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
Download the SAS program here: nutrient7.sas.
View the video on using SAS for Hotellings \(T^2\).
Using Minitab
At this time Minitab does not support this procedure.
Analysis
MU0 | YBAR |
---|---|
1 | 0.6240493 |
1 | 0.7419933 |
1 | 1.096724 |
1 | 1.0495442 |
1 | 1.0523793 |
S | ||||
---|---|---|---|---|
0.1578294 | 0.0626726 | 0.1012636 | 0.1280139 | 0.0893549 |
0.062626 | 0.1591579 | 0.1267311 | 0.1985961 | 0.1223751 |
0.1012636 | 0.1267311 | 0.259688 | 0.1527094 | 0.1060444 |
0.1280139 | 0.1985961 | 0.1527094 | 4.1694568 | 0.3677208 |
0.0893549 | 0.1223751 | 0.1060444 | 0.3677208 | 0.9628914 |
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:
download the SAS Program here: nutrient8.sas
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.
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.