An article gave a scatter plot, along with the least squares line, of x = rainfa

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 simple linear regression model provides a very good fit to data on rainfall and runoff volume (n = 15) given below. The equation of the least squares line is

y = ?1.242 + 0.83162x,

r2 = 0.974, and s = 5.44.

  ,

m3

Does the resulting interval suggest that precise information about the value of runoff for this future observation is available? Explain your reasoning.

Yes, precise information is available because the resulting interval is very wide.Yes, precise information is available because the resulting interval is very narrow.    No, precise information is not available because the resulting interval is very wide.No, precise information is not available because the resulting interval is very narrow.

(b) Calculate a 95% PI for runoff when rainfall is 50. (Round your answers to two decimal places.)

  ,

m3

x 5 12 14 17 23 30 40 47 55 67 72 81 96 112 127 y 4 10 14 15 14 24 26 47 39 47 53 70 82 100 100

Explanation / Answer

Run Regression in R:

rcode:

x <- c( 5 ,12 ,14 ,17 ,23 ,30 ,40 ,47 ,55 ,67, 72 ,81 ,96, 112 ,127)

y <- c( 4, 10, 14, 15, 14, 24 ,26 ,47, 39, 47, 53, 70, 82, 100 ,100)

rmod=lm(y~x)

coefficients(rmod)

newdata = data.frame(x=40)

predict(rmod, newdata, interval="predict")

OUTPUT:

> x <- c( 5 ,12 ,14 ,17 ,23 ,30 ,40 ,47 ,55 ,67, 72 ,81 ,96, 112 ,127)

> y <- c( 4, 10, 14, 15, 14, 24 ,26 ,47, 39, 47, 53, 70, 82, 100 ,100)

> rmod=lm(y~x)

> coefficients(rmod)

(Intercept) x

-1.242024 0.831617

> newdata = data.frame(x=40)

> predict(rmod, newdata, interval="predict")

fit lwr upr

1 32.02266 19.83676 44.20855

REGRESSION EQ IS

Y= -1.242024 + 0.831617 x

when rainfall =40mm^3 95% PI for runoff is

lower limit=19.84

upper limit=44.21

b) Calculate a 95% PI for runoff when rainfall is 50. (Round your answers to two decimal places.)

Rcode:

rmod=lm(y~x)
coefficients(rmod)
newdata1 = data.frame(x=50)
predict(rmod, newdata1, interval="predict")

output:

(Intercept) x
-1.242024 0.831617
> newdata1 = data.frame(x=50)
> predict(rmod, newdata1, interval="predict")
fit lwr upr
1 40.33883 28.19816 52.47949

when rainfall=50,95% PI for runoff is

lower limt=28.2

upper imit=52.48

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