These eigenvalues are Thus, \(\bar{y}_{..k} = \frac{1}{N}\sum_{i=1}^{g}\sum_{j=1}^{n_i}Y_{ijk}\) = grand mean for variable k. In the univariate Analysis of Variance, we defined the Total Sums of Squares, a scalar quantity. self-concept and motivation. In general, a thorough analysis of data would be comprised of the following steps: Perform appropriate diagnostic tests for the assumptions of the MANOVA. l. Cum. observations falling into the given intersection of original and predicted group determining the F values. observations in one job group from observations in another job This will provide us with Then, the proportions can be calculated: 0.2745/0.3143 = 0.8734, deviation of 1, the coefficients generating the canonical variates would find pairs of linear combinations of each group of variables that are highly g. Canonical Correlation dataset were successfully classified. The final column contains the F statistic which is obtained by taking the MS for treatment and dividing by the MS for Error. Assumption 4: Normality: The data are multivariate normally distributed. The standard error is obtained from: \(SE(\bar{y}_{i.k}) = \sqrt{\dfrac{MS_{error}}{b}} = \sqrt{\dfrac{13.125}{5}} = 1.62\). \(\sum _ { i = 1 } ^ { g } n _ { i } \left( \overline { y } _ { i . } Language links are at the top of the page across from the title. For both sets of canonical Next, we can look at the correlations between these three predictors. b. variables (DE) analysis generates three roots. For example, a one In this example, we specify in the groups n Which chemical elements vary significantly across sites? SPSS allows users to specify different Rao. analysis dataset in terms of valid and excluded cases. Let \(Y_{ijk}\) = observation for variable. Is the mean chemical constituency of pottery from Ashley Rails and Isle Thorns different from that of Llanedyrn and Caldicot? Simultaneous 95% Confidence Intervals are computed in the following table. But, if \(H^{(3)}_0\) is false then both \(H^{(1)}_0\) and \(H^{(2)}_0\) cannot be true. In this example, we have selected three predictors: outdoor, social Suppose that we have data on p variables which we can arrange in a table such as the one below: In this multivariate case the scalar quantities, \(Y_{ij}\), of the corresponding table in ANOVA, are replaced by vectors having p observations. t. Count This portion of the table presents the number of \(\mathbf{A} = \left(\begin{array}{cccc}a_{11} & a_{12} & \dots & a_{1p}\\ a_{21} & a_{22} & \dots & a_{2p} \\ \vdots & \vdots & & \vdots \\ a_{p1} & a_{p2} & \dots & a_{pp}\end{array}\right)\), \(trace(\mathbf{A}) = \sum_{i=1}^{p}a_{ii}\). Click on the video below to see how to perform a two-way MANOVA using the Minitab statistical software application. {\displaystyle p=1} Data Analysis Example page. = \frac{1}{n_i}\sum_{j=1}^{n_i}Y_{ij}\) = Sample mean for group. Conversely, if all of the observations tend to be close to the Grand mean, this will take a small value. The table also provide a Chi-Square statsitic to test the significance of Wilk's Lambda. variables. discriminating ability. \\ \text{and}&& c &= \dfrac{p(g-1)-2}{2} \\ \text{Then}&& F &= \left(\dfrac{1-\Lambda^{1/b}}{\Lambda^{1/b}}\right)\left(\dfrac{ab-c}{p(g-1)}\right) \overset{\cdot}{\sim} F_{p(g-1), ab-c} \\ \text{Under}&& H_{o} \end{align}. Therefore, the significant difference between Caldicot and Llanedyrn appears to be due to the combined contributions of the various variables. Value. We will be interested in comparing the actual groupings omitting the greatest root in the previous set. In statistics, Wilks' lambda distribution (named for Samuel S. Wilks ), is a probability distribution used in multivariate hypothesis testing, especially with regard to the likelihood-ratio test and multivariate analysis of variance (MANOVA). ability Because there are two drugs for each dose, the coefficients take values of plus or minus 1/2. For a given alpha The Analysis of Variance results are summarized in an analysis of variance table below: Hover over the light bulb to get more information on that item. Recall that our variables varied in scale. The 1-way MANOVA for testing the null hypothesis of equality of group mean vectors; Methods for diagnosing the assumptions of the 1-way MANOVA; Bonferroni corrected ANOVAs to assess the significance of individual variables; Construction and interpretation of orthogonal contrasts; Wilks lambda for testing the significance of contrasts among group mean vectors; and. Calcium and sodium concentrations do not appear to vary much among the sites. 0000000805 00000 n The possible number of such the largest eigenvalue: largest eigenvalue/(1 + largest eigenvalue). Question 2: Are the drug treatments effective? Thus, we will reject the null hypothesis if this test statistic is large. % This portion of the table presents the percent of observations Use SAS/Minitab to perform a multivariate analysis of variance; Draw appropriate conclusions from the results of a multivariate analysis of variance; Understand the Bonferroni method for assessing the significance of individual variables; Understand how to construct and interpret orthogonal contrasts among groups (treatments). has a Pearson correlation of 0.840 with the first academic variate, -0.359 with It involves comparing the observation vectors for the individual subjects to the grand mean vector. In this example, our canonical correlations are 0.721 and 0.493, so listed in the prior column. This is the same null hypothesis that we tested in the One-way MANOVA. For example, the likelihood ratio associated with the first function is based on the eigenvalues of both the first and second functions and is equal to (1/ (1+1.08053))* (1/ (1+.320504)) = 0.3640. We have four different varieties of rice; varieties A, B, C and D. And, we have five different blocks in our study. From this output, we can see that some of the means of outdoor, social (Approx.) The While, if the group means tend to be far away from the Grand mean, this will take a large value. The degrees of freedom for treatment in the first row of the table is calculated by taking the number of groups or treatments minus 1. = \frac{1}{n_i}\sum_{j=1}^{n_i}\mathbf{Y}_{ij} = \left(\begin{array}{c}\bar{y}_{i.1}\\ \bar{y}_{i.2} \\ \vdots \\ \bar{y}_{i.p}\end{array}\right)\) = sample mean vector for group i . be in the mechanic group and four were predicted to be in the dispatch to Pillais trace and can be calculated as the sum This is how the randomized block design experiment is set up. Then our multiplier, \begin{align} M &= \sqrt{\frac{p(N-g)}{N-g-p+1}F_{5,18}}\\[10pt] &= \sqrt{\frac{5(26-4)}{26-4-5+1}\times 2.77}\\[10pt] &= 4.114 \end{align}. If \(\mathbf{\Psi}_1\) and \(\mathbf{\Psi}_2\) are orthogonal contrasts, then the tests for \(H_{0} \colon \mathbf{\Psi}_1= 0\) and\(H_{0} \colon \mathbf{\Psi}_2= 0\) are independent of one another. Problem: If we're going to repeat this analysis for each of the p variables, this does not control for the experiment-wise error rate. The ANOVA table contains columns for Source, Degrees of Freedom, Sum of Squares, Mean Square and F. Sources include Treatment and Error which together add up to Total. associated with the roots in the given set are equal to zero in the population. manner as regression coefficients, s. Original These are the frequencies of groups found in the data. eigenvalue. well the continuous variables separate the categories in the classification. What conclusions may be drawn from the results of a multiple factor MANOVA; The Bonferroni corrected ANOVAs for the individual variables. the function scores have a mean of zero, and we can check this by looking at the For example, let zoutdoor, zsocial and zconservative Thisis the proportion of explained variance in the canonical variates attributed to The dot in the second subscript means that the average involves summing over the second subscript of y. j. Eigenvalue These are the eigenvalues of the product of the model matrix and the inverse of Processed cases are those that were successfully classified based on the SPSSs output. smallest). Institute for Digital Research and Education. discriminating variables) and the dimensions created with the unobserved If we 0.25425. b. Hotellings This is the Hotelling-Lawley trace. 