7.1  Basic
7.1  BasicIn this section we will introduce the standard statistical methods for making inferences on multivariate population mean.
7.1.1  An Application of OneSample Hotelling’s TSquare
7.1.1  An Application of OneSample Hotelling’s TSquareWomen’s Health Survey: OneSample Hotelling's TSquare
In 1985, the USDA commissioned a study of women’s nutrition. Nutrient intake was measured for a random sample of 737 women aged 2550 years. Five nutritional components were measured: calcium, iron, protein, vitamin A and vitamin C. In previous analyses of these data, the sample mean vector was calculated. The table below shows the recommended daily intake and the sample means for all the variables:
Variable  Recommended Intake \((\mu_{o})\)  Mean 
Calcium  1000 mg  624.0 mg 
Iron  15mg  11.1 mg 
Protein  60g  65.8 g 
Vitamin A  800 μg  839.6 μg 
Vitamin C  75 mg  78.9 mg 
One of the questions of interest is whether women meet the federal nutritional intake guidelines. If they fail to meet the guidelines, then we might ask for which nutrients the women fail to meet the guidelines.
The hypothesis of interest is that women meet nutritional standards for all nutritional components. This null hypothesis would be rejected if women fail to meet nutritional standards on any one or more of these nutritional variables. In mathematical notation, the null hypothesis is the population mean vector \(μ\) equals the hypothesized mean vector \(\mu_{o}\) as shown below:
\(H_{o}\colon \mu = \mu_{o}\)
Let us first compare the univariate case with the analogous multivariate case in the following tables.
Focus of Analysis
Measuring only a single nutritional component (e.g. Calcium).
Data: scalar quantities \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\)
Measuring multiple (say p) nutritional components (e.g. Calcium, Iron, etc).
Data: p × 1 random vectors
\(\mathbf{X} _ { 1 } , \mathbf{X} _ { 2 } , \ldots , \mathbf{X} _ { n }\)
Assumptions Made In Each Case
Distribution
The data all have a common mean \(\mu\) mathematically, \(E \left( X _ { i } \right) = \mu ; i = 1,2 , \dots , n\) This implies that there is a single population of subjects and no subpopulations with different means.
The data have a common mean vector \(\boldsymbol{\mu}\) ; i.e., \(E \left( \boldsymbol { X } _ { i } \right) =\boldsymbol{\mu} ; i = 1,2 , . , n \) This also implies that there are no subpopulations with different mean vectors.
Homoskedasticity
The data have common variance \(\sigma^{2}\) ; mathematically, \(\operatorname { var } \left( X _ { i } \right) = \sigma ^ { 2 } ; i = 1,2 , . , n\)
The data for all subjects have common variancecovariance matrix \(Σ\) ; i.e., \(\operatorname { var } \left( \boldsymbol{X} _ { i } \right) = \Sigma ; i = 1,2 , \dots , n\)
Independence
The subjects are independently sampled.
The subjects are independently sampled.
Normality
The subjects are sampled from a normal distribution
The subjects are sampled from a multivariate normal distribution.
Hypothesis Testing in Each Case
Consider hypothesis testing:
\(H _ { 0 } \colon \mu = \mu _ { 0 }\)
against alternative
\(H _ { \mathrm { a } } \colon \mu \neq \mu _ { 0 }\)
Consider hypothesis testing:
\(H _ { 0 } \colon \boldsymbol{\mu} = \boldsymbol{\mu _ { 0 }}\) against \(H _ { \mathrm { a } } \colon \boldsymbol{\mu} \neq \boldsymbol{\mu _ { 0 }}\)
Here our null hypothesis is that mean vector \(\boldsymbol{\mu}\) is equal to some specified vector \(\boldsymbol{\mu_{0}}\). The alternative is that these two vectors are not equal.
We can also write this expression as shown below:
\(H_0\colon \left(\begin{array}{c}\mu_1\\\mu_2\\\vdots \\ \mu_p\end{array}\right) = \left(\begin{array}{c}\mu^0_1\\\mu^0_2\\\vdots \\ \mu^0_p\end{array}\right)\)
The alternative, again is that these two vectors are not equal.
\(H_a\colon \left(\begin{array}{c}\mu_1\\\mu_2\\\vdots \\ \mu_p\end{array}\right) \ne \left(\begin{array}{c}\mu^0_1\\\mu^0_2\\\vdots \\ \mu^0_p\end{array}\right)\)
Another way of writing this null hypothesis is shown below:
\(H_0\colon \mu_1 = \mu^0_1\) and \(\mu_2 = \mu^0_2\) and \(\dots\) and \(\mu_p = \mu^0_p\)
The alternative is that μ_{j} is not equal to \(\mu^0_j\) for at least one j.
\(H_a\colon \mu_j \ne \mu^0_j \) for at least one \(j \in \{1,2, \dots, p\}\)
Univariate Statistics: \(t\)test
In your introductory statistics course you learned to test this null hypothesis with a tstatistic as shown in the expression below:
\(t = \dfrac{\bar{x}\mu_0}{\sqrt{s^2/n}} \sim t_{n1}\)
Under \(H _ { 0 } \) this tstatistic is has a t distribution with n1 degrees of freedom. We reject \(H _ { 0 } \) at level \(α\) if the absolute value of the test statistic t is greater than the critical value from the ttable, evaluated at \(α/2\) as shown below:
\(t > t_{n1, \alpha/2}\)
7.1.2  A Naive Approach
7.1.2  A Naive ApproachFollowing the univariate method, a naive approach for testing a multivariate hypothesis is to compute the ttest statistics for each individual variable; i.e.,
\(t_j = \dfrac{\bar{x}_j\mu^0_j}{\sqrt{s^2_j/n}}\)
Thus we could define \(t_{j}\), which would be the tstatistic for the \(j^{th}\) variable as shown above. We may then reject \(H_0\colon \boldsymbol{\mu} = \boldsymbol{\mu_0}\) if \(t_j > t_{n1, \alpha/2}\) for at least one variable \(j \in \{1,2,\dots, p\}\) .
Problem with Naive Approach
The basic problem with this naive approach is that it does not control the familywise error rate. By definition, the familywise error rate is the probability of rejecting at least one of the null hypotheses \(H^{(j)}_0\colon \mu_j = \mu^0_j\) when all of the \(H_{0}\)’s are true.
To understand the familywise error rate suppose that the experimental variancecovariance matrix is diagonal. This implies zero covariances between the variables. If the data are multivariate normally distributed then this would mean that all of the variables are independently distributed. In this case, the family wide error rate is
\(1(1\alpha)^p > \alpha\)
where p is the dimension of the multivariate date and \(α\) is the level of significance. By definition, the familywide error rate is equal to the probability that we reject \(H _ { 0 } ^ { ( j ) }\) for at least one j, given that \(H _ { 0 } ^ { ( j ) }\) is true for all j. Unless p is equal to one, the error will be strictly greater than \(α\).
Consequence
The naive approach yields a liberal test. That is, we will tend to reject the null hypothesis more often than we should.
Bonferroni Correction
Under the Bonferroni Correction we would reject the null hypothesis that mean vector \(\boldsymbol{\mu}\) is equal to our hypothesized mean vector \(\boldsymbol{\mu_{0}}\) at level \(\alpha\) if, for at least one variable j, the absolute value of \(t_{j}\) is greater than the critical value from the ttable with n1 degrees of freedom evaluated at \(\alpha /(2p)\) for at least one variable j between 1 and p.
Note! For independent data, this yields a familywide error rate of approximately \(\alpha\). For example, the familywide error rate is shown in the table below for different values of p. You can see that these are all close to the desired level of \(\alpha = 0.05\).
p  FamilyWide Error rate 

