A regression of y calcium content gL on x dissolved materi

A regression of y = calcium content (g/L) on x = dissolved material (mg/cm²) was reported. The equation of the estimated regression line was y = 3.612 + 0.142x, with r ² = 0.84, based on n = 18.

(a) Calculate a point estimate of the true average calcium content when the amount of dissolved material is 51 mg/cm². (Give answer accurate to 2 decimal places.)


(b) The value of total sum of squares was SST = 317.770. Calculate an estimate of the error of standard deviation ? in the simple linear model. (Give answer accurate to 2 decimal places.)

Solution

a)
The estimated slope of 0.142 tells us that if the concentration of dissolved material were to increase by 1 mg/cm^2 then the calcium content would increase by 0.142 g/L.

The coefficient of determination tells us that the proportion of the observed variability in calcium content that is explained by taking account of the concentration of dissolved material is 84%. This is fairly high.

b) The point estimate is 3.612 + 0.142 x 51 = 10.854 g/L.

c) The coefficient of determination is 1 - RSS/SST, where RSS is the residual sum of squares.
The formula for the estimated standard deviation is
sqrt[RSS/(n-2)].

So, 0.84 = 1 - RSS/SST = 1 - RSS/317.77

Therefore, RSS = (1 - 0.84) x 317.77 = 50.84.

Therefore, the estimated standard deviation is

sqrt (50.84/16) = 1.783