In: Statistics and Probability
A statistical program is recommended.
A highway department is studying the relationship between traffic flow and speed. The following model has been hypothesized:
y = β0 + β1x + ε
where
The following data were collected during rush hour for six highways leading out of the city.
Traffic Flow (y) |
Vehicle Speed (x) |
---|---|
1,256 | 35 |
1,328 | 40 |
1,228 | 30 |
1,337 | 45 |
1,349 | 50 |
1,122 | 25 |
Develop an estimated regression equation for the data of the form
ŷ = b0 + b1x + b2x2.
(Round b0 to the nearest integer and b1 to two decimal places and b2 to three decimal places.)
ŷ =
Find the value of the test statistic. (Round your answer to two decimal places.)
Find the p-value. (Round your answer to three decimal places.)
p-value =
Base on the model predict the traffic flow in vehicles per hour at a speed of 38 miles per hour. (Round your answer to two decimal places.) vehicles per hou
question 2-
A statistical program is recommended.
Car manufacturers produced a variety of classic cars that continue to increase in value. Suppose the following data is based upon the Martin Rating System for Collectible Cars, and shows the rarity rating (1–20) and the high price ($1,000) for 15 classic cars.
Model | Rating | Price ($1,000) |
---|---|---|
A | 16 | 225.0 |
B | 16 | 375.0 |
C | 19 | 1,325.0 |
D | 18 | 1,625.0 |
E | 19 | 4,025.0 |
F | 17 | 400.0 |
G | 15 | 102.5 |
H | 14 | 87.0 |
I | 17 | 450.0 |
J | 17 | 140.0 |
K | 19 | 2,675.0 |
L | 18 | 1,000.0 |
M | 18 | 350.0 |
N | 16 | 100.0 |
O | 13 | 95.0 |
Develop an estimated multiple regression equation with x = rarity rating and
x2
as the two independent variables. (Round b0 and b1 to the nearest integer and b2 to one decimal place.)
ŷ =
Consider the nonlinear relationship shown by equation (16.7):
E(y) = β0β1x
Use logarithms to develop an estimated regression equation for this model. (Round b0 to three decimal places and b1 to four decimal places.)
log(ŷ) =
1)
Excel > Data > Data Analysis > Regression
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.989505956 | |||||||
R Square | 0.979122037 | |||||||
Adjusted R Square | 0.965203395 | |||||||
Standard Error | 16.25305832 | |||||||
Observations | 6 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 2 | 37165.51429 | 18582.75714 | 70.34609006 | 0.0030167 | |||
Residual | 3 | 792.4857143 | 264.1619048 | |||||
Total | 5 | 37958 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | 421.2857143 | 144.9818614 | 2.905782214 | 0.062209261 | -40.11127487 | 882.6827034 | -40.11127487 | 882.6827034 |
Vehicle Speed(x) | 38.01571429 | 8.017835518 | 4.741393634 | 0.017792499 | 12.49938326 | 63.53204531 | 12.49938326 | 63.53204531 |
X^2 | -0.39 | 0.106401243 | -3.665370717 | 0.035113239 | -0.728616242 | -0.051383758 | -0.728616242 | -0.051383758 |
Regression equation:
Y = 421+38.02*X-0.390*X^2
Fstat = 70.35
P value = 0.003
If X = 38 miles per hour
Y = 421+38.02*X-0.390*X^2
Y = 421+38.02*38-0.390*38^2 = 1302.60 vehicles per hour
2)
a) Quadratic regression:
Y = 34745-4683*X+157.0*X^2
Exponential regression:
Price ($1,000)(Y) | Ln(y) | Rating(X) |
225 | 5.4161 | 16 |
375 | 5.926926 | 16 |
1325 | 7.189168 | 19 |
1625 | 7.393263 | 18 |
4025 | 8.30028 | 19 |
400 | 5.991465 | 17 |
102.5 | 4.629863 | 15 |
87 | 4.465908 | 14 |
450 | 6.109248 | 17 |
140 | 4.941642 | 17 |
2675 | 7.891705 | 19 |
1000 | 6.907755 | 18 |
350 | 5.857933 | 18 |
100 | 4.60517 | 16 |
95 | 4.553877 | 13 |
Log(Y) = -4.175+0.6064*X
Y = e^(-4.175+0.6064*X)
Y = 0.0154*1.8338^X