Sample size: 117 R (correlation coefficient) = 0.8448 R-sq = 0.7136788 Estimate of error standard deviation: 204.45117 Parameter estimates (rounded): Parameter Intercept Slope O-52.649 and the variety of T listed below: O 0.5282 O 0.5012 Estimate 0.5383 47.819 O 0.5582 0.6237 Std. Err. 62.855 0.0519 DF T.01,115= 2.35921, T.02,115-2.07731, T.01,116-2.35892 T.02,116=2.07710, T.01,117-2.35864 What is the approximate lower level of the 98% confidence interval on the slope? T-Stat 0.76079 16.2187 P-Value 0.4483 <0.0001
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- 2. Another variable the enthusiast recorded was the weight of the car in pounds. The regression with horsepower is given below. Simple linear regression results: Dependent Variable: Weight Independent Variable: HP Weight=2084.29 + 6.938 HP Sample size: 425 R (correlation coefficient) = 0.647 R-sq = 0.418 Estimate of error standard deviation: 581.01 Parameter estimates: Parameter Estimate Std. Err. Alternative Intercept 2084.29 90.14 +0 Slope 6.938 0.398 #0 DF T-Stat P-value 423 23.12 <0.0001 XXXX 423 XXXX a. Interpret the slope of this regression line if possible. If not, explain. b. Interpret the intercept of this regression line if possible. If not, explain. c. If the amount of power produced by the engine is 190 HP, what is the expected weight for this car? d. If the weight is 3,400 pounds for a car with engine power of 190 HP, what is the residual of weight? e. What percentage of the variability in weight can be explained by the regression line relationship with engine power? f. If…find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. In each case, there is sujficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. Altitude and Temperature Listed below are altitudes (thousands of feet) and outside air temperatures (°F) recorded by the author during Delta Flight 1053 from New Orleans to Atlanta. For the prediction interval, use a 95% confidence level with the altitude of 6327 ft (or 6.327 thousand feet).Simple linear regression results: Dependent Variable: Heart. DiseaseIndependent Variable: BikingHeart. Disease = 18.115809 - 0.20845321 BikingSample size: 68R (correlation coefficient) = -0.94616452R-sq = 0.8952273Estimate of error standard deviation: 1.5273175 Correlation between Biking and Heart. Disease is: -0.94616452. Correlation between Smoking and Heart. Disease is: 0.32421. 1) State r 2 (i.e., the coefficient of determination) for “Biking” and “Heart.Disease” and explain what this value means in context of the data set.
- The service regresses the number of complaints lodged against an employee last year on the hourly wage of the employee for the year. The analyst ran a simple linear regression shown below. Table 7: Model Summary Model R R Square Adjusted R Square Std. Error of the Estimate 1 .854a .730 .695 6.6235 a. Predictors: (Constant), Hourly Wage Table 8: ANOVA ANOVAb Model Sum of Squares df Mean Square F Sig. 1 Regression 1918.458 1 1918.458 129.783 .000a Residual 709.567 48 14.782 Total 2628.025 49 a. Predictors: (Constant), Hourly Wage b. Dependent Variable: Number of Complaints Table 9: Coefficients Coefficientsa Model Unstandardized Coefficients Standardized Coefficients t Sig. B Std. Error Beta 1 (Constant) 20.2 4.357 4.636 .000 Hourly Wage -1.20 .088 -.946 -13.636 .000 a. Dependent Variable: Number of…The service regresses the number of complaints lodged against an employee last year on the hourly wage of the employee for the year. The analyst ran a simple linear regression shown below. Table 7: Model Summary Model R R Square Adjusted R Square Std. Error of the Estimate 1 .854a .730 .695 6.6235 a. Predictors: (Constant), Hourly Wage Table 8: ANOVA ANOVAb Model Sum of Squares df Mean Square F Sig. 1 Regression 1918.458 1 1918.458 129.783 .000a Residual 709.567 48 14.782 Total 2628.025 49 a. Predictors: (Constant), Hourly Wage b. Dependent Variable: Number of Complaints Table 9: Coefficients Coefficientsa Model Unstandardized Coefficients