Determine the values of the three principal components for the following observation. Age, Experience, and Income ($1000) of a bank customer were used in a principal component analysis. The mean of the original variables and thecoefficients of the three principal components are given in the table below.VariableMeanPC1PC2PC3Age280.8440.145-0.04Experience-0.12290.9150.022Income ($1000)1900.05-0.0070.824Determine the values of the three principal components for the following observation.VariableValueAge66Experience22Income140Answer:Z1 =Round your answer to 3 decimal places.Z2 =Round your answer to 3 decimal places.Z3 =Round your answer to 3 decimal places.

Answered: Determine the values of the three…

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter4: Equations Of Linear Functions

Section4.5: Correlation And Causation

Problem 11PPS

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Question

Determine the values of the three principal components for the following observation.

Age, Experience, and Income ($1000) of a bank customer were used in a principal component analysis. The mean of the original variables and the
coefficients of the three principal components are given in the table below.
Variable
Mean
PC1
PC2
PC3
Age
28
0.844
0.145
-0.04
Experience
-0.122
9
0.915
0.022
Income ($1000)
190
0.05
-0.007
0.824
Determine the values of the three principal components for the following observation.
Variable
Value
Age
66
Experience
22
Income
140
Answer:
Z1 =
Round your answer to 3 decimal places.
Z2 =
Round your answer to 3 decimal places.
Z3 =
Round your answer to 3 decimal places.

Expert Solution

Step 1

principle components is the new variable which is linear combination of initial variable .

In the above example initial variables are Age , Experience and Income based on that three principle components are given as follows :

$z_{1} = (0.844 \times A g e) + (- 0.122 \times E x p e r i e n c e) + (0.05 \times I n c o m e) z_{1} = (0.844 \times 66) + (- 0.122 \times 22) + (0.05 \times 140) z_{1} = 55.704 - 2.684 + 7 z_{1} = 60.020 z_{2} = (0.145 \times A g e) + (0.915 \times E x p e r i e n c e) + (- 0.007 \times I n c o m e) z_{2} = (0.145 \times 66) + (0.915 \times 22) + (- 0.007 \times 140) z_{2} = 9.57 + 20.13 - 0.98 z_{2} = 28.720 z_{3} = (- 0.04 \times A g e) + (0.022 \times E x p e r i e n c e) + (0.824 \times I n c o m e) z_{3} = (- 0.04 \times 66) + (0.022 \times 22) + (0.824 \times 140) z_{3} = - 2.64 + 0.484 + 115.36 z_{3} = 113.204$

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