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

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 $j \in \{1,2,\dots, p\}$

Using SAS

To see these, consider the SAS program below

View the video on using SAS for Hotellings $T^2$.

Using Minitab

At this time Minitab does not support this procedure.

Note! The results are identical to the previously obtained values for Hotelling's $T^{2}$ statistic, F degrees of freedom and the p-value.

Analysis

Hotelling T2 - Women's Nutrition Data

MU0 YBAR
1 0.6240493
1 0.7419933
1 1.096724
1 1.0495442
1 1.0523793
 0.157829 0.0626726 0.101264 0.128014 0.0893549 0.062626 0.159158 0.126731 0.198596 0.122375 0.101264 0.126731 0.259688 0.152709 0.106044 0.128014 0.198596 0.152709 4.16946 0.367721 0.0893549 0.122375 0.106044 0.367721 0.962891
T2 F DF1 DF2 P
1758.5413 349.7968 5 732 0

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 null hypothesis that the mean of these differences is equal to 0.

$H_0\colon \boldsymbol{\mu}_D = \mathbf{0}$

Using SAS

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

View the video for an explanation of the SAS code.

Using Minitab

At this time Minitab does not support this procedure.

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
 0.191621 -0.071018 0.0451147 -0.037562 -0.071018 0.165384 -0.178844 0.029556 0.0451147 -0.178844 4.12373 -3.75507 -0.037562 0.029556 4.12373 -3.75507
T2 F DF1 DF2 P
1030.7953 256.64843 4 733 0

Example 7-14: Women’s Health Survey

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.

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