2  0.049375 
3  0.049171 
4  0.049070 
5  0.049010 
10  0.048890 
20  0.048830 
50  0.048794 
The problem with the Bonferroni Correction, however, is that it can be quite conservative when there is a correlation among the variables, particularly when doing a large number of tests. In a multivariate setting, the assumption of independence between variables is likely to be violated.
Consequence
When using the Bonferroni adjustment for many tests with correlated variables, the true familywide error rate can be much less than \(\alpha\). These tests are conservative and there is low power to detect the desired effects.
7.1.3  Hotelling’s TSquare
7.1.3  Hotelling’s TSquareA more preferable test statistic is Hotelling’s \(T^2\) and we will focus on this test.
To motivate Hotelling's \(T^2\), consider the square of the tstatistic for testing a hypothesis regarding a univariate mean. Recall that under the null hypothesis t has a distribution with n1 degrees of freedom. Now consider squaring this test statistic as shown below:
\[t^2 = \frac{(\bar{x}\mu_0)^2}{s^2/n} = n(\bar{x}\mu_0)\left(\frac{1}{s^2}\right)(\bar{x}\mu_0) \sim F_{1, n1}\]
When you square a tdistributed random variable with n1 degrees of freedom, the result is an Fdistributed random variable with 1 and n1 degrees of freedom. We reject \(H_{0}\) at level \(α\) if \(t^2\) is greater than the critical value from the Ftable with 1 and n1 degrees of freedom, evaluated at level \(α\).
\(t^2 > F_{1, n1,\alpha}\)
 Hotelling's TSquare

