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- In Galton’s height data (Figure 7.1, in Section 7.1), the least-squares line for predictingforearm length (y) from height (x) is y = −0.2967 + 0.2738x.a) Predict the forearm length of a man whose height is 70 in.b) How tall must a man be so that we would predict his forearm length to be 19 in.?c) All the men in a certain group have heights greater than the height computed in part(b). Can you conclude that all their forearms will be at least 19 in. long? Explain.In a study of the causes of bearing wear, a machine was run 24 times, with various loads (denoted x1), oil viscosities (x2), and ambient temperatures (x3). The wear, denoted y, was modeled as y = β0 + β1x1 + β2x2 + β3x3 + ε. When this model was fit to the data, the sum of squares for error was SSE = 9.37. Then the reduced model y = β0 + β1x1 + β2x2 + β3x3 was fit, and the sum of squares for error was SSE = 27.49. Is it reasonable to use the reduced model, rather than the model containing all the interactions, to predict wear? Explain.When a least squares line is fit to the 12 observations in the labor cost data, we obtain SSE =656, 214.3589. Calculate s2 and s (isnt n-k-1)?
- Percentages of public school students in fourth and eighth grades who were above the proficiency level were given for 8 western states 4th Grade Proficiency 8th Grade Proficiency Arizona 15 21 California 11 18 Hawaii 16 16 Montana 22 37 New Mexico 13 13 Oregon 21 32 Utah 23 26 Wyoming 19 25 Construct a scatterplot and comment on any interesting characteristics Find the least squares equation. If Nevada had a 4th grade proficiency of 14%, what would you expect the 8th grade proficiency to be?Various doses of a poisonous substance were givento groups of 25 mice and the following results wereobserved: Dose (mg) Number of deathsx y4 16 38 610 812 1414 1616 20 (a) Find the equation of the least squares line fit tothese data.(b) Estimate the number of deaths in a group of 25 micethat receive a 7-milligram dose of this poison.Rework problem 23 in section 6.3 of your textbook (page 276) except use the following data instead of the data in the textbook: Assume that the production of 1 unit of magnesium requires 0.2 units of magnesium and 0.3 units of aluminum, and that the production of 1 unit of aluminum requires 0.2 units of magnesium and 0.8 units of aluminum.Find the production schedule that satisfies an external demand for 30 units of magnesium and 60 units of aluminum. Rework problem 23 in section 6.3 of your textbook (page 276) except use the following data instead of the data in the textbook: Assume that the production of 1 unit of magnesium requires 0.2 units of magnesium and 0.3 units of aluminum, and that the production of 1 unit of aluminum requires 0.2 units of magnesium and 0.8 units of aluminum. Please look at attched image and show steps to solve for solution
- The following are the scores that 12 students obtainedon the midterm and final examinations in a course instatistics:Midterm examination Final examinationx y71 8349 6280 7673 7793 8985 7458 4882 7864 7632 5187 7380 89 (a) Find the equation of the least squares line that willenable us to predict a student’s final examination score inthis course on the basis of his or her score on the midtermexamination.(b) Predict the final examination score of a student whoreceived an 84 on the midterm examination.Is each of these True or False? (i) The least squares regression line is the line that minimizes the sum of the squares of the residuals. (ii) The least squares regression line must go through as many data points as possible. (iii) If a data point falls on the least squares regression line, then its residual is 0. (iv) The least squares regression line is the line that makes the square of the correlation as small as possible.In the manufacture of synthetic fiber, the fiber is often “set” by subjecting it to high temperatures. The object is to improve the shrinkage properties of the fiber. In a test of 23 yarn specimens, the relationship between temperature in °C (x) and shrinkage in % (y) was summarized by the least-squares line y = −12.789 + 0.133x. The total sum of square was ∑ni=1(yi−y⎯⎯)2∑i=1n(yi−y¯)2 = 57.313, and the estimated error variance was s2 = 0.0670. Compute the coefficient of determination r 2. Round the answer to three decimal places.
- In a study of copper bars, the relationship between shear stress in ksi (x) and shear strain in % (y) was summarized by the least-squares line y = −20.00 + 2.56x. There were a total of n = 18 observations, and the coefficient of determination was r 2 = 0.9111. If the total sum of squares was ∑ni=1(yi−y⎯⎯)2∑i=1n(yi−y¯)2 = 234.19, compute the estimated error variance s2. Round the answer to three decimal places.Problem 2:Examine the relationship between amount of time spent studying for an exam (X) in hours andthe score that person makes on an exam (Y)X Y2 653 703 754 705 856 857 90 using spss Give the following:1. Null hypothesis2. Alternative hypothesis3. Statistical test4. Computation5. Decision6. ConclusionProblem 3. Suppose a test for detecting a certain rare disease has been perfected that is capable of discovering the disease in 97% of all afflicted individuals. Suppose further that when it is tried on healthy individuals, 5% of them are incorrectly diagnosed as having the disease. Finally, suppose that when it is tried on individuals who have certain other milder diseases, 10% of them are incorrectly diagnosed. It is known that the percentages of individuals of the three types being considered here in the populations at large are 1%, 96%, and 3%, respectively. Calculate the probability that an individual, selected at random from the population at large and tested for the rare disease, actually has the disease if the test indicates he is so afflicted.