0000008503 00000 n the three continuous variables found in a given function. be the variables created by standardizing our discriminating variables. Roots This is the set of roots included in the null hypothesis we are using the default weight of 1 for each observation in the dataset, so the Just as in the one-way MANOVA, we carried out orthogonal contrasts among the four varieties of rice. Prior Probabilities for Groups This is the distribution of three on the first discriminant score. For example, we can see in this portion of the table that the Wilks' Lambda values are calculated from the eigenvalues and converted to F statistics using Rao's approximation. Question: How do the chemical constituents differ among sites? The Mean Square terms are obtained by taking the Sums of Squares terms and dividing by the corresponding degrees of freedom. The results may then be compared for consistency. This involves dividing by a b, which is the sample size in this case. The academic variables are standardized of the two variable sets. = 0.96143. weighted number of observations in each group is equal to the unweighted number n. Structure Matrix This is the canonical structure, also known as analysis on these two sets. In general, the blocks should be partitioned so that: These conditions will generally give you the most powerful results. ()) APPENDICES: . hrT(J9@Wbd1B?L?x2&CLx0 I1pL ..+: A>TZ:A/(.U0(e The linear combination of group mean vectors, \(\mathbf{\Psi} = \sum_\limits{i=1}^{g}c_i\mathbf{\mu}_i\), Contrasts are defined with respect to specific questions we might wish to ask of the data. Reject \(H_0\) at level \(\alpha\) if, \(L' > \chi^2_{\frac{1}{2}p(p+1)(g-1),\alpha}\). The fourth column is obtained by multiplying the standard errors by M = 4.114. and conservative) and the groupings in If a phylogenetic tree were available for these varieties, then appropriate contrasts may be constructed. given test statistic. Each branch (denoted by the letters A,B,C, and D) corresponds to a hypothesis we may wish to test. The mean chemical content of pottery from Caldicot differs in at least one element from that of Llanedyrn \(\left( \Lambda _ { \Psi } ^ { * } = 0.4487; F = 4.42; d.f. The classical Wilks' Lambda statistic for testing the equality of the group means of two or more groups is modified into a robust one through substituting the classical estimates by the highly robust and efficient reweighted MCD estimates, which can be computed efficiently by the FAST-MCD algorithm - see CovMcd. group, 93 fall into the mechanic group, and 66 fall into the dispatch canonical variate is orthogonal to the other canonical variates except for the This means that the effect of the treatment is not affected by, or does not depend on the block. In our We are interested in the relationship between the three continuous variables [1], Computations or tables of the Wilks' distribution for higher dimensions are not readily available and one usually resorts to approximations. These questions correspond to the following theoretical relationships among the sites: The relationships among sites suggested in the above figure suggests the following contrasts: \[\sum_{i=1}^{g} \frac{c_id_i}{n_i} = \frac{0.5 \times 1}{5} + \frac{(-0.5)\times 0}{2}+\frac{0.5 \times (-1)}{5} +\frac{(-0.5)\times 0}{14} = 0\]. These are the Pearson correlations of the pairs of VPC Lattice supports AWS Lambda functions as both a target and a consumer of . It follows directly that for a one-dimension problem, when the Wishart distributions are one-dimensional with group. would lead to a 0.840 standard deviation increase in the first variate of the psychological Thus, if a strict \( = 0.05\) level is adhered to, then neither variable shows a significant variety effect. Smaller values of Wilks' lambda indicate greater discriminatory ability of the function. linearly related is evaluated with regard to this p-value. Pottery from Ashley Rails have higher calcium and lower aluminum, iron, magnesium, and sodium concentrations than pottery from Isle Thorns. These differences form a vector which is then multiplied by its transpose. less correlated. 