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	5	12	14	18	23	30	40	46	55	67	72	80	96	112	127
y	4	10	13	15	15	25	27	46	38	46	53	72	82	99	101

(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 47. (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

a)

Yes, the scatterplot shows a reasonable linear relationship.

b)

slope =0.8349

intercept =-1.2953

c)

predicted val=-1.2953+47*0.8349=

37.9450

(please try 37.9458 if this comes wrong)

d)

SSE =Syy-(Sxy)2/Sxx=

390.975

a	s2 =SSE/(n-2)=	30.0750
std error σ              =	=se =√s2=	5.48

e)

SST=Syy=		14,662.9333
SSE =Syy-(Sxy)2/Sxx=		390.975
SSR =(Sxy)2/Sxx =		14,271.9588

Coeffficient of determination R^2 =SSR/SST=

0.9733