\f\f1) Using the SILVERSPRING data set, develop a multiple linear regression model to predict asking price based on assessed value, size and number of bedrooms. a) Interpret the meaning of the slope coefficients. b) Use the regression model to find the predicted asking price of a house with an assessed value of 600,000, 3,000 square feet and 5 bedrooms. Also find the corresponding prediction and confidence interval. c) Interpret the meaning of the coefficient of multiple regression. d) At the 0.05 significance level, determine if there is a significant relationship between asking price and the three independent variables. e) At the 0.05 significance level, determine whether each independent variable makes a significant contribution to the regression model. 2) Do 14.81 from textbook, parts a, b, c, e, g, h ( just look at r2), i 3) The following is the linear regression model predicting college GPA based on weekly study hours x1, high school GPA x2 and SAT score x3. y = 0.02x1 + 0.75x2 + 0.0001x3 a) Interpret the meaning of the slope coefficients. b) Use the model to predict the college GPA of a student who studies 20 hours per week, had a high school GPA of 4.2 and a combined SAT of 1350. 4) Assume a simple linear regression model. Given the predicted value of y when x equals 5 is 10 and the predicted value of y when x equals 6 is 13, find the value of the slope coefficient b1. 4) Assume a simple linear regression model. Given b0 = 95.6, find the value of y when x = 0. \fRegression Analysis: Wins versus E.R.A., Runs Scored Analysis of Variance Source Regression E.R.A. Runs Scored Error Total DF 2 1 1 27 29 Adj SS 3818.8 2655.1 1093.3 311.2 4130.0 Adj MS 1909.42 2655.05 1093.32 11.52 F-Value 165.68 230.38 94.87 P-Value 0.000 0.000 0.000 Model Summary S 3.39478 R-sq 92.47% R-sq(adj) 91.91% R-sq(pred) 90.97% Coefficients Term Constant E.R.A. Runs Scored Coef 83.04 -19.09 0.1063 SE Coef 9.27 1.26 0.0109 T-Value 8.96 -15.18 9.74 P-Value 0.000 0.000 0.000 VIF 1.00 1.00 Regression Equation Wins = 83.04 - 19.09 E.R.A. + 0.1063 Runs Scored a. Regression Equation Wins = 83.04 - 19.09 E.R.A. + 0.1063 Runs Scored b. B1=19.09 is change in number of wins due to a unit change in E.R.A holding runs scored as a constant. B2=0.1063 is change in number of wins due to a unit change in runs scored holding ERA as a constant c. Prediction for Wins Regression Equation Wins = 83.04 - 19.09 E.R.A. + 0.1063 Runs Scored Variable E.R.A. Runs Scored Fit 76.8595 Setting 4.5 750 SE Fit 1.03185 95% CI (74.7423, 78.9767) 95% PI (69.5793, 84.1397) Answer=76.8595 77 wins d. Residual Plots for Wins Normal Probability Plot Versus Fits 99 10 Residual Percent 90 50 5 0 10 -5 1 -10 -5 0 5 10 60 70 Residual 80 90 100 Fitted Value Histogram Versus Order 10 Residual Frequency 12 9 6 3 0 5 0 -5 -6 -4 -2 0 2 4 6 8 2 4 Residual 6 8 10 12 14 16 18 20 22 24 26 28 30 Observation Order From the normal probability plot of residuals t the assumption that the residuals are normally distributed is not violated since the normal probability plot of the residuals is approximately following a straight line From residuals versus order plot the assumption that the residuals are independent from one another is not violated since from the plot residuals show no trends or patterns. The Histogram is used to test the assumption that data come from normal distribution. From our plot above, the histogram is almost symmetric hence confirming this assumption. e. Yes, f. P-values=0.0001 This is the probability that the regression model is not significant. g. R-squared= 92.47%. Means that 92.47% variation in number of wins is explained by the model R-sq(adj)= 91.91% i. Coefficients Term Coef SE Coef Constant 83.04 9.27 E.R.A. -19.09 1.26 Runs Scored 0.1063 0.0109 h. T-Value 8.96 -15.18 9.74 P-Value 0.000 0.000 0.000 VIF 1.00 1.00 Both of the coefficients are significant since their p-values is less than 0.05 \f