For the time series in the problem, fit a regression line by least squares; that is, find an equation of the linear regression line of y on x. Let the first year in the table correspond to x = 1. Production of Product A, 2002-2006 (thousands of units) Year 2002 2003 2004 2005 2006 Thousands 10 16 17 19 22 The least squares line is y = %3D (Do not round until the final answer. Then round to one decimal place as needed.)

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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Production of Product A, 2002-2006
(thousands of units)
For the time series in the problem, fit a regression
line by least squares; that is, find an equation of the
linear regression line of y on x. Let the first year in
the table correspond to x 1.
Year
2002 2003 2004 2005 2006
Thousands 10
16
17
19
22
The least squares line is y =.
%3D
(Do not round until the final answer. Then round to one decimal place as needed.)
Enter your answer in the answer box.
Up n
Use r
FEB
24
МaсB
esc
000
F1
F2
F3
F4
F5
#:
2$
Transcribed Image Text:Production of Product A, 2002-2006 (thousands of units) For the time series in the problem, fit a regression line by least squares; that is, find an equation of the linear regression line of y on x. Let the first year in the table correspond to x 1. Year 2002 2003 2004 2005 2006 Thousands 10 16 17 19 22 The least squares line is y =. %3D (Do not round until the final answer. Then round to one decimal place as needed.) Enter your answer in the answer box. Up n Use r FEB 24 МaсB esc 000 F1 F2 F3 F4 F5 #: 2$
Expert Solution
Step 1

Linear regression: Suppose (x1, y1), (x2, y2)---(xn, yn) are n pairs of observations on variables X and Y.

we assume that Y a dependent variable, which can be expressed in terms of x. The simplest form is the linear relation. Suppose y = β0+β1X. however, when we observe the numerical values of x and y the relation may not be observed perfectly. we assume the model.

y = β01X+ε

y = Response variable,

x = Explanatory variable.

β0 and β1 are constants.

ε error component.

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