Consider the last term in the above expression for \(t^2\). In the expression for Hotelling's \(T^2\), the difference between the sample mean and \(\mu_{0}\) is replaced with the difference between the sample mean vector and the hypothesized mean vector \(\boldsymbol{\mu _{0}}\). The inverse of the sample variance is replaced by the inverse of the sample variancecovariance matrix S, yielding the expression below:
\(T^2 = n\mathbf{(\overline{X}\mu_0)'S^{1}(\overline{X}\mu_0)}\)
Notes on \(\mathbf{T^2}\)
For large n, \(T^2\) is approximately chisquare distributed with p degrees of freedom.
If we replace the sample variancecovariance matrix, S, by the population variancecovariance matrix, \(Σ\)
\(n\mathbf{(\overline{X}\mu_0)'\Sigma^{1}(\overline{X}\mu_0)},\)
then the resulting test is exactly chisquare distributed with p degrees of freedom when the data are normally distributed.
For small samples, the chisquare approximation for \(T^2\) does not take into account variation due to estimating \(Σ\) with the sample variancecovariance matrix S.
Better results can be obtained from the transformation of the Hotelling \(T^2\) statistic as below:
\[F = \frac{np}{p(n1)}T^2 \sim F_{p,np}\]
Under null hypothesis, \(H_{0}\colon \boldsymbol{\mu} = \boldsymbol{\mu_{0}}\), this will have a F distribution with p and np degrees of freedom. We reject the null hypothesis, \(H_{0}\),_{ }at level \(α\) if the test statistic F is greater than the critical value from the Ftable with p and np degrees of freedom, evaluated at level \(α\).
\(F > F_{p, np, \alpha}\)
To illustrate the Hotelling's \(T^2\) test we will return to the USDA Women’s Health Survey data.
7.1.4  Example: Women’s Survey Data and Associated Confidence Intervals
7.1.4  Example: Women’s Survey Data and Associated Confidence IntervalsExample 71: Woman's Survey Data (Hotelling's \(T^{2}\) Test)
The data are stored in the file that can be downloaded: nutrient.txt
Using SAS
The Hotelling's \(T^{2}\) test is calculated using the SAS program as shown below.
It turns out that SAS does not have any procedures for calculating Hotelling's \(T^{2}\). So, in this case, we are going to have to rely on the simple procedure, which can carry out matrix manipulation. And in this case, it can be used for carrying out Hotelling's \(T^{2}\) test.
Download the SAS Program: nutrient4.sas
To use this program, I recommend copying the program here and then making the necessary changes to use it with your dataset on your homework assignment. Again, only three entries are all we really need to change to use this with any dataset. First the specified value of μ_{o}. Second, which dataset you want to use, and third, the specification of what variable you want to analyze.
Download the results printed in the output: nutrient4.lst.
View the video below to see how to find the Hotelling's \(T^2\) value using the SAS statistical software application.Using Minitab
View the video below to see how to find the Hotelling's \(T^2\) value using the Minitab statistical software application.
Analysis
The recommended intake and sample mean are given below
Variable  Recommended Intake (\(μ_{o}\))  Mean 
Calcium  1000 mg  624.0 mg 
Iron  15mg  11.1 mg 
Protein  60g  65.8 g 
Vitamin A  800 μg  839.6 μg 
Vitamin C  75 mg  78.9 mg 
as well as the sample variancecovariance matrix:
\(S = \left(\begin{array}{rrrrr}157829.4 & 940.1 & 6075.8 & 102411.1 & 6701.6 \\ 940.1 & 35.8 & 114.1 & 2383.2 & 137.7 \\ 6075.8 & 114.1 & 934.9 & 7330.1 & 477.2 \\ 102411.1 & 2383.2 & 7330.1 & 2668452.4 & 22063.3 \\ 6701.6 & 137.7 & 477.2 & 22063.3 & 5416.3 \end{array}\right)\)
Hotelling’s Tsquare comes out to be:
\(T^2 = 1758.54\)
The Fstatistic is:
\(F = 349.80 > 3.042 = F_{5,732,0.01}\)
For an 0.01 level test, the critical value is approximately 3.02. Because 349.80 is greater than this value, we can reject the null hypothesis that the average dietary intake meets recommendations.
\((T^2 = 1758.54; F = 349.80; d.f. = 5,732; p < 0.0001)\)
The SAS program reports the pvalue as 0.00. In this case, the pvalues can never equal zero. It is preferable to state that the pvalue is less than 0.0001.
Conclusion
For all women between 25 and 50 years old, the average dietary intake does not meet recommended standards. Returning to the table of sample means and recommended dietary intake, it appears that women fail to meet nutritional standards for calcium and iron, and perhaps exceed intakes for protein, vitamin A and vitamin C.
Such a statement, however, is not entirely backed up by the evidence.
A Question Emerges...
For which nutrients do women fall significantly below recommended nutritional intake levels? Or, conversely, for what nutrients do women fall significantly above recommended nutritional intake levels?
A naive approach to addressing the above is to calculate Confidence Intervals for each of the nutritional intake levels, oneatatime, using the univariate method as shown below:
\(\bar{x}_j \pm t_{n1, \alpha/2} \sqrt{s^2_j/n}\)
If we consider only a single variable, we can say with \((1  α) × 100\%\) confidence that the interval includes the corresponding population mean.
Example 72: Women’s Health Survey:
A oneatatime 95% confidence interval for calcium is given by the following where values are substituted into the formula and the calculations are worked out as shown below:
\(624.04925 \pm t_{736, 0.025}\sqrt{157829.4/737}\)
\(624.04925 \pm 1.96 \times 14.63390424\)
\(624.04925 \pm 28.68245\)
\((595.3668, 652.7317)\)
The oneatatime confidence intervals are summarized in the table below:
Variable  \(μ_{0}\)  95% Confidence Interval 
Calcium  1000 mg  595.3, 652.8 
Iron  15mg  10.7, 11.6 
Protein  60g  63.6, 68.0 
Vitamin A  800 μg  721.5, 957.8 
Vitamin C  75 mg  73.6, 84.2 
Looking at this table, it appears that the average daily intakes of calcium and iron are too low (because the intervals fall below the recommended intakes of these variables), and the average daily intake of protein is too high (because the interval falls above the recommended intake of protein).
Problem: The problem with these oneatatime intervals is that they do not control for familywide error rate.
Consequence: We are less than 95% confident that all of the intervals simultaneously cover their respective means.
To fix this problem we can calculate a \((1  \alpha) × 100\%\) Confidence Ellipse for the population mean vector \(\boldsymbol{\mu}\). To calculate this confidence ellipse we must recall that for independent random observations from a multivariate normal distribution with mean vector \(\mu\) and variancecovariance matrix \(Σ\), the Fstatistic, (shown below), is Fdistributed with p and np degrees of freedom:
\(F = n\mathbf{(\bar{x}  \boldsymbol{\mu})'S^{1}(\bar{x}  \boldsymbol{\mu})}\dfrac{(np)}{p(n1)} \sim F_{p,np}\)
This next expression says that the probability that n times the squared Mahalanobis distance between the sample mean vector, \(\boldsymbol{\bar{x}}\), and the population mean vector \(\boldsymbol{\mu}\) is less than or equal to p times n1 times the critical value from the Ftable divided by np is equal to \(1  α\).
\(\text{Pr}\{n\mathbf{(\bar{x}  \boldsymbol{\mu})'S^{1}(\bar{x}  \boldsymbol{\mu})} \le \dfrac{p(n1)}{(np)}F_{p,np,\alpha}\} = 1\alpha\)
Here the squared Mahalanobis distance between \(\bar{\mathbf{x}}\) and \(\boldsymbol{\mu}\) is being used.
Note! A closelyrelated equation for a hyperellipse is:
\(n\mathbf{(\bar{x} \boldsymbol{\mu})'S^{1}(\bar{x}  \boldsymbol{\mu})} = \dfrac{p(n1)}{(np)}F_{p,np, \alpha}\)
In particular, this is the \((1  \alpha) × 100%\) confidence ellipse for the population mean, \(\boldsymbol{\mu}\).
Interpretation
The \((1  \alpha) × 100\%\) confidence ellipse is very similar to the prediction ellipse that we discussed earlier in our discussion of the multivariate normal distribution. A \((1  \alpha) × 100\%\) confidence ellipse yields simultaneous \((1  \alpha) × 100\%\) confidence intervals for all linear combinations of the variable means. Consider linear combinations of population means as below:
\(c_1\mu_1 + c_2\mu_2 + \dots c_p \mu_p = \sum_{j=1}^{p}c_j\mu_j = \mathbf{c'\boldsymbol{\mu}}\)
The simultaneous \((1  \alpha) × 100\%\) confidence intervals for the above are given by the expression below
\(\sum_{j=1}^{p}c_j\bar{x}_j \pm \sqrt{\frac{p(n1)}{(np)}F_{p, np, \alpha}}\sqrt{\frac{1}{n}\sum_{j=1}^{p}\sum_{k=1}^{p}c_jc_ks_{jk}}\)
In terms of interpreting the \((1  \alpha) × 100\%\) confidence ellipse, we can say that we are \((1  \alpha) × 100%\) confident that all such confidence intervals cover their respective linear combinations of the treatment means, regardless of what linear combinations we may wish to consider. In particular, we can consider the trivial linear combinations which correspond to the individual variables. So this says that we going to be also \((1  \alpha) × 100\%\) confident that all of the intervals given in the expression below:
\(\bar{x}_j \pm \sqrt{\frac{p(n1)}{(np)}F_{p, np, \alpha}}\sqrt{\frac{s^2_j}{n}}\)
cover their respective treatment population means. These intervals are called simultaneous confidence intervals.
Example 73: Women’s Health Survey
Simultaneous confidence intervals are computed using hand calculations. For calcium, we substituted the following values: The sample mean was 624.04925. We have p = 5 variables, a total sample size of n = 737, and if we look up in the Ftable for 5 and 732 degrees of freedom for alpha = 0.05, the critical value is 2.21. The standard error of the sample mean for calcium is equal to the square root of the sample variance for calcium, 157,829.44, divided by the sample size, 737. The math is carried out to obtain an interval running from 575.27 to approximately 672.83 as shown below:
\(624.04925 \pm \sqrt{\frac{5(7371)}{7375}\underset{2.21}{\underbrace{F_{5,7375,0.05}}}}\sqrt{157829.4/737}\)
\(624.04925 \pm 3.333224 \times 14.63390424\)
\(624.04925 \pm 48.77808\)
\((575.27117, 672.82733)\)
Using SAS
These calculations may also be carried out using the SAS program.
SAS Program can be downloaded: nutrient5.sas
In terms of using this program, there is only a couple of things you need to change: the value of p in line five and what appears in the data step at the top of the page.
View the video below to see how to find the Hotelling's \(T^2\) value using the SAS statistical software application.In terms of using the program with other datasets, basically what you need to do is create a separate line starting with the variable for each variable that is included in the analysis. In this case, we have five nutritional variables so we have five of these lines in the first data step. Then you set this equal to, in quotes, the name you wish to give that variable. After the semicolon, we set x = to the name that we specified for that variable in the input statement and finally after another semicolon, we type in "output;".
The output is available to download here: nutrient5.lst.
The output contains the sample means for each of the variables, gives the results of the calculations under data step "b".
Confidence intervals are given by the columns for "losim" and "upsim"
Using Minitab
Minitab does not support this function
The following table gives the confidence intervals:
Variable  \(μ_{0}\)  95% Confidence Interval 
Calcium  1000 mg  575.1, 673.0 
Iron  15 mg  10.4, 11.9 
Protein  60 g  62.0, 69.6 
Vitamin A  800 μg  638.3, 1040.9 
Vitamin C  75 mg  69.9, 88.0 
Looking at these simultaneous confidence intervals we can see that the upper bound of the interval for calcium falls below the recommended daily intake of 1000 mg. Similarly, the upper bound for iron also falls below the recommended daily intake of 15 mg. Conversely, the lower bound for protein falls above the recommended daily intake of 60 g for protein. The intervals for both vitamin A and C both contain the recommended daily intake for these two vitamins.
Therefore, we can conclude that the average daily intakes of calcium and iron fall below the recommended levels, and the average daily intake of protein exceeds the recommended level.
Problem with Simultaneous Confidence Intervals
The problem with the simultaneous confidence intervals is that if we are not interested in all possible linear combinations of variables or anything other than the means by themselves, then the simultaneous confidence interval may be too wide, and hence, too conservative. As an alternative to the simultaneous confidence intervals, we can use the Bonferroni intervals.
Bonferri Intervals
The Bonferroni intervals are given in the expression below:
\(\bar{x}_j \pm t_{n1, \frac{\alpha}{2p}}\sqrt{s^2_j/n}\)
An example from the USDA Women’s Health Survey data will illustrate this calculation.
Example 74: Women’s Health Survey
Here, the 95% confidence interval for calcium under the Bonferroni correction is calculated:
\(624.04925 \pm t_{7371, \frac{0.05}{2 \times 5}}\sqrt{157829.4/737}\)
\(624.04925 \pm 2.576 \times 14.63390424\)
\(624.04925 \pm 37.69694\)
\((586.35231, 661.74619)\)
This calculation uses the values for the sample mean for calcium, 624.04925, the critical value from the ttable, with n1 degrees of freedom, evaluated at alpha over 2 times p, (0.05 divided 2 times 5, or .005). This critical value turns out to be 2.576 from the ttable. The standard error is calculated by taking the square root of the sample variance, (157,829.44), divided by the sample size, 737.
Carrying out the math, the interval goes from 586.35 to 661.75.
Using SAS
These calculations can also be obtained in the SAS program with the downloadable code below. The calculations of the upper and lower Bonferroni intervals are given by "lobon" and "upbon" at the end of data step "b". They involve the calculations:
lobon=xbartb*sqrt(s2/n);
upbon=xbar+tb*sqrt(s2/n);
SAS Program can be downloaded here: nutrient5.sas
Using Minitab
At this time Minitab does not support this procedure.
The table below shows the results of the computation
Variable  \(\mu_{0}\)  95% Confidence Interval 
Calcium  1000 mg  586.3, 661.8 
Iron  15mg  10.6, 11.7 
Protein  60g  62.9, 68.7 
Vitamin A  800 μg  684.2, 995.0 
Vitamin C  75 mg  71.9, 85.9 
When compared to the simultaneous intervals, we see that the Bonferroni intervals are narrower. However, in this case, the conclusions will be the same. The confidence intervals for both vitamin A and C both cover their respective recommended daily intakes. Intervals for calcium and iron fall below the recommended intake, while the interval for protein falls above it.
7.1.5  Profile Plots
7.1.5  Profile PlotsIf the data is of a very large dimension, tables of simultaneous or Bonferroni confidence intervals are hard to grasp at a cursory glance. A better approach is to visualize the coverage of the confidence intervals through a profile plot.
Procedure
A profile plot is obtained with the following three step procedure:
Steps
 Standardize each of the observations by dividing them by their hypothesized means. So the \(i^{th}\) observation of the \(j^{th}\) variable, \(X_{ij}\), is divided by its hypothesized mean for\(j^{th}\) variable \(\mu_0^j\). We will call the result \(Z_{ij}\) as shown below:
\(Z_{ij} = \dfrac{X_{ij}}{\mu^0_j}\)

Compute the sample mean for the \(Z_{ij}\)'s to obtain sample means corresponding to each of the variables j, 1 to p. These sample means, \(\bar{z_j}\), are then plotted against the variable j.

Plot either simultaneous or Bonferroni confidence bands for the population mean of the transformed variables,
Simultaneous \((1  α) × 100\%\) confidence bands are given by the usual formula, using the z's instead of the usual x's as shown below:
\(\bar{z}_j \pm \sqrt{\dfrac{p(n1)}{(np)}F_{p,np,\alpha}}\sqrt{\dfrac{s^2_{Z_j}}{n}}\)
The same substitutions are made for the Bonferroni \((1  α) × 100\%\) confidence band formula:
\(\bar{z}_j \pm t_{n1,\frac{\alpha}{2p}}\sqrt{\dfrac{s^2_{Z_j}}{n}}\)
Example 75: Women's Health Survey (Profile Plots)
Using SAS
The profile plots are computed using the SAS program.
Download the SAS Program: nutrient6.sas
View the video below to see how to create a profile plot using the SAS statistical software application.
Using Minitab
View the video below to see how to create a profile plot using the Minitab statistical software application.
Analysis
From this plot, the results are immediately clear. We can easily see that the confidence intervals for calcium and iron fall below 1 indicating that the average daily intakes for these two nutrients are below recommended levels. The protein confidence interval falls above the value 1 indicating that the average daily intake for protein exceeds the recommended level. The confidence intervals for vitamin A and C both contain 1 showing no significant evidence against the null hypothesis and suggesting that they meet the recommended intake of these two vitamins.
7.1.6  Paired Hotelling's TSquare
7.1.6  Paired Hotelling's TSquarePaired Samples occur in a number of different situations. For example:
 Pairs of similar individuals are selected from a population. These are selected in such a way that the two members of each pair are more similar to one another than they are to different observations in the dataset. Under this setting, one treatment may be applied to a randomly selected member of each pair, while the other treatment is applied to the remaining member.
 For a single sample of individuals, measurements may be taken both before and after treatment. For example, the mineral content of six bones is measured before and after one year of diet and exercise treatments.
 Cat eye experiment: At the University of Georgia an experiment was conducted to test the effect of a treatment for glaucoma. In this experiment, a treatment for glaucoma is applied to a randomly selected eye of each cat, while the remaining eye received a placebo. Many measurements were taken on each eye including pressure, dilation, etc.
The example that we are going to consider in this lesson involves spouse data. In this case, we have husband and wife pairs.
Example 76: Spouse Data
A sample of husband and wife pairs are asked to respond to each of the following questions:
 What is the level of passionate love you feel for your partner?
 What is the level of passionate love your partner feels for you?
 What is the level of companionate love you feel for your partner?
 What is the level of companionate love your partner feels for you?
A total of 30 married couples were questioned. Responses were recorded on the fivepoint scale. Responses included the following values:
 None at all
 Very little
 Some
 A great deal
 Tremendous amount
We will try to address two separate questions from these data.
 Do the husbands respond to the questions in the same way as their wives?
 Do the husbands and wives accurately perceive the responses of their spouses?
We shall address Question 1 first...
7.1.7  Question 1: The Univariate Case
7.1.7  Question 1: The Univariate CaseExample 77: Spouse Data (Question 1)
Question 1: Do the husbands respond to the questions in the same way as their wives?
Before considering the multivariate case let's review the univariate approach to answering this question. In this case we will compare the responses to a single question.
Univariate Paired ttest Case: Consider comparing the responses to a single question.
Notation will be as follows:
 \(X _ { 1 i }\) = response of husband i  the first member of pair i
 \(X _ { 2 i }\) = response of wife i  the second member of pair i
 \(\mu _ { 1 }\) = population mean for the husbands  the first population
 \(\mu _ { 2 }\) = population mean for the wives  the second population
Our objective here is to test the null hypothesis that the population means are equal against the alternative hypothesis that means are not equal, as described in the expression below:
\(H_0\colon \mu_1 =\mu_2 \) against \(H_a\colon \mu_1 \ne \mu_2\)
In the univariate course you learned that the null hypothesis is tested as follows. First we define \(Y _ { i }\) to be the difference in responses for the \(i^{th}\) pair of observations. In this case, this will be the difference between husband i and wife i. Likewise we can also define \(\mu _ { Y }\) to be the population mean of these differences, which is the same as the difference between the population means \(\mu _ { 1 }\) and \(\mu _ { 2 }\), both as noted below:
\(Y_i = X_{1i}X_{2i}\) and \(\mu_Y = \mu_1\mu_2\)
Testing the null hypothesis for the equality of the population means is going to be equivalent to testing the null hypothesis that \(\mu _ { Y }\) is equal to 0 against the general alternative that \(\mu _ { Y }\) is not equal to 0.
\(H_0\colon \mu_Y =0 \) against \(H_a\colon \mu_Y \ne 0\)
This hypothesis is tested using the paired ttest.
We will define \(\bar{y}\) to be the sample mean of the \(Y _ { i }\)'s:
\(\bar{y} = \dfrac{1}{n}\sum_{i=1}^{n}Y_i\)
We will also define \(s_{2}Y\) to be the sample variance of the \(Y _ { i }\)'s:
\(s^2_Y = \dfrac{\sum_{i=1}^{n}Y^2_i  (\sum_{i=1}^{n}Y_i)^2/n}{n1}\)
We will make the usual four assumptions in doing this:
 The \(Y _ { i }\)'s have common mean \(\mu _ { Y }\)
 Homoskedasticity. The \(Y _ { i }\)'s have common variance \(\sigma^2_Y\).
 Independence. The \(Y _ { i }\)'s are independently sampled.
 Normality. The \(Y _ { i }\)'s are normally distributed.
The test statistic is a tstatistic which is, in this case, equal to the sample mean divided by the standard error as shown below:
\[t = \frac{\bar{y}}{\sqrt{s^2/n}} \sim t_{n1}\]
Under the null hypothesis, \(H _ { o }\) this test statistic is going to be tdistributed with n1 degrees of freedom and we will reject \(H _ { o }\) at level \(α\) if the absolute value of the tvalue exceeds the critical value from the tdistribution with n1 degrees of freedom evaluated at \(α\) over 2.
\(t > t_{n1, \alpha/2}\)
7.1.8  Multivariate Paired Hotelling's TSquare
7.1.8  Multivariate Paired Hotelling's TSquareExample 78: Spouse Data (Multivariate Case)
Now let's consider the multivariate case. All scalar observations will be replaced by vectors of observations. As a result we will use the notation that follows:
Husband:
\(\mathbf{X}_{1i} = \left(\begin{array}{c}X_{1i1}\\ X_{1i2}\\\vdots\\X_{1ip}\end{array}\right)\) = vector of observations for the \(i ^{ th } \text{ husband}\)
\(X _{ 1i1 }\) will denote the response of the \(i ^{ th }\) husband to the first question. \(X _{ 1i2 }\) will denote the response of the \(i ^{ th }\) husband to the second question and so on...
Wife:
\(\mathbf{X}_{2i} = \left(\begin{array}{c}X_{2i1}\\ X_{2i2}\\\vdots\\X_{2ip}\end{array}\right)\) = vector of observations for the \(i ^{ th } \text{ wife}\)
\(X _{ 2i1 }\) will denote the response of the \(i ^{ th }\) wife to the first question. \(X _{ 2i2 }\) will denote the response of the \(i ^{ th }\) wife to the second question and so on...
The scalar population means are replaced by population mean vectors so that \(\mu _{ 1 }\) = population mean vector for husbands and \(\mu _{ 2 }\) = population mean vector for the wives.
Here we are interested in testing the null hypothesis that the population mean vectors are equal against the general alternative that these mean vectors are not equal.
\(H_0\colon \boldsymbol{\mu_1} = \boldsymbol{\mu_2}\) against \(H_a\colon \boldsymbol{\mu_1} \ne \boldsymbol{\mu_2}\)
Under the null hypothesis the two mean vectors are equal element by element. As with the onesample univariate case we are going to look at the differences between the observations. We will define the vector \(Y _{ 1 }\) for the \(i ^{ th }\) couple to be equal to the vector \(X _{ 1i }\) for the \(i ^{ th }\) husband minus the vector \(X _{ 2i }\) for the \(i ^{ th }\) wife. Then we will also, likewise, define the vector \(\mu _{ Y }\) to be the difference between the vector \(\mu _{ 1 }\) and the vector \(\mu _{ 2 }\).
\(\mathbf{Y}_i = \mathbf{X}_{1i}\mathbf{X}_{2i}\) and \(\boldsymbol{\mu}_Y = \boldsymbol{\mu}_1\boldsymbol{\mu}_2\)
Testing the above null hypothesis is going to be equivalent to testing the null hypothesis that the population mean vector \(\mu _{ Y }\) is equal to 0. That is, all of its elements are equal to 0. This is tested against the alternative that the vector \(\mu _{ Y }\) is not equal to 0, that is at least one of the elements is not equal to 0.
\(H_0\colon \boldsymbol{\mu}_Y = \mathbf{0}\) against \(H_a\colon \boldsymbol{\mu}_Y \ne \mathbf{0}\)
This hypothesis is tested using the paired Hotelling's \(T^{ 2 }\) test.
As before, we will define \(\mathbf{\bar{Y}}\) to denote the sample mean vector of the vectors \(Y_{ i }\).
\(\mathbf{\bar{Y}} = \dfrac{1}{n}\sum_{i=1}^{n}\mathbf{Y}_i\)
And, we will define \(S _{ Y }\) to denote the sample variancecovariance matrix of the vectors \(Y_{ i }\).
\(\mathbf{S}_Y = \dfrac{1}{n1}\sum_{i=1}^{n}\mathbf{(Y_i\bar{Y})(Y_i\bar{Y})'}\)
The assumptions are similar to the assumptions made for the onesample Hotelling's Tsquare test:
 The vectors \(Y _{ 1 }\)'s have common population mean vector \(\mu _{ Y }\), which essentially means that there are no subpopulations with mean vectors.
 The vectors \(Y _{ 1 }\)'s have common variancecovariance matrix \(\Sigma_Y\).
 Independence. The \(Y _{ 1 }\)'s are independently sampled. In this case, independence among the couples in this study.
 Normality. The \(Y _{ 1 }\)'s are multivariate normally distributed.
Paired Hotelling's TSquare test statistic is given by the expression below:
\(T^2 = n\bar{\mathbf{Y}}'\mathbf{S}_Y^{1}\mathbf{\bar{Y}}\)
It is a function of sample size n, the sample mean vectors, \(\mathbf{\bar{Y}}\), and the inverse of the variancecovariance matrix \(\mathbf{S} _{ Y }\).
Then we will define an Fstatistic as given in the expression below:
\(F = \dfrac{np}{p(n1)}T^2 \sim F_{p, np}\)
Under the null hypothesis, \(H _{ o } \colon \mu _{ Y } = 0\), this will have an Fdistribution with p and np degrees of freedom. We will reject \(H _{ o }\) at level \(\alpha\) if the Fvalue exceeds the value from Fvalue with p and np degrees of freedom, evaluated at level \(\alpha\).
\(F > F_{p,np,\alpha}\)
Let us return to our example...
7.1.9  Example: Spouse Data
7.1.9  Example: Spouse DataExample 79: Spouse Data (Paired Hotelling's)
Download the text file containing the data here: spouse.txt
Using SAS
The Spouse Data may be analyzed using the SAS program as shown below:
Download the SAS Program: spouse.sas
View the video below to see how to compute the Paired Hotelling's \(T^2\) using the SAS statistical software application.
The downloadable output is given by spouse.lst.
The first page of the output just gives a list of the raw data. You can see all of the data are numbers between 1 and 5. In fact if you look at them closely, you can see that they are mostly 4's and 5's, a few 3's. I don't think we see a 1 in there are all.
You can also see the columns for d1 through d4, and you should be able to confirm that they are indeed equal to the differences between the husbands and wives responses to the four questions.
The second page of the output gives the results of the iml procedure. First, it gives the hypothesized values of the population mean under the null hypothesis. In this case, it is just a column of zeros. The sample means of the differences are given in the next column. So the mean of the differences between husband and wives response to the first question is 0.0666667. This is also copied into the table below. The differences for the next three questions follow.
Following the sample mean vector is the sample variancecovariance matrix. The diagonal elements give the sample variances for each of the questions. So, for example, the sample variance for the first question is 0.8229885, which we have rounded off to 0.8230 and copied into the table below as well. The second diagonal element gives the sample variance for second question, and so on.
The results of the Hotelling's Tsquare statistic are given at the bottom of the output page.
Using Minitab
View the video below to see how to compute the Paired Hotelling's \(T^2\) using the Minitab statistical software application.
Analysis
The sample variancecovariance matrix from the output of the SAS program:
\(\mathbf{S} =\left(\begin{array}{rrrr}0.8230 & 0.0782 & 0.0138 & 0.0598 \\ 0.0782 & 0.8092 & 0.2138 & 0.1563 \\ 0.0138 & 0.2138 & 0.5621 & 0.5103 \\ 0.0598 & 0.1563 & 0.5103 & 0.6023 \end{array}\right)\)
Sample means and variances of the differences in responses between the husbands and wives.
Question  Mean \((\bar {y})\)  Variance \((s_{Y}^{2})\) 
1  0.0667  0.8230 
2  0.1333  0.8092 
3  0.3000  0.5621 
4  0.1333  0.6023 
Here we have \(T^{2}\) = 13.13 with a corresponding Fvalue of 2.94, with 4 and 26 degrees of freedom. The 4 corresponds with the number of questions asked of each couple. The 26 comes from subtracting the sample size of 30 couples minus the 4 questions. The pvalue for the test is 0.039.
The results of our test are that we can reject the null hypothesis that the mean difference between husband and wife responses is equal to zero.
Conclusion
Husbands do not respond to the questions in the same way as their wives (\(T^{2}\)= 13.13; F = 2.94; d.f. = 4, 26; p = 0.0394).
This indicates that husbands respond differently on at least one of the questions from their wives. It could be one question or more than one question.
The next step is to assess on which question do the husband and wives differ in their responses. This step of the analysis will involve the computation of confidence intervals.
7.1.10  Confidence Intervals
7.1.10  Confidence IntervalsConfidence intervals for the Paired Hotelling's Tsquare are computed in the same way as for the onesample Hotelling's Tsquare, therefore, the notes here will not be as detailed as for the onesample. Let's review the basic procedures:
Simultaneous (1  \(\alpha\)) × 100% Confidence Intervals for the mean differences are calculated using the expression below:
\(\bar{y}_j \pm \sqrt{\frac{p(n1)}{np}F_{p,np,\alpha}}\sqrt{\frac{s^2_{Y_j}}{n}}\)
Bonferroni (1  \(\alpha\)) × 100% Confidence Intervals for the mean differences are calculated using the following expression:
\(\bar{y}_j \pm t_{n1, \frac{\alpha}{2p}}\sqrt{\frac{s^2_{Y_j}}{n}}\)
As before, simultaneous intervals will be used if we are potentially interested in confidence intervals for linear combinations among the variables of the data. Bonferroni intervals should be used if we want to simply focus on the means for each of the individual variables themselves. In this case, the individual questions.
Example 710: Spouse Data (Bonferroni CI)
Using SAS
The simultaneous Bonferroni Confidence intervals may be computed using the SAS program that can be downloaded below:
Download the Spouse 1 SAS Program: spouse1a.sas
In this output losim and upsim give the lower and upper bounds for the simultaneous intervals, and lobon and upbon give the lower and upper bounds for the Bonferroni interval which are copied into the table below. You should be able to find where all of these numbers are obtained.
Using Minitab
At this time Minitab does not support this procedure.
Analysis
95 % Confidence Intervals


