The following estimated regression equation is based on 30 observations. ŷ = 18.5 + 3.7x1 – 2.2x2 + 7.4x3 + 2.9x4 The values of SST and SSR are 1,802 and 1,765, respectively. a. Compute R² (to 3 decimals). b. Compute R (to 3 decimals). c. Comment on the goodness of fit. The estimated regression equation - Select your answer -
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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.The following estimated regression equation based on 30 observations was presented. ŷ = 17.6 + 3.8x1 − 2.3x2 + 7.6x3 + 2.7x4 The values of SST and SSR are 1,808 and 1,780, respectively. (a) Compute R2. (b) Compute Ra2. (c) Comment on the goodness of fit.The following estimated regression equation is based on 30 observations. y=17.4 - 4.0x 1- 2.3x2 +7.3x32.9x4 The values of SST and SSR are 1,808 and 1,760, respectively. Compute R2 (to 3 decimals). Compute Ra2 (to 3 decimals). How good is the fit provided by the estimated regression equation?
- The following estimated regression equation is based on 30 observations. ŷ = 18.3 + 3.9x1 − 2.2x2 + 7.5x3 + 2.5x4 The values of SST and SSR are 1,805 and 1,762, respectively. a. Compute R2 = (to 3 decimals). b. Compute Ra2 = (to 3 decimals).Using this regression equation, Ŷ = .45X+ 9, find the SSresidual you would need to calculate the standard error of the estimate. X Y 15 3 12 6 17 2 11 5The following estimated regression equation based on 10 observations was presented. ŷ = 29.1260 + 0.5306x1 + 0.4680x2 The values of SST and SSR are 6,728.125 and 6,215.375, respectively. (a) Find SSE. SSE = (b) Compute R2. (Round your answer to three decimal places.) R2 = (c) Compute Ra2. (Round your answer to three decimal places.) Ra2 = (d) Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) The estimated regression equation provided a good fit as a small proportion of the variability in y has been explained by the estimated regression equation.The estimated regression equation did not provide a good fit as a small proportion of the variability in y has been explained by the estimated regression equation. The estimated regression equation provided a good fit as a large proportion of the variability in y has been explained by the estimated regression equation.
- The following estimated regression equation based on 30 observations was presented. ŷ = 17.6 + 3.8x1 − 2.3x2 + 7.6x3 + 2.7x4 The values of SST and SSR are 1,807 and 1,757, respectively. (a) Compute R2. (Round your answer to three decimal places.) R2 = (b) Compute Ra2. (Round your answer to three decimal places.) Ra2 = (c) Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) The estimated regression equation did not provide a good fit as a large proportion of the variability in y has been explained by the estimated regression equation. The estimated regression equation provided a good fit as a large proportion of the variability in y has been explained by the estimated regression equation. The estimated regression equation provided a good fit as a small proportion of the variability in y has been explained by the estimated regression equation. The estimated regression equation did not provide a good…The following estimated regression equation based on 30 observations was presented. ŷ = 17.6 + 3.8x1 − 2.3x2 + 7.6x3 + 2.7x4 The values of SST and SSR are 1,801 and 1,758, respectively. (a)Compute R2. (Round your answer to three decimal places.) R2 = (b)Compute Ra2.(Round your answer to three decimal places.) Ra2 = (c) Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) The estimated regression equation provided a good fit as a small proportion of the variability in y has been explained by the estimated regression equation.The estimated regression equation did not provide a good fit as a large proportion of the variability in y has been explained by the estimated regression equation. The estimated regression equation did not provide a good fit as a small proportion of the variability in y has been explained by the estimated regression equation.The estimated regression equation provided a good fit as a…The following multiple regression printout can be used to predict a person's average annual salary given his or her years of employment and number of years of education beyond high school. Regression Analysis: Salary Versus YrsEm, Educ Coefficients Term Coef SE Coef T-Value P-Value Constant 23,175 1,771 13.09 0.000 YrsEm 671 141 4.76 0.000 Educ 1,911 378 5.06 0.000 (a) Is the regression coefficient of education (Educ) statistically significant? (Use ? = 0.05.) Given this output, the regression coefficient for education (is/is not) statistically significant. (b) Does the variable education belong in the model? Given this output, the variable education (may/may not) belong in the model. (c) Given this output, which of the following is the correct interpretation for Education in this model? -For each additional year of education beyond high school, the estimated change in the average annual salary is $23,175, controlling for years of employment. -For each…
- Run a regression analysis on the following data set, where yy is the final grade in a math class and xx is the average number of hours the student spent working on math each week. hours/weekx Gradey 5 59 6 54.4 6 56.4 8 59.2 11 68.4 12 78.8 13 75.2 14 89.6 14 89.6 16 87.4 State the regression equation y=m⋅x+by=m⋅x+b, with constants accurate to two decimal places. What is the predicted value for the final grade when a student spends an average of 15 hours each week on math?Grade = Round to 1 decimal place.Consider the set of ordered pairs shown below. Assuming that the regression equation is y= 2.814 + 0.406x and the SSE = 5.313, construct a 90% prediction interval for x= 3. x 6 0 4 4 4 2 y 5 3 6 5 3 3 Calculate the upper and lower limits of the prediction interval. UPL=LPL= Someone please help me solve this! Thank you!Consider the following data on x = weight (pounds) and y = price ($) for 10 road-racing bikes. These data provided the estimated regression equation ŷ = 28,240 − 1,419x. For these data, SSE = 7,209,342.96 and SST = 50,969,800. Use the F test to determine whether the weight for a bike and the price are related at the 0.05 level of significance. -Find the value of the test statistic. (Round your answer to two decimal places.)
