true average degree of polymerization. (Round your answers to two decimal places.)The following data pertain to x, the amount of fertil-izer (in pounds) that a farmer applies to his soil, and y, his yield of wheat (in bushels per acre): xy xy xy112 33 88 24 37 2792 28 44 17 23 972 38 132 36 77 3266 17 23 14 142 38112 35 57 25 37 1388 31 111 40 127 2342 8 69 29 88 31126 37 19 12 48 3772 32 103 27 61 2552 20 141 40 71 1428 17 77 26 113 26 Assuming that the data can be looked upon as a randomsample from a bivariate normal population, calculate rand test its significance at the 0.01 level of significance.Also, draw a scattergram of these paired data and judgewhether the assumption seems reasonable.Wild irises are beautiful flowers found throughout the United States, Canada, and northern Europe. This problem concerns the length of the sepal (leaf-like part covering the flower) of different species of wild iris. Data are based on information taken from an article by R. A. Fisher in Annals of Eugenics (Vol. 7, part 2, pp. 179 -188). Measurements of sepal length in centimeters from random samples of Iris setosa (I), Iris versicolor (II), and Iris virginica (III) are as follows below. I II III 5.9 5.5 6.7 4.1 6.8 5.8 5.2 6.5 4.5 5.4 4.5 7.2 4.1 5.4 5.2 5.2 6.2 6.9 5.6 5.2 6.7 Shall we reject or not reject the claim that there are no differences among the population means of sepal length for the different species of iris? Use a 10% level of significance. (a) What is the level of significance? 0.01 State the null and alternate hypotheses. O H,: H1 = H, = Hzi H;: All three means are different. O H,: H1 = H2 = Hzi H;: Exactly two means are equal. 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Does there appear to be an outlier in the data? Does the assumption…3. For the data provided in the given table Thermocouple 821 835 816 (x) 820 840 836 825 840 833 827 IR 818 834 824 821 845 830 819 843 832 835 measurement (y) a) Fit a simple linear regression model. b) Calculate the correlation coefficient.The following data were collected during an experiment in which 10 laboratory animals were inoculated with a pathogen. The variables are Time after inoculation (X, in minutes) and Temperature (Y, in Celsius degrees). Х, Time Y, Temperature (0C) Minutes) 24 38.8 28 39.5 32 40.3 36 40.7 40 41.0 44 41.1 48 41.4 52 41.6 56 41.8 60 41.9 i. Draw a Scatter Diagram to show the association, if any, between these two variables (correct scale is not very important); Can you draw any conclusion/observation without doing any calculation? ii. iii. Calculate the Coefficient of Correlation. 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