Question

Simultaneous

Bonferroni

1

0.5127, 0.6460

0.3744, 0.5078

2

0.7078, 0.4412

0.5707, 0.3041

3

0.7788, 0.1788

0.6645, 0.0645

4

0.6290, 0.3623

0.5107, 0.2440

The simultaneous confidence intervals are plotted using Profile Plots.
Using SAS
The downloadable SAS program for the profile plot can be obtained here: spouse1.sas.
(Which in this case is analogous to the earlier SAS program Download here: nutrient6.sas)
Using Minitab
At this time Minitab does not support this procedure.
Analysis
Note: The plot is given in plot1 shown below:
Here we can immediately notice that all of the simultaneous confidence intervals include 0 suggesting that the husbands and wives do not differ significantly in their responses to any of the questions. So what is going on here? Earlier the Hotelling's \(T^{2}\) test told us that there was a significant difference between the husband and wives in their responses to the questions. But the plot of the confidence intervals suggests that there are no differences.
Basically, the significant Hotelling's Tsquare result is achieved through the contributions from all of the variables. It turns out that there is going to be a linear combination of the population means of the form:
\[\Psi = \sum_{j=1}^{n}c_j\mu_{Y_j}\]
whose confidence interval will not include zero.
The profile plot suggests that the largest difference occurs in response to question 3. Here, the wives respond more positively than their husbands to the question: "What is the level of companionate love you feel for your partner?"
7.1.11  Question 2: Matching Perceptions
7.1.11  Question 2: Matching PerceptionsExample 711: Spouse Data (Question 2)
Next we will return to the second question posed earlier in this lesson.
Question 2: Do the husbands and wives accurately perceive the responses of their spouses?
To understand this question, let us return to the four questions asked of each husband and wife pair. The questions were:
 What is the level of passionate love you feel for your partner?
 What is the level of passionate love your partner feels for you?
 What is the level of companionate love you feel for your partner?
 What is the level of companionate love your partner feels for you?
Notice that these questions are all paired. The odd numbered questions ask about how each person feel about their spouse, while the even numbered questions ask how each person thinks their spouse feels towards them. The question that we are investigating now asks about perception, so here we are trying to see if the husbands accurately perceive the responses of their wives and conversely to the wives accurately perceive the responses of their husbands.
In more detail, we may ask:
In this case we are asking if the wife accurately perceives the amount of passionate love her husband feels towards her.
Here, we are asking if the husband accurately perceives the amount of passionate love his wife feels towards him.
 Does the husband's answer to question 1 match the wife's answer to question 2.
 Secondly, does the wife's answer to question 1 match the husband's answer to question 2.
 Similarly, does the husband's answer to question 3 match the wife's answer to question 4, and
 Does the wife's answer to question 3 match the husband's answer to question 4.
To address the research question we need to define four new variables as follows:
 \(Z_{i1} = X_{1i1}  X_{2i2}\)  (for the \(i^{th}\) couple, the husbands response to question 1 minus the wives response to question 2.)
 \(Z_{i2} = X_{1i2}  X_{2i1}\)  (for the \(i^{th}\) couple, the husbands response to question 2 minus the wives response to question 1.)
 \(Z_{i3} = X_{1i3}  X_{2i4}\)  (for the \(i^{th}\) couple, the husbands response to question 3 minus the wives response to question 4.)
 \(Z_{i4} = X_{1i4}  X_{2i3}\)  (for the \(i^{th}\) couple, the husbands response to question 4 minus the wives response to question 3.)
These Z's can then be collected into a vector. We can then calculate the sample mean for that vector...
\(\mathbf{\bar{z}} = \dfrac{1}{n}\sum_{i=1}^{n}\mathbf{Z}_i\)
as well as the sample variancecovariance matrix...
\(\mathbf{S}_Z = \dfrac{1}{n1}\sum_{i=1}^{n}\mathbf{(Z_i\bar{z})(Z_i\bar{z})'}\)
Here we will make the usual assumptions about the vector \(Z_{i}\) containing the responses for the \(i^{th}\) couple:
 The \(Z_{i}\)'s have common mean vector \(\mu_{Z}\)
 The \(Z_{i}\)'s have common variancecovariance matrix \(\Sigma_{Z}\)
 Independence. The \(Z_{i}\)'s are independently sampled.
 Multivariate Normality. The \(Z_{i}\)'s are multivariate normally distributed.
Question 2 is equivalent to testing the null hypothesis that the mean \(\mu_{Z}\) is equal to 0, against the alternative that \(\mu_{Z}\) is not equal to 0 as expressed below:
\(H_0\colon \boldsymbol{\mu}_Z = \mathbf{0}\) against \(H_a\colon \boldsymbol{\mu}_Z \ne \mathbf{0}\)
We may then carry out the Paired Hotelling's TSquare test using the usual formula with the sample mean vector zbar replacing the mean vector ybar from our previous example, and the sample variancecovariance matrix \(S_{Z}\) replacing the the sample variancecovariance matrix \(S_{Y}\) also from our previous example:
\(n\mathbf{\bar{z}'S^{1}_Z\bar{z}}\)
We can then form the Fstatistic as before:
\(F = \frac{np}{p(n1)}T^2 \sim F_{p, np}\)
And, under \(H_{o} \colon \mu_{Z} = 0\) we will reject the null hypothesis \(H_{o}\) at level \(\alpha\) if this Fstatistic exceeds the critical value from the Ftable with p and np degrees of freedom evaluated at \(α\).
\(F > F_{p, np, \alpha}\)
Using SAS
The analysis may be carried out using the SAS program as shown below:
The SAS program hopefully resembles the program spouse.sas used to address question #1.
Download the SAS Program: spouse2.sas
View the video below to see how to find Hotelling's \(T^{2}\) using the SAS statistical software application.The output contains on its first page a printing of the data for each of the matrixes, including the transformations d1 through d4.
Page two contains the output from the iml procedure which carries out the Hotelling's \(T^{2}\) test. Again, we can see that \(\mu_{0}\) is defined to be a vector of 0's. The sample mean vectors is given under YBAR.
S is our sample variancecovariance matrix for the Z's.
Using Minitab
At this time Minitab does not support this procedure.
Analysis
The Hotelling's \(T^{2}\) statistic comes out to be 6.43 approximately with a corresponding F of 1.44, with 4 and 26 degrees of freedom. The pvalue is 0.24 which exceeds 0.05, therefore we do not reject the null hypothesis at the 0.05 level.
Conclusion
In conclusion, we can state that there is no statistically significant evidence against the hypothesis that the husbands and wives accurately perceive the attitudes of their spouses. Our evidence includes the following statistics: ( \(T^{2}\) = 6.43; F = 1.44; d.f. = 4, 26; p = 0.249).
7.1.12  TwoSample Hotelling's TSquare
7.1.12  TwoSample Hotelling's TSquareIntroduction
This test is the multivariate analog of the two sample ttest in univariate statistics. These two sample tests, in both cases are used to compare two populations. Two populations may correspond to two different treatments within an experiment.
For example, in a completely randomized design, the two treatments are randomly assigned to the experimental units. Here, we would like to distinguish between the two treatments. Another situation occurs where the observations are taken from two distinct populations. In either case, there is no pairing of the observations.
Example 712: Swiss Bank Notes
An example where we are sampling from two distinct populations occurs with 1000 franc Swiss Bank Notes.
1. The first population is the population of Genuine Bank Notes
2. The second population is the population of Counterfeit Bank Notes
While the diagram shows a more recent version issue of the 1000 franc note, it depicts the different measurement locations taken in this study. For both population of bank notes the following measurements were taken:
 Length of the note
 Width of the LeftHand side of the note
 Width of the RightHand side of the note
 Width of the Bottom Margin
 Width of the Top Margin
 Diagonal Length of Printed Area
Objective: To determine if counterfeit notes can be distinguished from the genuine Swiss bank notes.
This is essential for the police if they wish to use these kinds of measurements to determine if a bank notes is genuine or not and help solve counterfeiting crimes.
Before considering the multivariate case, we will first consider the univariate case...
7.1.13  The Univariate Case
7.1.13  The Univariate CaseSuppose we have data from a single variable from two populations:
 The data will be denoted in Population 1 as: \(X_{11}\),\(X_{12}\), ... , \(X_{1n_{1}}\)
 The data will be denoted in Population 2 as: \(X_{21}\), \(X_{22}\), ... , \(X_{2n_{2}}\)
For both populations, the first subscript will denote which population the note is from. The second subscript will denote which observation we are looking at from each population.
Here we will make the standard assumptions:
 The data from population i is sampled from a population with mean \(\mu_{i}\). This assumption simply means that there are no subpopulations to note.
 Homoskedasticity: The data from both populations have common variance \(σ^{2}\)
 Independence: The subjects from both populations are independently sampled.
 Normality: The data from both populations are normally distributed.
Here we are going to consider testing, \(H_0\colon \mu_1 = \mu_2\) against \(H_a\colon \mu_1 \ne \mu_2\), that the populations have equal means, against the alternative hypothesis that the means are not equal.
