# 11.1.9 - Confidence Intervals

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Confidence intervals for the Paired Hotelling's T-square are computed in the same way as for the one-sample Hotelling's T-square, therefore, the notes here will not be in quite as detailed as they were previously. But let's review the basic procedures:

Simultaneous (1 - α) × 100% Confidence Intervals for the mean differences are calculated using the expression below:

$\bar{y}_j \pm \sqrt{\frac{p(n-1)}{n-p}F_{p,n-p,\alpha}}\sqrt{\frac{s^2_{Y_j}}{n}}$

Bonferroni (1 - α) × 100% Confidence Intervals for the mean differences are calculated using the following expression:

$\bar{y}_j \pm t_{n-1, \frac{\alpha}{2p}}\sqrt{\frac{s^2_{Y_j}}{n}}$

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: Spouse Data

The simultaneous Bonferroni Confidence intervals may be computed using the SAS program spouse1a.sas as shown below:

Note that this SAS program is similar in format to nutrient5.sas which we considered earlier so I will go through this a little more quickly this time.

In this output losim and up sim 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.

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At this time Minitab Version 16 does not support this procedure.

 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 may be plotted using Profile Plots.

The SAS program for plot is spouse1.sas.

(Which in this case is analogous to the earlier SAS program nutrient6.sas.)

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At this time Minitab Version 16 does not support this procedure.

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 T2 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 interval suggests that are no differences.

Basically, the significant Hotelling's T-square 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?"