>> % Loading the data into matlab
>> load temp.dat
>> % determining the number of data points
>> [n,m] = size(temp)

n =

   101


m =

     1

>> % Based on the summary statistics and shape of histogram
>> % one possible distribution that might fit the data is the
>> % Normal density - determining the MLE's for the Normal
>> % density parameters.
>> mu = mean(temp)

mu =

    5.2876

>> v = var(temp)

v =

   22.2008

>> foo = v * (n-1)/n

foo =

   21.9810

>> sigma = sqrt(foo)

sigma =

    4.6884

>> % Hence from the data the Normal (5.28, 4.688) is the most likely fit
>> % Comparing summary stats 
>> % for data in temp.dat
>> %mean 
>> mu 

mu =

    5.2876

>> % variance
>> v

v =

   22.2008

>> irange = iqr(temp)

irange =

    6.4281

>> % From the Normal have 
>> mu

mu =

    5.2876

>> vn = sigma^2

vn =

   21.9810


>> nirange = norminv(.75, mu,sigma) - norminv(.25,mu,sigma)

nirange =

    6.3245

>> % Coefficient of variation
>> cov = std(temp)/mu

cov =

    0.8911

>> %Normal cov
>> ncov = sigma/mu

ncov =

    0.8867

>> % Constructing a P-P plot
>> sorttemp = sort(temp);
>> ptheory = normcdf(sorttemp,mu,sigma);
>> for i = 1:n
   ptemp(i) = (i-.5)/n;
   end
>> plot (ptemp,ptheory,ptemp,ptemp)
>> title('P-P plot of Normal versus sample data')
>> xlabel('Sample Data probabilities')
>> ylabel('Normal Distribution probabilities')
>> print p-p.ps
>> % Constructing a Q-Q plot
>> xtheory = norminv(ptemp,mu,sigma);
>> plot (sorttemp,xtheory,sorttemp,sorttemp)
>> title('Q-Q plot of Normal quantiles versus data quantiles')
>> xlabel('Sample Data quantiles')
>> ylabel('Normal Distribution quantiles')
>> print q-q.ps
>> % Notice that botht he Q-Q plot and P-P plot indicate that 
>> % the Normal is a pretty good fit to the data 
>> % An overlay plot of the Normal CDF and data CDF is given by
>> plot(sorttemp,ptemp,sorttemp,ptheory)
>> title('Plot of Normal Distribution and Data Distribution')
>> xlabel('x')
>> ylabel('F(x)')
>> print distplot.ps
>> % plot of f(x) and histogram of data
>> k = 7;
>> y = normpdf(sorttemp,mu,sigma);
>> [nc,x] = hist(temp,k);
>> nc = nc/n;
>> bar(x,nc)
>> hold on 
>> plot(sorttemp,y)
>> print  overlay.ps
>> % Plot histogram of Normal and data
>> nnc(1) = normcdf(x(1),mu,sigma);
>> for j = 2:k
nnc(j) = normcdf(x(j),mu,sigma) - normcdf(x(j-1),mu,sigma);
end
>> hold off
>> bar(x,nc)
>> hold on
>> bar(x,nnc,'+')
>> bar(x,nnc,'-')
>> title('Histograms of Normal and Sample Data')
>> xlabel('x')
>> ylabel('Frequencies')
>> print histc2.ps
>> % Performing ChiSquare test
>> % Determining the number of observations in each cell of histogram
>> nc = nc*n;
>> nnc = nnc*n;
>> nc

nc =

    3.0000    6.0000   16.0000   26.0000   25.0000   18.0000    7.0000

>> nnc

nnc =

    0.9073    3.7819   11.4640   22.0936   27.0855   21.1263   10.4817

>> % Note that this is a poor choice of cells for the Chi-Square test
>> % since there are cells with less than 5 observations expected 
>> % I will use this cell width anyway to illustrate the computations
>> chisum = 0;
>> for i = 1:k
   chisum = chisum + ((nc(i)-nnc(i))^2)/nnc(i);
end
>> chisum

chisum =

   10.3931

>> % comparing with chi-squared test value 
>> % there are 7 - 2- 1 = 4 degrees of freedom 
>> test = chi2inv(.95,4)

test =

    9.4877

>> % Notice that the chi-squard value is less then the error sum 
>> % value thus one WOULD REJECT the hypothesis that the data is
>> % Normally distributed - however the values are close and 
>> % if one goes back and  recomputes with differenct cell widths
>> % it may pass - also would be a good idea to  collects a LARGER
>> % Sample of the data and then retest
>> % may pass the Chi-Squared test - (for this example it will!)
>> exit

 31049 flops.