Standardized Coefficients t Sig. B Std. Error Beta 1 (Constant) 20.2 4.357 4.636 .000 Hourly Wage -1.20 .088 -.946 -13.636 .000 a. Dependent Variable: Number of…A study on how the time of exercise affects heart rate had the following outputSimple Linear regression results:Dependent Variable: Heart RateIndependent Variable: Exercise timeSample size: 81R (correlation coefficient) = 0.4188R-sq = 0.1754Estimate of error standard deviation: 0.32416Parameter estimates: Parameter Estimate Std. Err. DF T-Stat P-Value Intercept 1.80938 1.9352 79 0.935 0.3524 Slope 0.4752395 0.11594 79 4.099 <0.0001 According to the output, if I exercise for time=163, what should be my heart rate?Use 4 decimal placesAfter exercising everybody has different heart rates, which means there is a lot of variability in heart rates. How much of that variability is explained by exercise time?Use 4 decimal places
- 3. The regression model below estimated the impact of tax on income using 41 observations. a. Is the slope parameters of variable of tax significant at 5 percent level? Explain. b. What the R-square in this regression indicates. const tax Model 1: High-Precision OLS, using observations 1-41 Dependent variable: income Coefficient 1.78191 7.00074 Mean dependent var S.D. dependent var Sum squared resid S.E. of regression R-squared Adjusted R-squared F(1,49) P-value(F) Std. Error 0.880324 0.0575306 T-ratioPvalue 2.024 121.7 0.0484 <0.0001 7.055 8.30 1.14 4.82 9.97 9.96 1.48 1.79 **In a sample of cars reviewed by Motor Trend magazine, the mean horsepower (hp) was 150 hp with a standard deviation of 36 hp. The mean weight (lbs) was 2500 lbs with a standard deviation of 720 lbs. Assume the relationship between weight and horsepower is linear and has a correlation of r = +0.55. What is the slope of the linear regression model predicting weight (y-variable) from horsepower (x-variable)? using the value of the slope you found, now compute the intercept of the linear regression model predicting weight (y-variable) from horsepower (x-variable). 850 550 1150 250Which of the variables is the indepenent variable and dependent variable for the following question. fit a simple linear regression model to predict latitudes using average monthly range lat= latitudes range= the average monthly range between mean montly maximum and minimum temperatures for a selected set of US cities.
- Assuming that there is a linear relationship bet- ween height (Xand weight (Y) of residents of a certain locality ; estimate the height of an individual whose weight is 150 1lbs, from the following data : Mean height Mean weight Standard deviation of height: =65 inches =160 lbs. %3| = 3 inches Standard deviation of weight Coefficient of correlation between height and weight 20 lbs. 0-60. Also, find the most likely weight of a person whose height is 68 inches.In a study of the performance of a new engine design, the weight of 12 cars (in pounds) and the top speed (in mph) were recorded. A regression line was generated and shown to be an appropriate description of the relationship. The results of the regression analysis are below. Depend Variable: Top Speed Variable Constant Coefficient 107.58 s.e. of Coeff t-ratio prob Weight 0.8710 11.12 0.4146 9.67 0.000 2.10 0.062 R squared = 30.6% R squared (adjusted) = 23.7% s = 10.42 with 12-2 = 10 degrees of freedom Part A: Provide the regression equation based off the analysis provided & explain it in context. Part B: List the conditions for inference that need to be verified. Assuming these conditions have been met, does the data provide convincing evidence of a relationship between weight and top speed? Part C: Assuming all conditions for inference have been verified, determine a 95% confidence interval estimate for the slope of the regression line.An agribusiness performed a regression of wheat yield (bushels per acre) using observations on 21 test plots with four predictors (rainfall, fertilizer, soil acidity, hours of sun). The standard error was 1.02 bushels.