Question

An article gave a scatter plot along with the least squares line of x = rainfall volume (m3) and y = runoff volume (m3) for a particular location. The accompanying values were read from the plot.

x 8 12 14 16 23 30 40 52 55 67 72 81 96 112 127

y 4 10 13 15 15 25 27 47 38 46 53 69 82 99 103

(a) Does a scatter plot of the data support the use of the simple linear regression model? Yes, the scatterplot shows a reasonable linear relationship. Yes, the scatterplot shows a random scattering with no pattern. No, the scatterplot shows a reasonable linear relationship. No, the scatterplot shows a random scattering with no pattern.

(b) Calculate point estimates of the slope and intercept of the population regression line. (Round your answers to five decimal places.) slope intercept

(c) Calculate a point estimate of the true average runoff volume when rainfall volume is 51. (Round your answer to four decimal places.) m3

(d) Calculate a point estimate of the standard deviation σ. (Round your answer to two decimal places.) m3

(e) What proportion of the observed variation in runoff volume can be attributed to the simple linear regression relationship between runoff and rainfall? (Round your answer to four decimal places.)

Answer 1

R codes:

a) a) Does a scatter plot of the data support the use of the simple linear regression model?

rainfall.vol <- c(8,12,14,16,23,30,40,52,55,67,72,81,96,112,127)
runoff.vol <- c(4,10,13,15,15,25,27,47,38,46,53,69,82,99,103)
plot(runoff.vol, rainfall.vol, xlab = "Rainfall Volume", ylab = "Runoff Volume")

Most of the data points on the scatter plot appear to lie on a straight line, supporting the use of the simple linear regression model.

b) Calculate point estimates of the slope and intercept of the population regression line.

fit <- lm(runoff.vol ~ rainfall.vol)
summary(fit)

Call:
lm(formula = runoff.vol ~ rainfall.vol)

Residuals:
Min 1Q Median 3Q Max
-8.304 -3.384 1.880 3.310 6.769

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.16451 2.16690 -0.999 0.336
rainfall.vol 0.84282 0.03329 25.315 1.91e-12 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4.748 on 13 degrees of freedom
Multiple R-squared: 0.9801,   Adjusted R-squared: 0.9786
F-statistic: 640.8 on 1 and 13 DF, p-value: 1.907e-12

The estimate of the slope is 0.84282 and the esimate of the intercept is -2.16451 for the population regression line.

c) Calculate a point estimate of the true average runoff volume when rainfall volume is 51

est <- c(fit$coefficients[1] + fit$coefficients[2] * 51)
> est
(Intercept)
40.81916
When rainfall volume is 51, the estimate of the true average runoff volume is 40.8191

d) Calculate a point estimate of the standard deviation \( \sigma \).

> summary(fit)$sigma
[1] 4.748162

The estimate of the standard deviation \( \sigma \) is 4.74

e)What proportion of the observed variation in runoff volume can be attributed to the simple linear regression relationship between runoff and rainfall?

summary(fit)$r.squared
[1] 0.9801176

The proportion of the observed variation in runoff volume that can be attributed to the simple linear regression relationship between runoff and rainfall is 0.9801