Question

A highway department is studying the relationship between traffic flow and speed. The following model has been hypothesized. ŷ = bo + b12+b222 where y=traffic flow in vehicles per hour I = vehicle speed in miles per hour The following data were collected during rush hour for six highways leading out of the city. Vehicle Speed (2) Traffic Flow (y) 1257 1330 1227 1336 1350 1125 a. Use the data to compute the coefficients of this estimated regression equation (to 4 decimals). (Enter negative value as negative number). bp = b1 = (Create the xa variable first using Data/Transform Data/Square.) b. Using a = .01, test for a significant relationship. F= (to 4 decimals) p-value = (to 4 decimals) The relationship - Select your answer - significant.

c. Estimate the traffic flow in vehicles per hour at a speed of 39 miles per hour (to 2 decimals).

The estimated value is = ____ (to 2 decimals)

99% Confidence interval = ( , ) (to 2 decimals)

99% Prediction interval = ( , ) (to 2 decimals)

Answer 1

Sol:

use lm fucntion in R to fit a linear model

predict function to get 99% confidence and prediction intervals

Rcode:

df1 =read.table(header = TRUE, text ="
Y   X2   X1
1257   1225   35
1330   1600   40
1227   1225   35
1336   2025   45
1350   3025   55
1125   900   30

"
)
df1
linmod <- lm(Y~X1+X2,data=df1)
round(coefficients(linmod),4)
attach(df)
summary(linmod)
newdata=data.frame(X1=39,X2=1521)
predict(linmod,newdata,interval="confidence",level=0.99)
predict(linmod,newdata,interval="predict",level=0.99)

Output::

Call:
lm(formula = Y ~ X1 + X2, data = df1)

Residuals:
1 2 3 4 5 6
18.199 16.248 -11.801 -20.513 4.529 -6.662

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -187.1382 236.5376 -0.791 0.4866
X1 63.2733 11.3875 5.556 0.0115 *
X2 -0.6438 0.1325 -4.860 0.0166 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 20.17 on 3 degrees of freedom
Multiple R-squared: 0.9674,   Adjusted R-squared: 0.9456
F-statistic: 44.47 on 2 and 3 DF, p-value: 0.005894

ANSWERS(a)

bo=  -187.1382

b1= 63.2733

b2=   -0.6438

ANSWERS(b)

F= 44.4709

p= 0.0059

P<0.01

The relationship is significant

Solution-c:

round(predict(linmod,newdata,interval="confidence",level=0.99),2)
round(predict(linmod,newdata,interval="predict",level=0.99),2)

Ouptut:

round(predict(linmod,newdata,interval="confidence",level=0.99),2)
fit lwr upr
1 1301.34 1234.21 1368.46
> round(predict(linmod,newdata,interval="predict",level=0.99),2)
fit lwr upr
1 1301.34 1165.76 1436.91

estimated value=1301.34

99% Confidence interval =( 1234.21, 1368.46)

99% Prediction interval =( 1165.76 1436.91)