clear
echo on
%Solution to Quiz #2
%We have the heights and weights (using cut and paste!)
data=[70 205 
71 235 
73 192 
74 220 
75 295 
74 248 
73 240 
74 210 
75 301 
77 290 
73 195 
72 195
74 215 
76 235 
76 280 
77 297 
78 242 
75 235 
68 160 
75 260 
74 247 
75 227 
75 230 
70 188 
74 211 
71 190 
74 195 
76 235 
75 283 
75 207 
71 190 
70 230 
73 185 
70 186 
77 250 
74 218 
72 198 
74 275 
71 245 
72 172 
76 285 
71 187 
76 225 
75 200 
76 283 
69 180 
77 290 
79 289 
74 225 
76 290 
76 215 
74 205 
74 238 
76 215 
73 239 
74 210 
74 191 
74 245 
72 300 
80 303 
75 350
77 297 
76 290 
74 227 
73 246 
71 185 
72 235 
72 232 
76 245 
69 175 
74 210 
71 198 
72 218 
77 295 
69 165 
77 292 
78 265 
73 176 
70 180 
76 245 
70 177 
74 243 
74 232 
72 195 
75 234 
76 208 
74 214 
73 215 
74 225 
77 351 
72 290 
78 236 
72 190 
74 245 
69 175 
71 196 
75 251 
75 210 
75 198 
72 190 
75 310 
76 240 
76 298 
73 285 
74 305 
72 210 
70 178];

ht=data(:,1);
wt=data(:,2);

%Question 1: the BMI...
bmi=703*wt./ht.^2;
n=length(ht);
avebmi=sum(bmi)/n
stdevbmi=norm(bmi-avebmi)/(n-1)^.5 %Which is another way to get it...

%You will note that (by clinical definition) the average bmi of a football player is
%borderline obese (BMI of 30) although they are obviously in good shape.  The extra
%weight is still tough on the heart, though - even if it is muscle rather than fat.
%Recent studies suggest that while the link between BMI and cardiovascular diseases
%isn't perfect (no surprise) there are still some correlations even for athletes.

%Question 2: The mean heights and weights...
aveht=sum(ht)/n
avewt=sum(wt)/n

%Question 3: The variances and covariance.  We'll do this as a matrix of covariance.  
%Let (1) be ht and (2) be wt...
covariance=zeros(2,2); %We intialize the matrix
for i=1:n
 covariance=covariance+[(ht(i)-aveht)^2,(ht(i)-aveht)*(wt(i)-avewt);(ht(i)-aveht)*(wt(i)-avewt),(wt(i)-avewt)^2];
 echo off
end
echo on
covariance=covariance/(n-1) %We finish it off...

%Note that the covariance (off diagonal elements) is positive but not all that large.
%For perfect covariance it would be the product of the square roots of the diagonal
%elements.  If they were negatively covarying, the off-diagonal element would have
%been negative.

%We can also do this one using the built in matlab command "cov":

cov2=cov(ht,wt)

%Which yields the same matrix...

%Question 4: Using error propagation to get the bmi standard deviation...
%The calculated bmi is easy:
calcbmi=703*avewt/aveht^2

%The calculated standard deviation is a little trickier.  We need the gradient in
%the bmi with respect to wt and ht (we do it analytically):

gradbmi=[-2*calcbmi/aveht,calcbmi/avewt]

%and thus the standard deviation:
calcstdevbmi=(gradbmi*covariance*gradbmi')^.5

%We compare to the directly computed values:

ratiobmi=calcbmi/avebmi

ratiostdev=calcstdevbmi/stdevbmi

%Which is pretty close...  They aren't exact because we had a non-linear function.

%Question 5:  We repeat the solution to question 4, but now we take the
%covariance to be zero.  This is easily done by using the "diag" command
%twice.  The first time it turns the diagonal of the matrix into a vector,
%and the second time it turns it back into a diagonal matrix:

newstdev=(gradbmi*diag(diag(covariance))*gradbmi')^.5
newratio=newstdev/stdevbmi

%As you can see, you get a much larger standard deviation.  That is because
%you are ignoring the large covariance in heights and weights.


echo off