Question

Problem 4: Using the data in the table below, answer the following: 0.16 0.22 0.30 0.30 0.32 0.33 0.35 0.37 0.37 0.38 3.30 11.70 13.60 14.30 14.40 16.00 16.70 17.80 19.10 21.00 0.50 0.53 0.60 0.60 0.61 0.63 0.73 0.76 0.84 0.85 22.20 123.70 27.50 27.90 28.50 28.60 28.70 29.20 31.10 31.50 (a) Calculate the least squares estimates of the slope and intercept (you may use Excel to aid in your calculations, but cannot have Excel fit the regression itself). What is the regression line? (b) Use your regression equation to predict the value of y if x = 0.47. (c) Find the residual for the observation where x = 0.32 and y = 14.40. Was the prediction fit by the model for x = 0.32 an overestimate, or an underestimate?

Answer 1

Using R code we have the answers.

CODE:

x <- c(0.16,0.22,0.3,0.3,0.32,0.33,0.35,0.37,0.37,0.38,0.5,0.53,0.6,0.6,0.61,0.63,0.73,0.76,0.84,0.85)
y <- c(3.3,11.7,13.6,14.3,14.4,16,16.7,17.8,19.1,21,22.2,23.7,27.5,27.9,28.5,28.6,28.7,29.2,31.1,31.5)
reg <- lm(y~x)
reg
predict(reg,data.frame(x = 0.47))
14.4 - predict(reg,data.frame(x = 0.47))

OUTPUT:

> x <- c(0.16,0.22,0.3,0.3,0.32,0.33,0.35,0.37,0.37,0.38,0.5,0.53,0.6,0.6,0.61,0.63,0.73,0.76,0.84,0.85)
> y <- c(3.3,11.7,13.6,14.3,14.4,16,16.7,17.8,19.1,21,22.2,23.7,27.5,27.9,28.5,28.6,28.7,29.2,31.1,31.5)
> reg <- lm(y~x)
> reg

Call:
lm(formula = y ~ x)

Coefficients:
(Intercept) x
3.844 35.888

> predict(reg,data.frame(x = 0.47))
1
20.71195
> 14.4 - predict(reg,data.frame(x = 0.47))
1
-6.311952

(a) The least squares estimates are,

slope = 35.888 and intercept = 3.844.

The regression line is Y = 3.844 + 35.888*X

(b) For X = 0.47, predicted Y = 20.71195.

(c) For X = 0.32, we have the residual -6.311952, which incidcates the prediction was overestimate.