**Univariate case**

As described in One Sample t-Test, the t-test can be used to test the null hypothesis that the population mean of a random variable *x* has a certain value, i.e. H_{0}: *μ* = *μ*_{0}. The test statistic is given by

The applicable univariate test of the null hypothesis is based on the fact that *t* ~ *T*(*n* – 1) provided the following assumptions are met:

- The population of
*x*has a unique mean: i.e. there are no distinct sub-populations with different means - The population of
*x*has a normal distribution - The sample is a random sample with each element in the sample taken independently

Regarding the normality assumption, if *n* is sufficiently large, the Central Limit Theorem holds, and we can proceed as if the population were normal. It turns out that the t-test is pretty robust for violations of the normality assumption provided the population is relatively symmetric about the mean.

The null hypothesis is rejected if |*t*| > *t _{crit}*. Also note that by Property 1 of F Distribution, an equivalent test can be made using the test statistic

*t*

^{2}and noting that

*t*

^{2}~

*F*(1,

*n*– 1).

Now *t*^{2} can be expressed as follows:

where *x̄* is the sample mean and *s* is the sample standard deviation.

**Multivariate case**

We now look at a multivariate version of the problem, namely that the population mean of the *k* × 1 random vector has a certain value, i.e. the null hypothesis H_{0}: *μ* = *μ*^{0} where *μ* and *μ*^{0} are vectors.

Since the null hypothesis is true when for all *i*, 1 ≤ *i* ≤ *k*, one way to carry out this test is to perform *k* separate univariate *t* tests (or the equivalent *F* tests). The null hypothesis is then rejected if any one of these *k* univariate tests rejects its null hypothesis.

As we observed in Experiment-wise Error Rate, this approach introduces experiment-wise error, namely if we use a given value of *α* for all *k* tests, then the probability of multivariate null hypothesis being rejected is much higher than *α*. For this reason we typically use a correction factor, namely either the Dunn/Sidák or Bonferroni correction factor, as described in Planned Comparisons*,* and so use either 1 –(1–*α*)^{1/k} or *α*/*k* instead of *α* for each of the *k* univariate tests.

This approach is perfectly reasonable when the random variables *x _{i}* in

*X*are independent, but when they are not independent then the Dunn/Sidák or Bonferroni correction factors over-correct and the resulting experiment-wise value for

*α*is lower than it needs to be, which results in a test with lower statistical power.

Since it is common to create experiments in which the random variables *x _{i}* in

*X*are not independent, it is better to use a different approach. In particular we will use the multivariate test based on the Hotelling’s T-square test statistic.

**Definition 1**: The Hotelling’s T-square test statistic is

where *S* is the covariance matrix of the sample for *X*, *X̄* is the mean of the sample, and where the sample for each random variable *x _{i}* in

*X*has

*n*elements.

Note the similarity between the expression for *T*^{2} and the expression for *t*^{2} given above.

**Theorem 1**:

**Corollary 1**: For *n* sufficiently large, *T*^{2} ~ *χ*^{2} (*k*)

**Observation**: For small *n*, *T*^{2} is not sufficiently accurate and a better estimate is achieved using the following theorem.

**Theorem 2**: Under the null hypothesis

If *F* > *F _{crit}* then we reject the null hypothesis.

**Example 1**: A shoe company evaluates new shoe models based on five criteria: style, comfort, stability cushioning and durability, with each of the first four criteria evaluated on a scale of 1 to 20 and the durability criteria evaluated on the scale of 1 to 10. Column I of Figure 1 shows the goals for each criteria expected from new products.

**Figure 1 – Product goals by criteria**

Based on the evaluations of 25 people about the company’s latest prototype (Model 1) shown in Figure 2, determine whether the shoe is ready for release to the market.

**Figure 2 – Sample data for Example 1**

The sample mean and standard deviation for each criteria is given in columns J and K of Figure 1. The sample covariance and correlation matrices are then calculated using the supplemental array formulas COV(B4:F28) and CORR(B4:F28), as shown in Figure 3.

**Figure 3 – Covariance and Correlation Matrices**

Using Definition 1 and the data in Figure 1, 2 and 3, we can calculate , as shown in cell I20 of Figure 4. Since the formula used to calculate is an array formula (even though it yields a numeric result), it is important to press Ctrl-Shft-Enter after entering the formula in cell I20.

