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Recall the data for the three populations. p1 = 0.576      n1 = 250 p2 = 0.48   n2 = 300 p3 = 0.45   n3 = 200 Find the critical value for the pairwise difference between populations i = 1 and j = 3,  CV13,  rounding the result to four decimal places. CVij  =    ??2    pi(1 − pi) ni +  pj(1 − pj) nj  CV13  =    ?20.05    p1(1 − p1) n1 +  p3(1 − p3) n3     =    5.991    0.576(1 − 0.576) 250 +  0.45(1 − 0.45)       =    Find the critical value for the pairwise difference between populations i = 2 and j = 3,  CV23,  rounding the result to four decimal places. CVij  =    ??2    pi(1 − pi) ni +  pj(1 − pj) nj  CV23  =    ?20.05    p2(1 − p2) n2 +  p3(1 − p3) n3     =    5.991    0.48(1 − 0.48) 300 +  0.45(1 − 0.45)       =

Recall the data for the three populations. p1 = 0.576      n1 = 250 p2 = 0.48   n2 = 300 p3 = 0.45   n3 = 200 Find the critical value for the pairwise difference between populations i = 1 and j = 3,  CV13,  rounding the result to four decimal places. CVij  =    ??2    pi(1 − pi) ni +  pj(1 − pj) nj  CV13  =    ?20.05    p1(1 − p1) n1 +  p3(1 − p3) n3     =    5.991    0.576(1 − 0.576) 250 +  0.45(1 − 0.45)       =    Find the critical value for the pairwise difference between populations i = 2 and j = 3,  CV23,  rounding the result to four decimal places. CVij  =    ??2    pi(1 − pi) ni +  pj(1 − pj) nj  CV23  =    ?20.05    p2(1 − p2) n2 +  p3(1 − p3) n3     =    5.991    0.48(1 − 0.48) 300 +  0.45(1 − 0.45)       =

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter7: Distance And Approximation
Section7.3: Least Squares Approximation
Problem 31EQ
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Recall the data for the three populations.
p1 = 0.576      n1 = 250
p2 = 0.48   n2 = 300
p3 = 0.45   n3 = 200
Find the critical value for the pairwise difference between populations i = 1 and j = 3, 
CV13,
 rounding the result to four decimal places.
CVij  = 
  ??2
 
 
pi(1 − pi)
ni
 + 
pj(1 − pj)
nj
 
CV13  = 
  ?20.05
 
 
p1(1 − p1)
n1
 + 
p3(1 − p3)
n3
 
   = 
  5.991
 
 
0.576(1 − 0.576)
250
 + 
0.45(1 − 0.45)
 
 
   =   
Find the critical value for the pairwise difference between populations i = 2 and j = 3, 
CV23,
 rounding the result to four decimal places.
CVij  = 
  ??2
 
 
pi(1 − pi)
ni
 + 
pj(1 − pj)
nj
 
CV23  = 
  ?20.05
 
 
p2(1 − p2)
n2
 + 
p3(1 − p3)
n3
 
   = 
  5.991
 
 
0.48(1 − 0.48)
300
 + 
0.45(1 − 0.45)
 
 
   =   
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