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- The data in the accompanying table include the appraised value, land area of the property in acres, and age, in years, for a small sample of 30 single-family homes in a small city. Perform a multiple regression analysis to predict appraised value based on land area of the property X1 and age, in years, X2 and determine the VIF for each independent variable in the model. Is there reason to suspect the existence of collinearity? Determine the VIF for each independent variable in the model.a) interpret the scatter plot. Give a reasonable estimate for the linear correlation r. b) use technology to find the equation of the least squares regression line describing the relationship between Year (t) and Global Temperature (G). Around to 0.0001. c) Plot the regression line on the scatterplot below. Clearly label three points including (t,G) on the LSRL. d) Clearly interpret the slope, intercept, and R^2 of the linear model on the context of the problem statement. Report with proper units. -slope: -intercept: -R^2: e) use the model to predict the Global Temperature in the year 2030. f) compute and mark the residual for the data point (2016, 14.80 degrees C) circled on the scatterplot. Data attached belowWrinkle recovery angle and tensile strength are the two most important characteristics for evaluating the performance of crosslinked cotton fabric. An increase in the degree of crosslinking, as determined by ester carboxyl band absorbance, improves the wrinkle resistance of the fabric (at the expense of reducing mechanical strength). The accompanying data on x = absorbance and y = wrinkle resistance angle was read from a graph in the paper "Predicting the Performance of Durable Press Finished Cotton Fabric with Infrared Spectroscopy".† x 0.115 0.126 0.183 0.246 0.282 0.344 0.355 0.452 0.491 0.554 0.651 y 334 342 355 363 365 372 381 392 400 412 420 Here is regression output from Minitab: Predictor Constant absorb S = 3.60498 Coef 321.878 156.711 SOURCE Regression Residual Error Total SE Coef 2.483 6.464 R-Sq = 98.5% DF 1 9 10 SS 7639.0 117.0 7756.0 T 129.64 24.24 0.000 0.000 R-Sq (adj) = 98.3% MS 7639.0 13.0 F P 587.81 (a) Does the simple linear regression model appear to be…
- Wrinkle recovery angle and tensile strength are the two most important characteristics for evaluating the performance of crosslinked cotton fabric. An increase in the degree of crosslinking, as determined by ester carboxyl band absorbance, improves the wrinkle resistance of the fabric (at the expense of reducing mechanical strength). The accompanying data on x = absorbance and y = wrinkle resistance angle was read from a graph in the paper "Predicting the Performance of Durable Press Finished Cotton Fabric with Infrared Spectroscopy".t 半 0.115 0.126 0.183 0.246 0.282 0.344 0.355 0.452 0.491 0.554 0.651 334 342 355 363 365 372 381 392 400 412 420 Here is regression output from Minitab: Predictor Coef SE Coef P Constant 321.878 2.483 129.64 0.000 absorb 156.711 6.464 24.24 0.000 S = 3.60498 R-Sq = 98.5% R-Są (adj) - 98.3% SOURCE DF MS F P Regression 1 7639.0 7639.0 587.81 0.000 Residual Error 9 117.0 13.0 Total 10 7756.0 (a) Does the simple linear regression model appear to be…Wrinkle recovery angle and tensile strength are the two most important characteristics for evaluating the performance of crosslinked cotton fabric. An increase in the degree of crosslinking, as determined by ester carboxyl band absorbance, improves the wrinkle resistance of the fabric (at the expense of reducing mechanical strength). The accompanying data on x = absorbance and y = wrinkle resistance angle was read from a graph in the paper "Predicting the Performance of Durable Press Finished Cotton Fabric with Infrared Spectroscopy".t x 0.115 0.126 0.183 0.246 0.282 0.344 0.355 0.452 0.491 0.554 0.651 y 334 342 355 363 365 372 381 400 392 412 420 Here is regression output from Minitab: Predictor Constant absorb S = 3.60498 Coef 321.878 156.711 SOURCE Regression Residual Error Total R-Sq= 98.5% DF SE Coef 2.483 6.464 1 9 10 SS 7639.0 117.0 7756..0 T 129.64 24.24 P 0.000 0.000. R-Sq (adj) 98.3% MS 7639.0 13.0 F 587.81 (a) Does the simple linear regression model appear to be appropriate?…In this image I just want the answers for h,I and J. Thanks in advance.
- Consider the regession equation: do gib + g(e) where 6-the average diference between the monthly retum on stock i and the monthly risk free rate bthe beta of stocki s(e)a measure of the nonsystematic variance of the stocki If you estimated this regression equation and the CAPM was valid, you would expect the estimated coefficient, g. has to be Da equal to the average dference between me monthly retum on the market portfolo and the monthly r tee rate Oh equal to the nsk-thren rate of retum. De. None of the options are correctArm circumferences (cm) and heights (cm) are measured from randomly selected adult females. The 139 pairs of measurements yield x = 31.99 cm, y = 163.33 cm, r= 0.032, P-value = 0.708, and y = 158 + 0.1703x. Find the best predicted value of y (height) given an adult female with an arm circumference of 35.0 cm. Let the predictor variable x be arm circumference and the response variable y be height. Use a 0.05 significance level. %3D ..... The best predicted value is cm. (Round to two decimal places as needed.)Find the new data point (x,y) in which x=2 from the data points (1.3) and (4.12)
- Q3) An experiment was carried out to investigate variation of solubility of chemical X in water. The quantities in kg that dissolved in 1 liter at various temperatures are show in the table (1). Table (1) Temperature C Mass of X 2.1 2.6 2.9 3.3 15 20 25 30 35 4 50 5.1 70 7 Use the proper methods to answer the following questions: a) Draw a scatter diagram to show the data. b) Estimate the temperature based on the mass of X. c) What quantity might be expected to dissolve at 42 C? Find the quantity that your cquation indicates would dissolve at 10 C and comment on your answer.Regression methods were used to analyze the data from a study investigating the relationship between roadway surface temperature (x) and pavement deflection (v). Summary quantities were 20 20 20 20 20 Σx 1478 , Σy- 12.75, Σ x-1432 15.8 , Σ.-8.86 and Σ xy 1083.67 n=1 n=1 n=1 n=1 n=1Can you please help with 4.30 sulfur, the ocean and the sun? Only part A which is make a scatter plot that shows how DMS responds to SRD.