A regression of y = calcium content (g/L) on x = dissolved material (mg/cm2) was reported in the article "Use of Fly Ash or Silica Fume to Increase the Resistance of Concrete to Feed Acids" (Mag. Concrete Res., 1997: 337–344). The equation of the estimated regression line was y = 3.678 + .144x, with r2 = .860, based on n = 23. a. Interpret the estimated slope .144 and the coefficient of determination .860.
b. Calculate a point estimate of the true average calcium content when the amount of dissolved material is 50 mg/cm2.
c. The value of total sum of squares was SST = 320.398. Calculate an estimate of the error standard deviation σ in the simple linear regression model.

a) The slope = 0.144 , means that an increase in dissolved material by 1 mg/cm^2 will lead to an increase in Calcium content by 0.144 g/l

while the coefficient of determination = 0.869 , means that 86% of the variation ( natural ) of calcium content is explainable using the Linear model.

b) 10.878

c) S ≈ 1.99

Step-by-step explanation:

a) The slope = 0.144 , means that an increase in dissolved material by 1 mg/cm^2 will lead to an increase in Calcium content by 0.144 g/l

while the coefficient of determination = 0.869 , means that 86% of the variation ( natural ) of calcium content is explainable using the Linear model.

B) Point estimate of true average calcium content

Given that dissolved material = 50 mg/cm^2

= 3.678 + 0.144 * dissolved material

=  3.678 + 7.2 =  10.878

C) Determine an estimate of error standard deviation

Given that ; sum of squares ( SST ) = 320.398

                   COD ( r )  = 0.86

first;  calculate the value of SSE = SST - SST ( r^2 )

SSE = 320.398 - 320.398 ( 0.86) ^2

       = 83.43

Finally ; determine the estimate of error standard deviation

S^2 = SSE / n -2

      = 83.43 / 23 - 2

∴ S ≈ 1.99

answer: y=4000x+14000

step-by-step explanation: