You can find an EXCEL file here that can be used to generate the logistic model for the United States population size (from the Census 1790-1940). I used Solver to obtain the logistic model. I showed how to use Solver to obtain an exponential model in a previous posting. In that posting, I used Solver to find values of two constants,a and b, for f(x)=ae^bx so that the sum (or average) of square differences is minimized. For the logistic model, you need to find constants a, b, and c for g(x)=c/(1+ae^-bx) so that the average (or sum) of the square differences between the raw data and g(x) is minimized. At some point in the future, I'll post a detailed tutorial on how to do this.

In the class project, we used an exponential model based on data from 1790-1860. If you want to see an EXCEL file that additionally considers an exponential model based on data from 1790-1940, click here.

I recently covered confidence intervals in my statistics class. While the formulas for constructing confidence intervals are fairly simple and should be learned by students, variaous forms of technology to automate the simple computations are available. One such form of technology is provided by the TI-83Plus graphing calculator. In this posting, I'll illustrate the process of constructing confidence intervals with the TI-83Plus graphing calculator, using examples from the course textbook, Essentials of Statistics 2nd Edition by Mario F. Triola, Pub. Pearson/Addison Wesley. In each example, I'll show how to obtain a confidence interval for the population mean, μ. Read more for the examples.

We then wish to test the null hypothesis




where k = 2, and h = 5.

We then need to compute the chi-square test statistic




where




and are the frequencies shown in the contingency table. We then reject the null hypothesis if





Now, to compute Q in MATLAB, we begin by storing the observed frequencies from the contingency table in a matrix

Y=[ 21 16 145 2 6 ; 14 4 175 13 4];

% Compute total number of trials
N=sum(sum(Y));
 
% Compute the totals for Attribute A (Gender)
nidot=sum(Y,2);
 
% Compute the totals for Attribute B (College)
ndotj=sum(Y);
 
% Compute the relative frequencies (probability estimates) 
% for Attribute B
pdotj = ndotj/N; 
 
% Compute the expected frequencies  (an outer product)
NP=nidot*pdotj;
 
% Compute the relative frequencies (probability estimates) 
% for Attribute A
pidot = nidot/N;
 
% Compute the chi-square statistic for the test of 
% independence of attributes
q=sum(sum(((Y-NP).^2)./NP));

Next we need to compare the resulting value of q to

% Compute the degrees of freedom for q:
[k h] = size(Y);
dof = (h-1)*(k-1);
 
% Find chi-square-subalpha (dof):
chiSquareSubAlpha = chiSquarePercentilesBisect(dof,alphaSig);

% Compute the p-value
pvalue=chiSquareProb(dof, q, inf);

alphaSig = 0.01;
whichtest=0;
[passORfail q chiSquareSubAlpha pvalue NP pidot pjdot nidot ndotj] ...
  = chiSquareIndependenceTest2Attr(Y, alphaSig, whichtest)

Input:
%  Y  -  k by h matrix (2-D array) with containing k events of attribute 
%        A and h events of attribute B. 
% alphaSig  - scalar significance level of test
% whichtest  -  scalar if 1, require p-value >= alphaSig for pass. 
%               Anything else will require chi-square test statistic
%               q <= chi-square_alpha[(k-1)(h-1)] 
%               for pass
%
% Output:
%  passORfail        - 1 if pass 0 if fail at alphaSig significance level
%  q                 - scalar chi-square test statistic
%  chisquareSubAlpha - scalar chi-square sub alpha
%  pvalue            - scalar p-value
%  NP                - array size of Y containing corresponding expected 
%                      frequencies
%  pidot             - vector containing relative frequencies (probability
%                      estimates) for Attribute A
%  pdotj             - vector containing relative frequencies (probability
%                      estimates) for Attribute B
%  nidot             - vector containing frequencies of Attribute A
%  ndotj             - vector containing frequencies of Attribute B

Fails independence test at 0.01 level of significance.
chi-square Test Statistic = 18.926482873851 > 13.276672 = chi^2_0.010(4).

passORfail =

     0


q =

   18.9265


chiSquareSubAlpha =

   13.2767


pvalue =

   8.1252e-04


NP =

   16.6250    9.5000  152.0000    7.1250    4.7500
   18.3750   10.5000  168.0000    7.8750    5.2500


pidot =

    0.4750
    0.5250


pjdot =

    0.0875    0.0500    0.8000    0.0375    0.0250


nidot =

   190
   210


ndotj =

    35    20   320    15    10

>> GoodnessOfFit4RandDigs(x,0.05)
Sequence fails Goodness of Fit test for randomness at the 0.05 level
 
             Frequency    Expected Number
-----------------------------------------
Same             0               5  
One away         8              10  
Other           42              35  
-----------------------------------------

>> x=floor(10*rand(51,1));
>> GoodnessOfFit4RandDigs(x,0.05)
Sequence passes Goodness of Fit test for randomness at the 0.05 level
 
             Frequency    Expected Number
-----------------------------------------
Same             5               5  
One away         9              10  
Other           36              35  
-----------------------------------------