**Figure 4 – Hotelling T ^{2} test for a single sample**

As we will see shortly, you can also obtain *T*^{2} by using the formula

=HotellingT2(B4:F28,I4:I8)

which employs the supplemental function **HotellingT2** found in the Real Statistics Resource Pack. Note that this formula is not an array formula and so it is sufficient to press Enter after entering the formula.

We can determine whether there is a significant difference between the sample means in the five categories and the goals (i.e. population means) by using Theorem 2. As can be seen from Figure 4, since p-value < .05 (or *F* > *F _{crit}*), we reject the null hypothesis and conclude there is a significant difference between the mean scores in the sample and the stated goals.

**Confidence intervals**

**Example 2**: For the prototype shoe in Example 1, determine which criteria meet the goals and which do not.

We can test each criteria using the One Sample t-Test. The results of this analysis is shown in Figure 5.

**Figure 5 – T-test for each criteria**

From Figure 5, we can conclude that Style and Cushioning are significantly below the goals, Durability is significantly higher than the goals and Comfort and Stability are within the goals.

Figure 5 includes the 95% confidence intervals for each criteria. In the univariate case, the 1 – α confidence interval for the population mean μ is based on

which nets out to the interior of the interval

The problem with this analysis is that we haven’t taken experiment-wise error into account. In fact the combined error rate is instead of .05.

If the random variables representing the 5 criteria were independent, then we could compensate for this by applying either the Dunn/Sidák or Bonferroni correction factor. As we can see from the correlation matrix in Figure 2, the variables are clearly not independent since some of the values off of the main diagonal are far from zero.

In order to control for experiment-wise error, we instead calculate the 95% confidence ellipse (using Theorem 2) which takes all 5 criteria into account simultaneously. In particular, we use the following modified version of the observation which follows Property 3 of Multivariate Normal Distribution Basic Concepts, noting that based on Definition 1 and Definition 3 of Multivariate Normal Distribution Basic Concepts, *T*^{2} is approximately *n* times the Mahalanobis distance between and *μ*^{0}.

Using Theorem 2, we have a 1 – *α* confidence hyper-ellipse for the population mean vector μ which is given by

Thus we are looking for values of μ which fall within the hyper-ellipse given by the equation

From the 1 – α confidence hyper-ellipse, we can also calculate simultaneous confidence intervals for any linear combination of the means of the individual random variables. For example, for the linear combination

the 1 – α simultaneous confidence interval is given by the expression

where the sample covariance matrix is *S* = [*s _{ij}*].

For the case where** ***c = μ _{i}*

**the 1 – α simultaneous confidence interval is given by the expression**

where *s _{i}* is

*s*= the standard deviation of the

_{ii}*x*. Since the 1 – α confidence intervals for all linear combinations are a manifestation of the 1 – α confidence hyper-ellipse, it follows that the following are simultaneously the 95% confidence intervals for all 5 criteria:

_{i}**Figure 6 – Simultaneous 95% confidence intervals**

Most of the cells in Figure 6 already appear in Figure 4. The value of t-crit (e.g. cell U29) is derived from F-crit using the formula =SQRT(U24*(U23-1)/(U23-U24)*U28). The value of the lower bound of the 95% confidence interval for Style is calculated by the formula =U22-U29*U30, and similarly for the upper and lower bounds of the other criteria.

We summarize the conclusions from this analysis in Figure 7.

**Figure 7 – Comparison of simultaneous intervals with goals**

From Figure 7, we conclude that Cushioning is significantly below the goals, Durability is significantly higher than the goals and the other criteria are within the goals set by the company.

The simultaneous confidence intervals handle all linear combination of the means, but since we are only interested in the individual means the stated confidence intervals may be too wide. In this case we might be better off using the Dunn/Sidák or Bonferroni correction factor. The calculations for the Bonferroni correction factor are similar to those in Figure 5, except that we use the experiment-wise value of alpha = α/k = .05/5 = .01. The results are shown in Figure 8.

**Figure 8 – Bonferroni confidence intervals**

The confidence intervals in Figure 8 are narrower than those in Figure 7 and the overall conclusions are a little different, as shown in Figure 9.

