A regression of y=calcium content (g/L) on x=dissolved material (mg/cm^2) was re

Question

A regression of y=calcium content (g/L) on x=dissolved material (mg/cm^2) was reported in an article. The eqn of the estimated regression line was y=3.678+.144x with r^2=.860, based on n=23.

a.) Interpret the estimated slope (.144) and the coeff. of determination .860
b.) Calculate a point estimate of the true avg calcium content when the amount of dissolved material is 50 mg/cm^2.

c.) The value of the total sum of squares was SST=320.398. Calculate an estimate of the error standard deviation in the simple linear regression model.

Explanation / Answer

a) The estimated slope of 0.144 tells us that if the concentration of dissolved material were to increase by 1 mg/cm^2 then the calcium content would increase by 0.144 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 86%. This is fairly high. b) The point estimate is 3.678 + 0.144 x 50 = 10.878 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, 0.86 = 1 - RSS/SST = 1 - RSS/320.398 Therefore, RSS = (1 - 0.86) x 320.398 = 44.856. Therefore, the estimated standard deviation is sqrt (44.856/21) = 1.462.

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