Question

3. No-fines conerete, made from a uniformly graded coarse aggregate and a cement-water paste, is beneficial in areas prone to excessive rainfall because of its excellent drainage properties. The article "Pavement Thickness for No-Fines Concrete Parking Lots" (J. of Transportation Engr., 1995: 476-484) evaluated the relationship between Y = porosity (%) and X unit weight (lb/ft) in concrete specimens. Summary data resulting from the study is provided below - - = -471.5 lb/ft2-% n 15 =1640.1 lb/ft SK6.1 lb/ft i=1 y 299.87% sy= 5.6% i-1 (a) Calculate the regression coefficients bo and bi for a simple linear model: y = bo + bix (b) Calculate the residual mean square error s^ (c) What proportion of the observed variability in the porosity can be explained by unit weight? (In other words, what is the value of R22) (d) Compute the adjusted-R2 value for this model (e) Given that a greater porosity is expected for smaller unit weights, perform a hypothesis test to determine if the slope parameter (B1) is significantly less than zero. What do you conclude at the 5% level?

Answer 1

The article "Pavement Thickness for No-Fines Concrete Parking Lots" evaluated the relationship between Y = porosity (%) and X =unit weight (15/ft) in concrete specimens. The summary of the data is given below: n=15 Sy = (x:-7)(y; - 5) = -471.5 16! ft 2x = 1640.1 10. HP S = 6.116! Žv = 299.87% s=5.6% The linear regression model is given as follows: y=bo +bix The formula for calculating regression coefficients b, and b is given below: by = ī-ba The mean of the unit weight in concrete specimens is calculated as follows: n21 1640.1 15 =109.34 Hence, 1 =109.34 The mean of the porosity in concrete specimens is calculated as follows: niet 299.8 15 = 19.9867 Hence. J = 19.9867 The value of S is calculated as follows: S = 6.1 (*: -F)* = 6.1 M52 (x; – 5)2 = 6.12 x14 S = 520.94 Hence. S: = 520.94 The value of S, is calculated as follows: sy = 5.6 Vis-i Ş0-5)=5.6 1)2 = 5.62x14 S, = 439.04 Hence, S. = 439.04. -0.905 The regression coefficients b, and is calculated as follows: 471.5 =-520.94 bo = 19.9867-(-0.905)(109.34)=118.9394 Hence, bo = 118.9394 and 5 = -0.905) 520.94 The residual mean square error is calculated as follows: 02SSR n-2 n-2 439.04-(-0.905) (-471.5) 15-2 12.3325 13 = 0.9486 Hence, the residual mean square error is Q2 =0.9486).

The proportion of the observed variability in the porosity can be explained by the unit weight is calculated as follows: R?=1-SSR -1_12.3325 439.04 =1-0.0280 =0.972 Hence, the proportion of the observed variability in the porosity can be explained by the unit weight is R2 =0.972 The adjusted R2 is calculated as follows: -1-[(1357-1)(-0.972) =1 -0.03015 = 0.96985 Hence, Adj R2 = 0.96985. To perform a hypothesis test to determine if the slope parameter is significantly less than zero. The hypothesis are given as follows: The null hypothesis is H.:B20. The alternative hypothesis is H:8 <0. The test statistic is given as follows: - Å B SEB Pankkee The standard error of A is calculated as follows: SE(A)VIS 10.94867 V 520.94 = 0.04156 Hence. S.E(B)=0.04156 The test statistic is calculated as follows: -0.905-0 0.04156 =-21.7757 Hence, tra=-21.7757 The degrees of freedom is df =n-1=15-1=14 The critical value of t-statistic at 5% level of significance at 14 degrees of freedom is * = 2.145 Decision rule: If |ta|>tras then reject the null hypothesis. Here. t = 21.7757 > = 2.145 then reject the null hypothesis. Therefore, it can be concluded that there is enough evidence to show that A is less than zero.