7.2.1 - Profile Analysis for One Sample Hotelling's T-Square

Let us revisit the original hypothesis of interest, as below

\(H_0\colon \boldsymbol{\mu} = \boldsymbol{\mu_0}\) against \(H_a\colon \boldsymbol{\mu} \ne \boldsymbol{\mu_a}\)

Note! It is equivalent to testing this null hypothesis:

\(H_0\colon \dfrac{\mu_1}{\mu^0_1} = \dfrac{\mu_2}{\mu^0_2} = \dots = \dfrac{\mu_p}{\mu^0_p} = 1\)

against the alternative that at least one of these ratios is not equal to 1, (below):

\(H_a\colon \dfrac{\mu_j}{\mu^0_j} \ne 1\) for at least one \(j \in \{1,2,\dots, p\}\)

Instead of testing the null hypothesis for the ratios of the means over their hypothesized means are all equal to one, profile analysis involves testing the null hypothesis that all of these ratios are equal to one another, but not necessarily equal to 1.

After rejecting

\(H_0\colon \dfrac{\mu_1}{\mu^0_1} = \dfrac{\mu_2}{\mu^0_2} = \dots = \dfrac{\mu_p}{\mu^0_p} = 1\)

we may wish to test

\(H_0\colon \dfrac{\mu_1}{\mu^0_1} = \dfrac{\mu_2}{\mu^0_2} = \dots = \dfrac{\mu_p}{\mu^0_p}\)

Profile Analysis can be carried out using the following procedure.

Step 1: Compute the differences between the successive ratios. That is we take the ratio of the j + 1-th variable over its hypothesized mean and subtract this from the ratio of jth variable over its hypothesized mean as shown below:

\(D_{ij} = \dfrac{X_{ij+1}}{\mu^0_{j+1}}-\dfrac{X_{ij}}{\mu^0_j}\)

We call this ratio \(D_{ij}\) for observation i.

Note! That, testing the null hypothesis that all of the ratios are equal to one another

\(H_0\colon \dfrac{\mu_1}{\mu^0_1} = \dfrac{\mu_2}{\mu^0_2} = \dots = \dfrac{\mu_p}{\mu^0_p}\)

is equivalent to testing the null hypothesis that all the mean differences are going to be equal to 0.

\(H_0\colon \boldsymbol{\mu}_D = \mathbf{0}\)

Step 2: Apply Hotelling’s \(T^{2}\) test to the data \(D_{ij}\) to test the null hypothesis that the mean of these differences is equal to 0.

\(H_0\colon \boldsymbol{\mu}_D = \mathbf{0}\)

Example
Example

This is carried out using the SAS program as shown below:

download the SAS Program here: nutrient8.sas

Explore the code for an explanation of the SAS code.

Note: In the upper right-hand corner of the code block you will have the option of copying () the code to your clipboard or downloading () the file to your computer.

options ls=78;
title "Profile Analysis - Women's Nutrition Data";

 /* After reading in the data, each of the original
  * variables is divided by its null value to convert 
  * to a common scale without specific units.
  * The differences are then defined for each successive
  * pair of variables.
  */

data nutrient;
  infile "D:\Statistics\STAT 505\data\nutrient.csv" firstobs=2 delimiter=','
  input id calcium iron protein a c;
  calcium=calcium/1000;
  iron=iron/15;
  protein=protein/60;
  a=a/800;
  c=c/75;
  diff1=iron-calcium;
  diff2=protein-iron;
  diff3=a-protein;
  diff4=c-a;
  run;

 /* The iml code to compute the hotelling t2 statistic
  * is similar to that for the one-sample t2 statistic
  * except that by working with differences of variable,
  * the null values are all 0s, and the corresponding
  * null hypothesis is that all variable means are equal
  * to each other.
  */

proc iml;
  start hotel;
    mu0={0,0,0,0};
    one=j(nrow(x),1,1);
    ident=i(nrow(x));
    ybar=x`*one/nrow(x);
    s=x`*(ident-one*one`/nrow(x))*x/(nrow(x)-1.0);
    print mu0 ybar;
    print s;
    t2=nrow(x)*(ybar-mu0)`*inv(s)*(ybar-mu0);
    f=(nrow(x)-ncol(x))*t2/ncol(x)/(nrow(x)-1);
    df1=ncol(x);
    df2=nrow(x)-ncol(x);
    p=1-probf(f,df1,df2);
    print t2 f df1 df2 p;
  finish;
  use nutrient;
  read all var{diff1 diff2 diff3 diff4} into x;
  run hotel;

Profile analysis

To test the hypothesis of equal mean ratios:

Analysis

The results yield a Hotelling's \(T^{2}\) of 1,030.7953 and an F value of 256.64843, 4, and 733 degrees of freedom, and a p-value very close to 0.

Profile Analysis - Women's Nutrition Data
MU0	YBAR
0	0.1179441
0	0.3547307
0	-0.04718
0	0.0028351

S
0.191621	-0.071018	0.0451147	-0.037562
-0.071018	0.1653837	-0.178844	0.029556
0.0451147	-0.178844	4.123726	-3.755071
-0.037562	0.029556	-3.75507	-3.755071

T2	F	DF1	DF2	P
1030.7953	256.64843	4	733	0

Example 7-14: Women’s Health Survey Section

Here we can reject the null hypothesis that the ratio of mean intake over the recommended intake is the same for each nutrient as the evidence has shown here:

\( T ^ { 2 } = 1030.80 ; F = 256.65 ; \mathrm { d.f. } = 4,733 ; p < 0.0001 \)

This null hypothesis could be true if, for example, all the women were taking in nutrients in their required ratios, but they were either eating too little or eating too much.