Homework 7

Solutions:

 

  Exercise 20  

 
To do this in Windows you should have followed these steps:
 Bring the data to the work window.
 Click on Statistics-Analysis of Variance-Fixed effects-Create
  formula-logarea~species.
 You should get something like

 
 species 
     
 Residuals 
  
 SS 
 
15.78
56.37

 
 df 
 1 
 292 


R.S.E.
0.4393

 And your ANOVA table should be like this

 
 d.f. 
     
 S.S. 
 M.S. 
 F-value 
  
 Pr(F) 
 
spgroup
7
1.67
0.23
1.24
0.27

 
 Residuals 
 285 
 54.70 
 0.19 

 To get the precise value of your F, you could do 1-pf(1.24,7,285)
  in S-Plus.
 Notice that the answer to question A is Sp=0.4318 =) Sp^2=0.1919,
  and this was the value of the Estimated pooled variance.
 This was direct with the information you had from the anova
  analysis in the computer, but you could have done it by hand:

 (((127-1)*(0.4979^2))+((44-1)*(0.4235^2))+((24-1)*(0.3955^2))+
((41-1)*(0.3183^2))+((18-1)*(0.3111^2))+((16-1)*(0.4649^2))+
((11-1)*(0.2963^2))+((7-1)*(0.3242^2))+((6-1)*(0.5842^2)))/((127-1)+
(44-1)+(24-1)+(41-1)+(18-1)+(16-1)+(11-1)+(7-1)+(6-1))

 

  Exercise 21  

 
 This problem just requires to apply the formula on Pg 116.
 95%CI=Estimate +/- t(343)*SE = Estimate +/- 1.96*SE
 p-value=2*(Pr(Estimate/SE)) = 2*(1-pt(Estimate/SE,df))
 Your table should, therefore, be something like

 
 Estimate 
     
 S.E. 
 Lower Bound 
 Upper Bound 
  
t-stat
 p-value 
 

NR50-NN85
9.6
1.2
7.248
11.952
8
about
0

 
RR50-NR50
0.6
1.2
-1.752
2.952
0.5
0.617
NR40-NR50
2.8
1.2
0.448
5.152
2.333
0.02
NR50
lopro-NR50
-2.6
1.2
-4.952
-0.248
2.1666
0.0308
NN85-NP
5.3
1.3
2.752
7.848
4.076
about
0



 

  Exercise 23  

 
 Bring the data from the file to the work window
 Change to factor your data. You can do this Choosing Data-Change
  Data Type-Column type=factor for "bonegrp."
 Statistics-Analysis of Variance-Fixed effects-Create
  Formula=Oxygen~bonegrp.
 You should obtain

 
 species 
     
 Residuals 
  
 SS 
 
6.06
2.97

 
 df 
11
240


R.S.E.
0.2725


 
 d.f. 
     
 S.S. 
 M.S. 
 F-value 
  
 Pr(F) 
 
spgroup
11
6.06
0.55
7.42
9.72e-007

 
 Residuals 
40
2.97
0.07

 You can do a nice plot by following these steps:
  A) Click in Graph-Multipanel-Boxplot
  B) Choose Column x=factor variable (bonegrp) and Column y=oxygen.


 

  Exercise 17  

 
 g=(-7.347/6)+(-7.368/6)+(-7.418/6)+(-7.487/6)+(-7.563/6)+(-7.568/6)+
  (8.214/3)+(8.272/3)+(8.297/3)=0.8025
 Sp=0.4381 (from problem 20)
 SD(g)=Sp*sqrt((1/(127*36))+(1/(44*36))+(1/(24*36))+(1/(41*36))
  +(1/(18*36))+(1/(16*36))+(1/(11*36))+(1/(11*9))+(1/(7*9))+(1/(6*9)))
 SD(g)=0.4381*0.2301783=0.10
 t(.975,285)=1.968
 CI(95%)=[0.8025 +/- 1.968*0.10]=[0.6057,1]


 

  Exercise 20  


You can solve this exercise in different ways. Here is one.
 Bring the code from your file to the work window
 Change to factor your variable code
 Recode your variable if you decide to group regardless of the
  causes of death. You can do this with the command Data-Recode
 Statistics-Analysis of Variance-Fixed
  Effects-Formula=Volume~Code.
 Your results (if you did it this way) should be like:

 
 Code 
     
 Residuals 
  
 SS 
 
44960.5
114187.1

 
 df 
2
38


R.S.E.
54.817


 
 d.f. 
     
 S.S. 
 M.S. 
 F-value 
  
 Pr(F) 
 
spgroup
2
44960.5
22480.25
7.481
0.0018

 
 Residuals 
38
114187.1
3004.92

 This 0.0018 indicates a low probability that this results happen
by chance. There are clear differences (especially when you use
t-tests) between homosexual and heterosexual male and between
heterosexual male and female, and there is no apparent difference
between homosexual male and heterosexual female.
 This is a non-random sample of the population, which hinders us
from a general extrapolation to the population. Cause-effect
relationships can't be drawn either.
 You can get a nice plot by following the steps in exercise 23.