19) Use the following linear regression equation to answer the questions.
x_{3} = −18.7 + 4.3x_{1} + 8.6x_{4} − 1.0x_{7}
(b) Which number is the constant term? List the coefficients with
their corresponding explanatory variables.
constant | |
x_{1} coefficient | |
x_{4} coefficient | |
x_{7} coefficient |
(c) If x_{1} = 5, x_{4} = -6, and
x_{7} = 4, what is the predicted value for
x_{3}? (Round your answer to one decimal
place.)
x_{3} =
(d)
Suppose x_{1} and x_{7} were held
at fixed but arbitrary values.
If x_{4} increased by 1 unit, what would we expect
the corresponding change in x_{3} to be?
If x_{4} increased by 3 units, what would be the
corresponding expected change in x_{3}?
If x_{4} decreased by 2 units, what would we
expect for the corresponding change in
x_{3}?
(e) Suppose that n = 18 data points were used to construct
the given regression equation and that the standard error for the
coefficient of x_{4} is 0.946. Construct a 90%
confidence interval for the coefficient of x_{4}.
(Round your answers to two decimal places.)
lower limit | |
upper limit |
(f) Using the information of part (e) and level of significance 1%,
test the claim that the coefficient of x_{4} is
different from zero. (Round your answers to two decimal
places.)
t | = | |
t critical | = ± |
b)
constant | -18.7 | |
x1 coeffiicent | 4.3 | |
x4 coefficient | 8.6 | |
x7 coeffiicent | -1 |
c)
predicted value =-18.7+4.3*5+8.6*(-6)-1*4 =-52.8
d)
If x_{4} increased by 1 unit, what would we expect the corresponding change in x_{3} to be =8.6
If x_{4} increased by 3 units, what would be the corresponding expected change in x_{3 =}3*8.6=25.8
If x_{4} decreased by 2 units, what would we expect for the corresponding change in x_{3} =-2*8.6 =-17.2
e)
sample size n= | 18 | ||||
number of independent variables p= | 3 | ||||
degree of freedom =n-p-1= | 14 | ||||
estimated slope b= | 8.6 | ||||
standard error of slope=s_{b}= | 0.9460 | ||||
for 90 % confidence and 14degree of freedom critical t= | 1.7610 | ||||
90% confidence interval =b1 -/+ t*standard error= | (6.93,10.27) |
lower limit =6.93 | |
upper limit=10.27 |
f)
t=8.6/0.946=9.09
tcritical= -/+ 2.98
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