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	4	12	14	18	23	30	40	47	55	67	72	80	96	112	127
y	4	10	13	14	15	25	27	44	38	46	53	69	82	99	100

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

slope	0.82834
intercept	-1.41247

(c) Calculate a point estimate of the true average runoff volume when rainfall volume is 51. (Round your answer to four decimal places.)
40.8329m³

(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.)

____________________

Please solve D and E, the rest of the answers are correct.

Answer 1

Sol;

with lm fucntion in R fit a linear model of Y on X

x <- c(   4   ,12   ,14   ,18,   23,   30,   40,   47,   55,   67,   72,   80,   96,   112,   127)
y <- c(4,   10,   13   ,14,   15,   25,   27,   44,   38,   46,   53,   69,   82,   99,   100)
lmodel <- lm(y~x)
coefficients(lmodel)
summary(lmodel)
anova(lmodel)

Output:

> coefficients(lmodel)
(Intercept) x
-1.4124839 0.8283403
> anova(lmodel)
Analysis of Variance Table

Response: y
Df Sum Sq Mean Sq F value Pr(>F)
x 1 14106 14105.6 566.04 4.206e-12 ***
Residuals 13 324 24.9
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> sqrt( 24.9)
[1] 4.98999
> summary(lmodel)

Call:
lm(formula = y ~ x)

Residuals:
Min 1Q Median 3Q Max
-8.086 -4.254 1.472 3.354 7.638

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.41248 2.25466 -0.626 0.542
x 0.82834 0.03482 23.792 4.21e-12 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4.992 on 13 degrees of freedom
Multiple R-squared: 0.9775,   Adjusted R-squared: 0.9758
F-statistic: 566 on 1 and 13 DF, p-value: 4.206e-12

ANSWER(D)

y^= -1.4124839 +0.8283403 *x

point estimate of the standard deviation=sqrt(MSE)=sqrt(24.9 )= 4.98999

=4.99(rounded to 2 decimals)

ANSWER(E)

proportion of the observed variation in runoff volume can be attributed to the simple linear regression relationship between runoff and rainfall is= R sq

=0.9775

ANSWERS(D)-4.99

ANSWERS(E)-0.9775