**Figure 9 – Comparison of Bonferroni intervals with goals**

**Observation**: Although we will generally use the hyper-ellipse based on *F* as described above, for large samples, based on Corollary 1, we could also use the following 1 – α confidence hyper-ellipse for any linear combination *c* of the means:

Where *c = μ _{i}*, this nets out to

I rewrite this in C++.

https://github.com/niitsuma/hotelling_t_square_test

May I use example data as test data?

https://github.com/niitsuma/hotelling_t_square_test/blob/master/hotelling_t2_test.cpp

Of course I need to protect the copyright on the software and website content, but I don’t have any problem with you using the test data for the hotelling T tests from my website.

Charles

Pingback: Hotelling T2 Excel | laatuhotellit

Charles – I am running through your example, and am running into an issue. When I input the covariance statement “=COV(B4:F28)”, it only returns one cell (4.3333) instead of the matrix you have in Figure 3. Is there something I am missing? (I am using Excel 2011 for Mac.)

Thanks for this great resource!

Nevermind – I figured it out. But for any other Mac users out there, it required highlighting the 5×5 cell range for the output and clicking Command+Shift+Enter (instead of Ctrl+Shift+Enter in Windows).

One other question – do you know the convention for reporting results of Hotelling’s T2 test? Do you report the T2 statistic or the F value?

I don’t know the convention, but based on how these tests are usually reported, you should report both the T2 statistic and the F value.

Charles

That makes sense. Thank you!

I am learning a great deal on this site that is helpful for my current research. Another question I am wondering about is as follows:

From what I understand, Hotelling’s T2 statistics will tell me if a linear composite of the different components differs from the constant (or goal). Suppose I am interested in an overall directional hypothesis, wondering if the components are not just different from the constant, but different in a specific direction.

Take your shoe company data as an example (and suppose that the goals for each of the categories was the same – 7 out of 10). What if I wanted a multivariate statistic to tell me whether, on average across categories, our shoe performs better than 7. Is there a multivariate test for that? Or would I simply take the overall mean across all categories, and conduct a univariate t-test?

I apologize for monopolizing this page, but this information has been profound helpful to me, and I thank you for it. If you have opportunity to respond, please do.

Thanks again!

Chris

Chris,

Perhaps I am not interpreting your question properly, but I would think the answer is no, since I tend to think of these linear composites as vectors and so the likelihood is that two vectors intersect and so one is not larger than the other. This is my initial intuition, but perhaps someone in the community has a different idea.

Charles

Hey Charles, I need your help for my final project.

the project data has 552 data points with each data record containing 209 values.

I need to perform phase 1 analysis, eliminate the Out of control points, determine the statistical parameters and then perform a Principal component analysis.

How can you help me in this?

Hi Nathan,

Please start by looking at the Factor Analysis webpages. See

Factor Analysis

Charles

I have a question please.

In a case where 7 human subjects were given an alcoholic drink and blood samples for glucose concentration for each sample, were taken from the 7 subjects at 1,2,3 &4 hrs intervals after the consumption of the alcoholic drink. If we intend to test whether blood glucose concentrations are the same for the four time points, what test statistics can I use?

Anie,

You can use repeated measures ANOVA or repeated measures MANOVA.

Charles

Dear Charles,

I created 10 small samples (500 instances) from a large dataset.

I wish to perform Hotelling’s T-squared test to check if my samples are representatives of the original dataset. Could I do two-sample test or do I need to perform one-sample test instead? Thank you

Nush,

I don’t understand why you would want to do such a thing if you know that these are random samples from the large data set. Also why do you want to use Hotelling’s T-square test to do this?

Charles

Dear Charles,

This dataset contains 60 features. Therefore, I need to perform a mutivariate analysis. My objective is to show that the selected random instances are of representative instances of the original dataset.

Thank you.

Dr. Zaiontz

I was experimenting with you RealStat Add-In to perform a One Sample Hotelling’s T-square test and then needed to calculate confidence intervals (simultaneous intervals and Bonferroni confidence intervals) . While I can do them by hand in excel your web site alludes to the fact that the Add-In can perform this using the One Sample t-Test, but I don’t fined that option in the Add-In.

PS. Really appreciate your work

Jeff,

The one-sample t test is an option on the T Test and Nonparametric Equivalents data analysis tool.

Charles