The article \"Characterization of Highway Runoff in Austin, Texas\" (J. of Envir
ID: 3223733 • Letter: T
Question
The article "Characterization of Highway Runoff in Austin, Texas" (J. of Envir. Engr., 1998;131-137) gave a scatterplot, along with the least squares line of x=rainfall volume (m^3) and y=runoff volume (m^3) for a particular location. We are trying to predict y=runoff volume from x=rainfall volume from the 15 data points below: Residual standard error: 5.24 on 13 degrees of freedom Multiple R-squared: 0.9753, Adjusted R-squared: 0.9734 F-statistic: 512.7 on 1 and 23 DF, p-value: 7.896e-12 Discuss the plots given. How do they relate to the linear regression model? write the equation of the linear model (least squares regression line) from the R output. Calculate a point estimate of the true average runoff volume when rainfall volume is 50 m^3. Use the data, the output, and t-table to verify by calculation the following values: (a) R^3 = 0.9753, (b) t = 22.642 (testing H_0: beta = 0 against H_a: beta notequalto 0), and (c) associated p-valueExplanation / Answer
Q-1
Y versus X plot is scatter plot which tells us that the points are close to the regression line indicating that the assumption of linearity is satisfied.
Residual versus fitted plot shows a funnel shape indicating violation of homoscedasticity of errors assumption.
Normal Q-Q plot shows points close to normal line indicating assumption of normality of errors is satisfied.
Q-2
Regression modlel is Y=-1.12830 + 0.82697 X
Estimated true average runoff volume =-1.12830 + 0.82697 * 50 =40.2202
Q-3
As t=22.642 has p-value=7.9e-12 <0.05 and F=512.65 has p-value=7.895e-12<0.05, so we reject the null hypothesis and conclude that X is significant predictor of Y.
Q-4
R-squar eis 0.9753 which means that X explained 97.53% of the variability in Y. So regression model quality is very good.
Q-5
97.53% of the variability in runoff volume can beattributed to simple regression model
Q-6
Standard error of regression is 5.24 which is the typical deviation from regression line.
Q-7
We have s2=5.242=27.4576=MSE=27.5
Q-8
Degree of freedom =13 (Error)
Critical t at 5% level = 2.1604
Predicted yhat==-1.12830 + 0.82697*1 =-0.3013
For 95% confidence interval =yhat± t(0.05,13)*s*sqrt(1/n+(x0-xbar)2/Sxx)
=-0.3013±2.1604*5.24*sqrt(1/15+(1-42.87)2/14079)
=(-5.37 4.76)
Q-1
Y versus X plot is scatter plot which tells us that the points are close to the regression line indicating that the assumption of linearity is satisfied.
Residual versus fitted plot shows a funnel shape indicating violation of homoscedasticity of errors assumption.
Normal Q-Q plot shows points close to normal line indicating assumption of normality of errors is satisfied.
Q-2
Regression modlel is Y=-1.12830 + 0.82697 X
Estimated true average runoff volume =-1.12830 + 0.82697 * 50 =40.2202
Q-3
As t=22.642 has p-value=7.9e-12 <0.05 and F=512.65 has p-value=7.895e-12<0.05, so we reject the null hypothesis and conclude that X is significant predictor of Y.
Q-4
R-squar eis 0.9753 which means that X explained 97.53% of the variability in Y. So regression model quality is very good.
Q-5
97.53% of the variability in runoff volume can beattributed to simple regression model
Q-6
Standard error of regression is 5.24 which is the typical deviation from regression line.
Q-7
We have s2=5.242=27.4576=MSE=27.5
Q-8
Degree of freedom =13 (Error)
Critical t at 5% level = 2.1604
Predicted yhat==-1.12830 + 0.82697*1 =-0.3013
For 95% confidence interval =yhat± t(0.05,13)*s*sqrt(1/n+(x0-xbar)2/Sxx)
=-0.3013±2.1604*5.24*sqrt(1/15+(1-42.87)2/14079)
=(-5.37 4.76)
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