TY  - CHAP
M1  - Book, Section
TI  - Solutions to Problems
A1  - Glantz, Stanton A.
A1  - Slinker, Bryan K.
A1  - Neilands, Torsten B.
Y1  - 2017
N1  - 
T2  - Primer of Applied Regression and Analysis of Variance, 3e
AB  - 2.1  Slope:  .444  g/cm;  standard  error  of  slope:  .0643  g/cm;  intercept:  −6.01  g;  standard  error  of  intercept:  2.39  g;  correlation  coefficient:  .925;  standard  error  of  the  estimate:  .964  g;  F:  47.7.2.2  Yes.  A  simple  regression  yields  r  =  .96  (t  =  7.8  with  5  degrees  of  freedom;  P  <  .001)  and  sy|x  =  3.9.2.3  Yes.  Regress  female  breast  cancer  rate  B  on  male  lung  cancer  rate  L  to  obtain  =  400  +  .282  L,  P  =  .02.2.4  The  regression  equation  is  Ĉ  =  1.36  log  CFU  −  .04  log  CFU/%  M,  with  R2  =  .649,  sy|x  =  .87  log  CFU,  and  F  =  96.1  with  1  and  52  degrees  of  freedom.  Both  the  slope  (t  =  −9.8;  P  <  .001;  95  percent  confidence  interval,  −.049  to  −.032)  and  intercept  (t  =  5.21;  P  <  .001;  95  percent  confidence  interval,  .84  to  1.88)  are  highly  significant.  This  linear  model  provides  a  significant  explanation  of  the  observed  variability  of  the  dependent  variable.2.5A.  Based  on  the  mean  data,  the  regression  equation  is  =  1.67  mmHg/(mL  ·  min  ·  kg)  −  .276  (mmHg  ·  h)/(mL  ·  min  ·  ng)Δr,  with  r  =  .957,  sy|x  =.97  mmHg/(mL  ·  min  ·  kg),  and  a  t  statistic  for  the  slope  of  −4.63  with  2  degrees  of  freedom  (P  =  .044).  Thus,  the  change  in  vascular  resistance  seems  to  be  significantly  related  to  the  change  in  renin  production  by  the  kidney.  B.  Using  the  raw  data  points,  we  find  the  similar  regression  equation  =  .183  mmHg/(mL  ·  min  ·  kg)  −.162  (mmHg  ·  h)/(mL  ·  min  ·  ng)Δr,  with  r  =  .632,  sy|x  =  2.15  mmHg/(mL  ·  min  ·  kg),  and  the  t  statistic  for  the  slope  is  −4.32  with  28  degrees  of  freedom  (P  <  .001).  Thus,  we  still  conclude,  based  on  the  t  statistic,  that  the  change  in  vascular  resistance  is  significantly  related  to  the  change  in  renin  production  by  the  kidney.  C.  Whereas  the  regression  equations  are  similar,  the  equation  computed  using  the  group  mean  values  had  a  much  higher  correlation  than  the  one  based  on  the  raw  data  (.957  versus  .632).  Likewise,  the  estimate  of  the  variability  about  the  regression  plane  sy|x  is  much  smaller  when  computed  using  the  mean  values  [.97  vs.  2.15  mmHg/(mL  ·  min  ·  kg)].  The  regression  equation  computed  from  the  group  means  seems  to  provide  a  much  better  prediction  of  the  data  than  the  one  computed  from  the  raw  data.  The  difference  in  the  results  of  the  two  regression  analyses  lies  in  the  difference  between  the  mean  values  and  the  raw  data.  By  computing  the  means  of  each  group,  then  analyzing  these  four  data  points  with  regression,  we  have  effectively  thrown  away  most  of  the  variability  in  the  observations.  Thus,  sy|x  is  artificially  reduced  and  is  no  longer  an  unbiased  estimate  of  the  population  variability  around  the  plane  of  means.  As  a  result,  r  is  inflated.  To  understand  why  r  and  sy|x  change  but  not  the  regression  equation,  you  need  to  consider  the  assumptions  behind  the  linear  regression  model.  The  first  ...
SN  - 
PB  - McGraw-Hill Education
CY  - New York, NY
Y2  - 2024/04/24
UR  - accessbiomedicalscience.mhmedical.com/content.aspx?aid=1141897489
ER  -