Compute the coefficients b₁ and b₂ for the regression a. rx,y=0.80, rx2y=0.60, rx₁x2 b. x₁y = -0.80, x2y c. rx₁y = 0.50, rx2y =0.80, x2y d. x₁y =200, Sx2 =0.40, Sx₁ y = 0.60, rx₁x2 = − 0.40, Sx₁ = 200, = 0.75, 5,5x1x2 =-0.20, rx₁x2 = -0.10, Sx₁ = 200, Sx2 model ŷ₁ = b0+b₁x₁₁ +b2x2 given the summary statistics shown below. = 100, sy = 500 Sx2 = 100, sy = 500 =0.70, Sx₁ =200, O, Sx2 = 100, sy = 500 = 100, Sy= = 500
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- The following fictitious table shows kryptonite price, in dollar per gram, t years after 2006. t= Years since 2006 0 1 2 3 4 5 6 7 8 9 10 K= Price 56 51 50 55 58 52 45 43 44 48 51 Make a quartic model of these data. Round the regression parameters to two decimal places.Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?Given are five observations for two variables, x and y. xi 1 2 3 4 5 yi 4 6 6 11 13 Develop the estimated regression equation by computing the values of b0 and b1 using b1 = Σ(xi − x)(yi − y) Σ(xi − x)2 and b0 = y − b1x. ŷ = (e) Use the estimated regression equation to predict the value of y when x = 2.
- Given are five observations for two variables, x and y. xi 3 12 6 20 14 yi 55 45 50 15 20 #1) Develop the estimated regression equation by computing the values of b0 and b1 using b1 = Σ(xi − x)(yi − y) Σ(xi − x)2 and b0 = y − b1x. y= #2) Use the estimated regression equation to predict the value of y when x = 13.Consider the following data for two variables, x and y. x 9 32 18 15 26 y 9 19 20 15 22 Develop an estimated regression equation for the data of the form ŷ = b0 + b1x + b2x2. (Round b0 to two decimal places and b1 to three decimal places and b2 to four decimal places.) ŷ = (c) Use the model from part (b) to predict the value of y when x = 20. (Round your answer to two decimal places.)The table below shows the number of state-registered automatic weapons and the murder rate for several Northwestern states. xx 11.7 8.3 7 3.8 2.6 2.6 2.7 0.9 yy 13.7 11.4 10 7.2 6.3 6 6.2 5.1 xx = thousands of automatic weaponsyy = murders per 100,000 residentsDetermine the regression equation in y = a + bx form and write it below. (Round to 2 decimal places) A) How many murders per 100,000 residents can be expected in a state with 2.4 thousand automatic weapons? Answer = Round to 4 decimal places. B) How many murders per 100,000 residents can be expected in a state with 2.3 thousand automatic weapons? Answer = Round to 4 decimal places.
- The birth lengths in cm (x) and birth weights in kg (y) of a sample of 50 newborn female babies are compared, yielding a correlation coefficient of r=0.578 and a linear regression equation of ŷ =−8.89+0.243x The babies all had lengths between 46.5 and 53.0 cm, and weights between 2.50 and 4.05 kg. Based on this, predict the birth weight of a newborn female baby with a birth length of 48.5 cm.Consider the following data for two variables, x and y x 9 32 18 15 26 y 9 20 22 17 23 A. Develop an estimated regression equation for the data of the form ŷ = b0 + b1x. (Round b0 to two decimal places and b1 to three decimal places.) B. Develop an estimated regression equation for the data of the form ŷ = b0 + b1x + b2x2. (Round b0 to two decimal places and b1 to three decimal places and b2 to four decimal places.) C. Use the model from part (b) to predict the value of y when x = 20. (Round your answer to two decimal places.) Please be as detailed as possible in the solution so i may follow along. Thank you for the help!The grades of a sample of 9 students on a prelim exam (x) and on the midterm exam (y) are shown in the excel worksheet. Find the regression equation
- The table below shows the number of state-registered automatic weapons and the murder rate for several Northwestern states. xx 11.7 8.6 6.7 3.5 2.3 2.6 2.2 0.7 yy 14 11 10 7 6.1 6.4 5.9 4.3 xx = thousands of automatic weaponsyy = murders per 100,000 residentsDetermine the regression equation in y = ax + b form and write it below. (Round to 2 decimal places) A) How many murders per 100,000 residents can be expected in a state with 6.5 thousand automatic weapons? Answer = Round to 3 decimal places. B) How many murders per 100,000 residents can be expected in a state with 11 thousand automatic weapons? Answer = Round to 3 decimal places.Which of the multivariate regression parameters listed below would be best interpreted as: the predicted value on the dependent variable when all of the independent variables in the model are equal to zero. a b1 X1 R2We wish to predict the salary for baseball players (y) using the variables RBI (x1) and HR (x2), then we use a regression equation of the form ˆy=b0+b1x1+b2x2y^=b0+b1x1+b2x2. HR - Home runs - hits on which the batter successfully touched all four bases, without the contribution of a fielding error. RBI - Run batted in - number of runners who scored due to a batters's action, except when batter grounded into double play or reached on an error Salary is in millions of dollars. The following is a chart of baseball players' salaries and statistics from 2016. Player Name RBI's HR's Salary (in millions) Adrian Beltre 104 32 18.000 Justin Smoak 34 14 3.900 Jean Segura 64 20 2.600 Justin Upton 87 31 22.125 Brandon Crawford 84 12 6.000 Curtis Granderson 59 30 16.000 Aaron Hill 38 10 12.000 Miquel Cabrera 108 38 28.050 Adrian Gonzalez 90 18 21.857 Jacoby Ellsbury 56 9 21.143 Mark Teixeira 44 15 23.125 Albert Pujols 119 31 25.000 Matt Wieters 66 17 15.800 Logan…