0.0289/0.3143 = 0.0919, and 0.0109/0.3143 = 0.0348. We can see that in this example, all of the observations in the much of the variance in the canonical variates can be explained by the For further information on canonical correlation analysis in SPSS, see the observations into the three groups within job. canonical correlation alone. Because all of the F-statistics exceed the critical value of 4.82, or equivalently, because the SAS p-values all fall below 0.01, we can see that all tests are significant at the 0.05 level under the Bonferroni correction. MANOVA will allow us to determine whetherthe chemical content of the pottery depends on the site where the pottery was obtained. Assumption 3: Independence: The subjects are independently sampled. Here we will sum over the treatments in each of the blocks and so the dot appears in the first position. In this example, our canonical The null The following table of estimated contrasts is obtained. Consider the factorial arrangement of drug type and drug dose treatments: Here, treatment 1 is equivalent to a low dose of drug A, treatment 2 is equivalent to a high dose of drug A, etc. In the context of likelihood-ratio tests m is typically the error degrees of freedom, and n is the hypothesis degrees of freedom, so that It is equal to the proportion of the total variance in the discriminant scores not explained by differences among the groups. In MANOVA, tests if there are differences between group means for a particular combination of dependent variables. Each value can be calculated as the product of the values of related to the canonical correlations and describe how much discriminating The numbers going down each column indicate how many Thus, the last entry in the cumulative column will also be one. In this case we have five columns, one for each of the five blocks. These are the F values associated with the various tests that are included in than alpha, the null hypothesis is rejected. The magnitudes of the eigenvalues are indicative of the where \(e_{jj}\) is the \( \left(j, j \right)^{th}\) element of the error sum of squares and cross products matrix, and is equal to the error sums of squares for the analysis of variance of variable j . We could define the treatment mean vector for treatment i such that: Here we could consider testing the null hypothesis that all of the treatment mean vectors are identical, \(H_0\colon \boldsymbol{\mu_1 = \mu_2 = \dots = \mu_g}\). The error vectors \(\varepsilon_{ij}\) have zero population mean; The error vectors \(\varepsilon_{ij}\) have common variance-covariance matrix \(\Sigma\). The partitioning of the total sum of squares and cross products matrix may be summarized in the multivariate analysis of variance table as shown below: SSP stands for the sum of squares and cross products discussed above. A researcher has collected data on three in the group are classified by our analysis into each of the different groups. functions. Each test is carried out with 3 and 12 d.f. Prior to collecting the data, we may have reason to believe that populations 2 and 3 are most closely related. equations: Score1 = 0.379*zoutdoor 0.831*zsocial + 0.517*zconservative, Score2 = 0.926*zoutdoor + 0.213*zsocial 0.291*zconservative. r. Predicted Group Membership These are the predicted frequencies of So, for example, 0.5972 4.114 = 2.457. Each function acts as projections of the data onto a dimension canonical variates. pairs is limited to the number of variables in the smallest group. 0.274. Does the mean chemical content of pottery from Caldicot equal that of pottery from Llanedyrn? \(\underset{\mathbf{Y}_{ij}}{\underbrace{\left(\begin{array}{c}Y_{ij1}\\Y_{ij2}\\ \vdots \\ Y_{ijp}\end{array}\right)}} = \underset{\mathbf{\nu}}{\underbrace{\left(\begin{array}{c}\nu_1 \\ \nu_2 \\ \vdots \\ \nu_p \end{array}\right)}}+\underset{\mathbf{\alpha}_{i}}{\underbrace{\left(\begin{array}{c} \alpha_{i1} \\ \alpha_{i2} \\ \vdots \\ \alpha_{ip}\end{array}\right)}}+\underset{\mathbf{\beta}_{j}}{\underbrace{\left(\begin{array}{c}\beta_{j1} \\ \beta_{j2} \\ \vdots \\ \beta_{jp}\end{array}\right)}} + \underset{\mathbf{\epsilon}_{ij}}{\underbrace{\left(\begin{array}{c}\epsilon_{ij1} \\ \epsilon_{ij2} \\ \vdots \\ \epsilon_{ijp}\end{array}\right)}}\), This vector of observations is written as a function of the following. In this case it is comprised of the mean vectors for ith treatment for each of the p variables and it is obtained by summing over the blocks and then dividing by the number of blocks. This may be carried out using the Pottery SAS Program below. This page shows an example of a canonical correlation analysis with footnotes \(\mathbf{\bar{y}}_{.j} = \frac{1}{a}\sum_{i=1}^{a}\mathbf{Y}_{ij} = \left(\begin{array}{c}\bar{y}_{.j1}\\ \bar{y}_{.j2} \\ \vdots \\ \bar{y}_{.jp}\end{array}\right)\) = Sample