Consider the fuel consumption data in Table 8.7 and the model y = beta_0 + beta_
ID: 3265125 • Letter: C
Question
Consider the fuel consumption data in Table 8.7 and the model y = beta_0 + beta_1 x_1 + beta_2 x_2 + epsilon which relates fuel consumption (y) to the average hourly temperature (x_1) and the chill index (x_2). a. Plot y versus x_1 and y versus x_2. Explain why the model y = beta_0 + beta_1 x_1 + beta_2 x_2 + epsilon might be a reasonable model relating y to x_1 and x_2. For this model it can be shown that Explained variation = 25.462472 SSE = 277528 b_0 = 12.917034, b_1 = -.087064, b_2 = 090221 s squareroot c_00 = 0.5492, s squareroot c_11 =.009035063, s squareroot c_22 = 0.014122 Using this information, do the following: b. Calculate F(model) and test the adequacy of the model with alpha = 05. That is, can we reject H_0: beta_1 = beta_2 = 0 with alpha = 0.5? Interpret the result of this test. c. Calculate s, R^2, and R. Interpret the value of R^2. d. Calculate an appropriate t statistic and use it to test the importance of the intercept with alpha = 05 and with alpha = 01. e. Calculate appropriate t statistics and use them to test the importance of the variables x_1 and x_2 with alpha = 01 and alpha = 05. What do you conclude about the importance of these variables? f. Calculate 95% confidence intervals for beta_0, beta_1, and beta_2. Interpret each of these intervals. g. Calculate a point prediction of y_0, an individual weekly fuel consumption when x_01 = 40 and x_02 = 10. h. When x_01, = 40 and x_02 = 10, it can be shown that s squareroot h_00 = 0.109411. Compute and interpret a 99% confidence interval for mu_0, mean fuel consumption when x_01 = 40 and x_02 = 10. i. Compute and interpret a 99% prediction interval for y_0, an individual fuel consumption when x_01 = 40 and x_02 = 10 (see part (h)).Explanation / Answer
Answer:
a).
it is reasonable to model y with x1 and x2.
b).Calculated F= 229.37, P=0.0000 which is < 0.01 level. The model is significant at 0.05 level.
c).
s=0.235596
R square =0.989218
R=0.994594.
98.92% of variation in y is explained by the model.
Regression Analysis
R²
0.989218
Adjusted R²
0.984905
n
8
R
0.994594
k
2
Std. Error
0.235596
Dep. Var.
y
ANOVA table
Source
SS
df
MS
F
p-value
Regression
25.462472
2
12.731236
229.37
1.21E-05
Residual
0.277528
5
0.055506
Total
25.740000
7
Regression output
confidence interval
variables
coefficients
std. error
t (df=5)
p-value
95% lower
95% upper
Intercept
12.917034
0.5492
23.520
2.59E-06
11.5053
14.3288
x1
-0.087064
0.0090
-9.636
.0002
-0.1103
-0.0638
x2
0.090221
0.0141
6.389
.0014
0.0539
0.1265
d).
calculated t=23.52, P=0.0000 which is < 0.05 and 0.01 level.
Intercept is significant from 0 at 0.05 and 0.01 level.
e).
for variable x1,
calculated t=-6.636, P=0.0002 which is < 0.05 and 0.01 level.
x1 is significant from 0 at 0.05 and 0.01 level.
for variable x2,
calculated t=6.389, P=0.0014 which is < 0.05 and 0.01 level.
X2 is significant from 0 at 0.05 and 0.01 level.
f).
Regression output
confidence interval
variables
coefficients
std. error
t (df=5)
p-value
95% lower
95% upper
Intercept
12.917034
0.5492
23.520
2.59E-06
11.5053
14.3288
x1
-0.087064
0.0090
-9.636
.0002
-0.1103
-0.0638
x2
0.090221
0.0141
6.389
.0014
0.0539
0.1265
We are 95% confidence that the true intercept and coefficients falls in the given interval.
g).
Predicted values for: y
99% Confidence Interval
99% Prediction Interval
x1
x2
Predicted
lower
upper
lower
upper
40
10
10.336694
9.895532
10.777856
9.289295
11.384092
predicted y=10.336694
h) 99% confidence interval = (9.895532, 10.777856)
Regression Analysis
R²
0.989218
Adjusted R²
0.984905
n
8
R
0.994594
k
2
Std. Error
0.235596
Dep. Var.
y
ANOVA table
Source
SS
df
MS
F
p-value
Regression
25.462472
2
12.731236
229.37
1.21E-05
Residual
0.277528
5
0.055506
Total
25.740000
7
Regression output
confidence interval
variables
coefficients
std. error
t (df=5)
p-value
95% lower
95% upper
Intercept
12.917034
0.5492
23.520
2.59E-06
11.5053
14.3288
x1
-0.087064
0.0090
-9.636
.0002
-0.1103
-0.0638
x2
0.090221
0.0141
6.389
.0014
0.0539
0.1265