The data shown in the following table represent the trust of a jet-turbine engin
ID: 3160084 • Letter: T
Question
The data shown in the following table represent the trust of a jet-turbine engine (y) and six candidate regressors: x1 = primary speed of rotation, x2 = secondary speed of rotation, x3 = fuel flow rate, x4 = pressure, x5 = exhaust temperature, and x6 = ambient temperature at the time of the test.
Round all intermediate calculations to four decimal places (e.g. 12.3456) and round the final answer to two decimal places (e.g. 98.76).
(a) Fit the model using y* = ln y as the response and x*3 = ln x3, x4, and x5 as the regressors.
Explanation / Answer
The data shown in the following table represent the trust of a jet-turbine engine (y) and six candidate regressors: x1 = primary speed of rotation, x2 = secondary speed of rotation, x3 = fuel flow rate, x4 = pressure, x5 = exhaust temperature, and x6 = ambient temperature at the time of the test.
Round all intermediate calculations to four decimal places (e.g. 12.3456) and round the final answer to two decimal places (e.g. 98.76).
Regression Analysis
R²
0.991
Adjusted R²
0.990
n
40
R
0.995
k
3
Std. Error
0.013
Dep. Var.
lny
ANOVA table
Source
SS
df
MS
F
p-value
Regression
0.6861
3
0.2287
1321.39
7.30E-37
Residual
0.0062
36
0.0002
Total
0.6923
39
Regression output
confidence interval
variables
coefficients
std. error
t (df=36)
p-value
99% lower
99% upper
Intercept
21.0682
9.6494
2.183
.0356
-5.1731
47.3095
lnx3
-1.4040
0.9656
-1.454
.1546
-4.0298
1.2218
x4
0.0055
0.00027515
19.951
4.74E-21
0.0047
0.0062
x5
0.00041786
0.00016489
2.534
.0158
-0.00003056
0.00086628
(a) Fit the model using y* = ln y as the response and x*3 = ln x3, x4, and x5 as the regressors.
lny = 21.0682-1.4040*lnx3+0.0055x4+0.00041786*x5
(b) Test for significance of regression using = 0.01.
Calculated F=1321.39, P=0.0000 which is < 0.01 level of significance.
The model is significant.
Calculate f0
© Use the t-statistic to test H0: j = 0 versus H0: bj 0. Use a = 0.01.
b*3 :
Calculate t0.
-1.45
Do not reject H0
b4 :
Calculate t0.
19.95
Reject H0
b5 :
Calculate t0.
2.53
Do not reject H0
Regression Analysis
R²
0.991
Adjusted R²
0.990
n
40
R
0.995
k
3
Std. Error
0.013
Dep. Var.
lny
ANOVA table
Source
SS
df
MS
F
p-value
Regression
0.6861
3
0.2287
1321.39
7.30E-37
Residual
0.0062
36
0.0002
Total
0.6923
39
Regression output
confidence interval
variables
coefficients
std. error
t (df=36)
p-value
99% lower
99% upper
Intercept
21.0682
9.6494
2.183
.0356
-5.1731
47.3095
lnx3
-1.4040
0.9656
-1.454
.1546
-4.0298
1.2218
x4
0.0055
0.00027515
19.951
4.74E-21
0.0047
0.0062
x5
0.00041786
0.00016489
2.534
.0158
-0.00003056
0.00086628
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