We shall define the sample means for each population using the following expression:
\(\bar{x}_i = \dfrac{1}{n_i}\sum_{j=1}^{n_i}X_{ij}\)
We will let \(s^2_i\) denote the sample variance for the \(i^{th}\) population, again calculating this using the usual formula below:
\(s^2_i = \dfrac{\sum_{j=1}^{n_i}X^2_{ij}(\sum_{j=1}^{n_i}X_{ij})^2/n_i}{n_i1}\)
Assuming homoskedasticity, both of these sample variances, \(s^2_1\) and \(s^2_2\), are estimates of the common variance \(σ^{2}\). A better estimate can be obtained, however, by pooling these two different estimates yielding the pooled variance as given in the expression below:
\(s^2_p = \dfrac{(n_11)s^2_1+(n_21)s^2_2}{n_1+n_22}\)
Our test statistic is the students' tstatistic which is calculated by dividing the difference in the sample means by the standard error of that difference. Here the standard error of that difference is given by the square root of the pooled variance times the sum of the inverses of the sample sizes as shown below:
\(t = \dfrac{\bar{x}_1\bar{x}_2}{\sqrt{s^2_p(\dfrac{1}{n_1}+\frac{1}{n_2})}} \sim t_{n_1+n_22}\)
Under the null hypothesis, \(H_{o} \)of the equality of the population means, this test statistic will be tdistributed with \(n_{1}\) + \(n_{2}\)  2 degrees of freedom.
We will reject \(H_{o}\) at level \(α\) if the absolute value of this test statistic exceeds the critical value from the ttable with \(n_{1}\) + \(n_{2}\)  2 degrees of freedom evaluated at \(α/2\).
\(t > t_{n_1+n_22, \alpha/2}\)
All of this should be familiar to you from your introductory statistics course.
Next, let's consider the multivariate case...
7.1.14  The Multivariate Case
7.1.14  The Multivariate CaseIn this case we are replacing the random variables \(X_{ij}\), for the \(j_{th}\) sample for the \(i_{th}\) population, with random vectors \(X_{ij}\), for the \(j_{th}\) sample for the \(i_{th}\) population. These vectors contain the observations from the p variables.
In our notation, we will have our two populations:
 The data will be denoted in Population 1 as: \(X_{11}\),\(X_{12}\), ... , \(X_{1n_{1}}\)
 The data will be denoted in Population 2 as: \(X_{21}\), \(X_{22}\), ... , \(X_{2n_{2}}\)
Here the vector \(X_{ij}\) represents all of the data for all of the variables for sample unit j, for population i.
\(\mathbf{X}_{ij} = \left(\begin{array}{c}X_{ij1}\\X_{ij2}\\\vdots\\X_{ijp}\end{array}\right)\)
This vector contains elements \(X_{ijk}\) where k runs from 1 to p, for p different observed variables. So, \(X_{ijk}\) is the observation for variable k of subject j from population i.
The assumptions here will be analogous to the assumptions in the univariate setting.
Assumptions
 The data from population i is sampled from a population with mean vector \(\mu_{i}\). Again, corresponding to the assumption that there are no subpopulations.
 Instead of assuming Homoskedasticity, we now assume that the data from both populations have common variancecovariance matrix \(Σ\).
 Independence. The subjects from both populations are independently sampled.
 Normality. Both populations are normally distributed.
Consider testing the null hypothesis that the two populations have identical population mean vectors. This is represented below as well as the general alternative that the mean vectors are not equal.
\(H_0: \boldsymbol{\mu_1 = \mu_2}\) against \(\boldsymbol{\mu_1 \ne \mu_2}\)
So here what we are testing is:
\(H_0\colon \left(\begin{array}{c}\mu_{11}\\\mu_{12}\\\vdots\\\mu_{1p}\end{array}\right) = \left(\begin{array}{c}\mu_{21}\\\mu_{22}\\\vdots\\\mu_{2p}\end{array}\right)\) against \(H_a\colon \left(\begin{array}{c}\mu_{11}\\\mu_{12}\\\vdots\\\mu_{1p}\end{array}\right) \ne \left(\begin{array}{c}\mu_{21}\\\mu_{22}\\\vdots\\\mu_{2p}\end{array}\right)\)
Or, in other words...
\(H_0\colon \mu_{11}=\mu_{21}\) and \(\mu_{12}=\mu_{22}\) and \(\dots\) and \(\mu_{1p}=\mu_{2p}\)
The null hypothesis is satisfied if and only if the population means are identical for all of the variables.
The alternative is that at least one pair of these means is different. This is expressed below:
\(H_a\colon \mu_{1k}\ne \mu_{2k}\) for at least one \( k \in \{1,2,\dots, p\}\)
To carry out the test, for each population i, we will define the sample mean vectors, calculated the same way as before, using data only from the\(i_{th}\) population.
\(\mathbf{\bar{x}}_i = \dfrac{1}{n_i}\sum_{j=1}^{n_i}\mathbf{X}_{ij}\)
Similarly, using data only from the \(i^{th}\) population, we will define the sample variancecovariance matrices:
\(\mathbf{S}_i = \dfrac{1}{n_i1}\sum_{j=1}^{n_i}\mathbf{(X_{ij}\bar{x}_i)(X_{ij}\bar{x}_i)'}\)
Under our assumption of homogeneous variancecovariance matrices, both \(S_{1}\) and \(S_{2}\) are estimators for the common variancecovariance matrix Σ. A better estimate can be obtained by pooling the two estimates using the expression below:
\(\mathbf{S}_p = \dfrac{(n_11)\mathbf{S}_1+(n_21)\mathbf{S}_2}{n_1+n_22}\)
Again, each sample variancecovariance matrix is weighted by the sample size minus 1.
7.1.15  The TwoSample Hotelling's TSquare Test Statistic
7.1.15  The TwoSample Hotelling's TSquare Test StatisticNow we are ready to define the Twosample Hotelling's TSquare test statistic. As in the expression below, you will note that it involves the computation of differences in the sample mean vectors. It also involves a calculation of the pooled variancecovariance matrix multiplied by the sum of the inverses of the sample size. The resulting matrix is then inverted.
\(T^2 = \mathbf{(\bar{x}_1  \bar{x}_2)}^T\{\mathbf{S}_p(\frac{1}{n_1}+\frac{1}{n_2})\}^{1} \mathbf{(\bar{x}_1  \bar{x}_2)}\)
For large samples, this test statistic will be approximately chisquare distributed with \(p\) degrees of freedom. However, as before this approximation does not take into account the variation due to estimating the variancecovariance matrix. So, as before, we will look at transforming this Hotelling's Tsquare statistic into an Fstatistic using the following expression.
Note! This is a function of the sample sizes of the two populations and the number of variables measured p.
\(F = \dfrac{n_1+n_2p1}{p(n_1+n_22)}T^2 \sim F_{p, n_1+n_2p1}\)
Under the null hypothesis, \(H_{o}\colon \mu_{1} = \mu_{2}\) this Fstatistic will be Fdistributed with p and \(n_{1} + n_{2}  p\) degrees of freedom. We would reject \(H_{o}\) at level \(α\) if it exceeds the critical value from the Ftable evaluated at \(α\).
\(F > F_{p, n_1+n_2p1, \alpha}\)
Example 713: Swiss Bank Notes (TwoSample Hotelling's)
Using SAS
The two sample Hotelling's \(T^{2}\) test can be carried out using the Swiss Bank Notes data using the SAS program as shown below:
Download the SAS Program: swiss10.sas
Download the output: swiss10.lst.
View the video below to see how to compute the Two Sample Hotelling's \(T^2\) using the SAS statistical software application.
At the top of the first output page you see that N1 is equal to 100 indicating that we have 100 bank notes in the first sample. In this case 100 real or genuine notes.
Using Minitab
View the video below to see how to compute the Two Sample Hotelling's \(T^2\) using the Minitab statistical software application.
Analysis
The sample mean vectors are copied into the table below:
Means  
Variable  Genuine  Counterfeit 
Length  214.969  214.823 
Left Width  129.943  130.300 
Right Width  129.720  130.193 
Bottom Margin  8.305  10.530 
Top Margin  10.168  11.133 
Diagonal  141.517  139.450 
The sample variancecovariance matrix for the real or genuine notes appears below:
\(S_1 = \left(\begin{array}{rrrrrr}0.150& 0.058& 0.057 &0.057&0.014&0.005\\0.058&0.133&0.086&0.057&0.049&0.043\\0.057&0.086&0.126&0.058&0.031&0.024\\0.057&0.057&0.058&0.413&0.263&0.000\\0.014&0.049&0.031&0.263&0.421&0.075\\0.005&0.043&0.024&0.000&0.075&0.200\end{array}\right)\)
The sample variancecovariance for the second sample of notes, the counterfeit note, is given below:
\(S_2 = \left(\begin{array}{rrrrrr}0.124&0.032&0.024&0.101&0.019&0.012\\0.032&0.065&0.047&0.024&0.012&0.005\\0.024&0.047&0.089&0.019&0.000&0.034\\0.101&0.024&0.019&1.281&0.490&0.238\\ 0.019&0.012&0.000&0.490&0.404&0.022\\0.012&0.005&0.034&0.238&0.022&0.311\end{array}\right)\)
This is followed by the pooled variancecovariance matrix for the two samples.
\(S_p = \left(\begin{array}{rrrrrr}0.137&0.045&0.041&0.022&0.017&0.009\\0.045&0.099&0.066&0.016&0.019&0.024\\0.041&0.066&0.108&0.020&0.015&0.005\\0.022&0.016&0.020&0.847&0.377&0.119\\0.017&0.019&0.015&0.377&0.413&0.049\\0.009&0.024&0.005&0.119&0.049&0.256\end{array}\right)\)
The twosample Hotelling's \(T^{2}\) statistic is 2412.45. The Fvalue is about 391.92 with 6 and 193 degrees of freedom. The pvalue is close to 0 and so we will write this as \(< 0.0001\).
In this case we can reject the null hypothesis that the mean vector for the counterfeit notes equals the mean vector for the genuine notes given the evidence as usual: (\(T_{2} = 2412.45\); \(F = 391.92\); \(d. f. = 6, 193\); \(p< 0.0001\))
Conclusion
The counterfeit notes can be distinguished from the genuine notes on at least one of the measurements.
After concluding that the counterfeit notes can be distinguished from the genuine notes the next step in our analysis is to determine upon which variables they are different.
7.1.16  Summary of Basic Material
7.1.16  Summary of Basic Material Application of Hotelling's Tsquare statistics to carry out inferences on a single sample mean, paired sample means, and twosample means
 How to set up a confidence interval for a population mean vector
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A more indepth analysis on some of these topics is done in the Advanced section.