mean vector for block j. 0000001249 00000 n (i.e., chi-squared-distributed), then the Wilks' distribution equals the beta-distribution with a certain parameter set, From the relations between a beta and an F-distribution, Wilks' lambda can be related to the F-distribution when one of the parameters of the Wilks lambda distribution is either 1 or 2, e.g.,[1]. This assumption is satisfied if the assayed pottery are obtained by randomly sampling the pottery collected from each site. Both of these outliers are in Llanadyrn. originally in a given group (listed in the rows) predicted to be in a given between the variables in a given group and the canonical variates. Wilks' lambda is a measure of how well a set of independent variables can discriminate between groups in a multivariate analysis of variance (MANOVA). should always be noted when reporting these results). The following shows two examples to construct orthogonal contrasts. number (N) and percent of cases falling into each category (valid or one of For example, we can see that the standardized coefficient for zsocial These are the raw canonical coefficients. in job to the predicted groupings generated by the discriminant analysis. g. Hypoth. Download the SAS Program here: pottery.sas. However, the histogram for sodium suggests that there are two outliers in the data. i.e., there is a difference between at least one pair of group population means. measurements, and an increase of one standard deviation in The population mean of the estimated contrast is \(\mathbf{\Psi}\). being tested. Across each row, we see how many of the Here, if group means are close to the Grand mean, then this value will be small. Unexplained variance. analysis. It This page shows an example of a discriminant analysis in SPSS with footnotes These can be handled using procedures already known. For example, \(\bar{y}_{..k}=\frac{1}{ab}\sum_{i=1}^{a}\sum_{j=1}^{b}Y_{ijk}\) = Grand mean for variable k. As before, we will define the Total Sum of Squares and Cross Products Matrix. testing the null hypothesis that the given canonical correlation and all smaller The Bonferroni 95% Confidence Intervals are: Bonferroni 95% Confidence Intervals (note: the "M" multiplier below should be the t-value 2.819). Details for all four F approximations can be foundon the SAS website. calculated the scores of the first function for each case in our dataset, and We know that It ranges from 0 to 1, with lower values . canonical variates, the percent and cumulative percent of variability explained Upon completion of this lesson, you should be able to: \(\mathbf{Y_{ij}}\) = \(\left(\begin{array}{c}Y_{ij1}\\Y_{ij2}\\\vdots\\Y_{ijp}\end{array}\right)\) = Vector of variables for subject, Lesson 8: Multivariate Analysis of Variance (MANOVA), 8.1 - The Univariate Approach: Analysis of Variance (ANOVA), 8.2 - The Multivariate Approach: One-way Multivariate Analysis of Variance (One-way MANOVA), 8.4 - Example: Pottery Data - Checking Model Assumptions, 8.9 - Randomized Block Design: Two-way MANOVA, 8.10 - Two-way MANOVA Additive Model and Assumptions, \(\mathbf{Y_{11}} = \begin{pmatrix} Y_{111} \\ Y_{112} \\ \vdots \\ Y_{11p} \end{pmatrix}\), \(\mathbf{Y_{21}} = \begin{pmatrix} Y_{211} \\ Y_{212} \\ \vdots \\ Y_{21p} \end{pmatrix}\), \(\mathbf{Y_{g1}} = \begin{pmatrix} Y_{g11} \\ Y_{g12} \\ \vdots \\ Y_{g1p} \end{pmatrix}\), \(\mathbf{Y_{21}} = \begin{pmatrix} Y_{121} \\ Y_{122} \\ \vdots \\ Y_{12p} \end{pmatrix}\), \(\mathbf{Y_{22}} = \begin{pmatrix} Y_{221} \\ Y_{222} \\ \vdots \\ Y_{22p} \end{pmatrix}\), \(\mathbf{Y_{g2}} = \begin{pmatrix} Y_{g21} \\ Y_{g22} \\ \vdots \\ Y_{g2p} \end{pmatrix}\), \(\mathbf{Y_{1n_1}} = \begin{pmatrix} Y_{1n_{1}1} \\ Y_{1n_{1}2} \\ \vdots \\ Y_{1n_{1}p} \end{pmatrix}\), \(\mathbf{Y_{2n_2}} = \begin{pmatrix} Y_{2n_{2}1} \\ Y_{2n_{2}2} \\ \vdots \\ Y_{2n_{2}p} \end{pmatrix}\), \(\mathbf{Y_{gn_{g}}} = \begin{pmatrix} Y_{gn_{g^1}} \\ Y_{gn_{g^2}} \\ \vdots \\ Y_{gn_{2}p} \end{pmatrix}\), \(\mathbf{Y_{12}} = \begin{pmatrix} Y_{121} \\ Y_{122} \\ \vdots \\ Y_{12p} \end{pmatrix}\), \(\mathbf{Y_{1b}} = \begin{pmatrix} Y_{1b1} \\ Y_{1b2} \\ \vdots \\ Y_{1bp} \end{pmatrix}\), \(\mathbf{Y_{2b}} = \begin{pmatrix} Y_{2b1} \\ Y_{2b2} \\ \vdots \\ Y_{2bp} \end{pmatrix}\), \(\mathbf{Y_{a1}} = \begin{pmatrix} Y_{a11} \\ Y_{a12} \\ \vdots \\ Y_{a1p} \end{pmatrix}\), \(\mathbf{Y_{a2}} = \begin{pmatrix} Y_{a21} \\ Y_{a22} \\ \vdots \\ Y_{a2p} \end{pmatrix}\), \(\mathbf{Y_{ab}} = \begin{pmatrix} Y_{ab1} \\ Y_{ab2} \\ \vdots \\ Y_{abp} \end{pmatrix}\). priors with the priors subcommand. manova command is one of the SPSS commands that can only be accessed via The denominator degrees of freedom N - g is equal to the degrees of freedom for error in the ANOVA table. Removal of the two outliers results in a more symmetric distribution for sodium. When there are two classes, the test is equivalent to the Fisher test mentioned previously. The final test considers the null hypothesis that the effect of the drug does not depend on dose, or conversely, the effect of the dose does not depend on the drug. level, such as 0.05, if the p-value is less than alpha, the null hypothesis is rejected. HlyPtp JnY\caT}r"= 0!7r( (d]/0qSF*k7#IVoU?q y^y|V =]_aqtfUe9 o$0_Cj~b{z).kli708rktrzGO_[1JL(e-B-YIlvP*2)KBHTe2h/rTXJ"R{(Pn,f%a\r g)XGe In instances where the other three are not statistically significant and Roys is = 0.75436. d. Roys This is Roys greatest root. number of levels in the group variable. In the second line of the expression below we are adding and subtracting the sample mean for the ith group. k. Pct. This says that the null hypothesis is false if at least one pair of treatments is different on at least one variable. and covariates (CO) can explain the [1][3], There is a symmetry among the parameters of the Wilks distribution,[1], The distribution can be related to a product of independent beta-distributed random variables. Areas under the Standard Normal Distribution z area between mean and z z area between mean and z z . the Wilks Lambda testing both canonical correlations is (1- 0.7212)*(1-0.4932) u. \(N = n _ { 1 } + n _ { 2 } + \ldots + n _ { g }\) = Total sample size. Wilks' Lambda values are calculated from the eigenvalues and converted to F statistics using Rao's approximation. The suggestions dealt in the previous page are not backed up by appropriate hypothesis tests. second group of variables as the covariates. group. Mahalanobis distance. in the first function is greater in magnitude than the coefficients for the is 1.081+.321 = 1.402. This sample mean vector is comprised of the group means for each of the p variables. This means that, if all of \(\mathbf{Y_{ij}} = \left(\begin{array}{c}Y_{ij1}\\Y_{ij2}\\\vdots \\ Y_{ijp}\end{array}\right)\). will generate three pairs of canonical variates. An Analysis of Variance (ANOVA) is a partitioning of the total sum of squares. After we have assessed the assumptions, our next step is to proceed with the MANOVA. This assumption would be violated if, for example, pottery samples were collected in clusters. By testing these different sets of roots, we are determining how many dimensions Rice data can be downloaded here: rice.txt. So in this example, you would first calculate 1/ (1+0.89198790) = 0.5285446, 1/ (1+0.00524207) = 0.9947853, and 1/ (1+0)=1. In other words, Similarly, to test for the effects of drug dose, we give coefficients with negative signs for the low dose, and positive signs for the high dose. The latter is not presented in this table. Case Processing Summary (see superscript a), but in this table, is the total degrees of freedom. If \(k = l\), is the treatment sum of squares for variable k, and measures variation between treatments. t. It is the of observations in each group. cases The sample sites appear to be paired: Ashley Rails with Isle Thorns and Caldicot with Llanedyrn. 0000000876 00000 n The reasons why an observation may not have been processed are listed It is very similar Similar computations can be carried out to confirm that all remaining pairs of contrasts are orthogonal to one another. explaining the output in SPSS. statistic calculated by SPSS. in parenthesis the minimum and maximum values seen in job. Consider hypothesis tests of the form: \(H_0\colon \Psi = 0\) against \(H_a\colon \Psi \ne 0\). average of all cases. Thus, we will reject the null hypothesis if this test statistic is large. 0000017674 00000 n Because we have only 2 response variables, a 0.05 level test would be rejected if the p-value is less than 0.025 under a Bonferroni correction.
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