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This example shows how to perform longitudinal
analysis using `mvregress`

.

Navigate to the folder containing sample data. Load the sample longitudinal data.

cd(matlabroot) cd('help/toolbox/stats/examples') load('longitudinalData')

The matrix `Y`

contains response data for 16
individuals. The response is the blood level of a drug measured at
five time points (*t* = 0, 2, 4, 6, and 8). Each
row of `Y`

corresponds to an individual, and each
column corresponds to a time point. The first eight subjects are female,
and the second eight subjects are male. This is simulated data.

Plot the data for all 16 subjects.

figure() t = [0,2,4,6,8]; plot(t,Y) hold on hf = plot(t,Y(1:8,:),'^'); hm = plot(t,Y(9:16,:),'o'); legend([hf(1),hm(1)],'Female','Male','Location','NorthEast') title('Longitudinal Response') ylabel('Blood Drug Level') xlabel('Time') hold off

Let *y _{ij}* denote the
response for individual

Consider fitting a quadratic longitudinal model, with a separate slope and intercept for each gender,

$${y}_{ij}={\beta}_{0}+{\beta}_{1}{G}_{i}+{\beta}_{2}{t}_{ij}+{\beta}_{3}{t}_{ij}^{2}+{\beta}_{4}{G}_{i}\times {t}_{ij}+{\beta}_{5}{G}_{i}\times {t}_{ij}^{2}+{\epsilon}_{ij},$$

To fit this model using `mvregress`

, the response
data should be in an *n*-by-*d* matrix. `Y`

is
already in the proper format.

Next, create a length-*n* cell array of *d*-by-*K* design
matrices. For this model, there are *K* = 6 parameters.

For individual *i*, the 5-by-6 design matrix
is

$$X\left\{i\right\}=\left(\begin{array}{cccccc}1& {G}_{i}& {t}_{i1}& {t}_{i1}^{2}& {G}_{i}\times {t}_{i1}& {G}_{i}\times {t}_{i1}^{2}\\ 1& {G}_{i}& {t}_{i2}& {t}_{i2}^{2}& {G}_{i}\times {t}_{i2}& {G}_{i}\times {t}_{i2}^{2}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 1& {G}_{i}& {t}_{i5}& {t}_{i5}^{2}& {G}_{i}\times {t}_{i5}& {G}_{i}\times {t}_{i5}^{2}\end{array}\right),$$

$$\beta =\left(\begin{array}{l}{\beta}_{0}\\ {\beta}_{1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\vdots \\ {\beta}_{5}\end{array}\right).$$

`X1`

has the design
matrix for a female, and `X2`

has the design matrix
for a male.Create a cell array of design matrices. The first eight individuals are females, and the second eight are males.

X = cell(8,1); X(1:8) = {X1}; X(9:16) = {X2};

Fit the model using maximum likelihood estimation. Display the estimated coefficients and standard errors.

[b,sig,E,V,loglikF] = mvregress(X,Y); [b sqrt(diag(V))]

ans = 18.8619 0.7432 13.0942 1.0511 2.5968 0.2845 -0.3771 0.0398 -0.5929 0.4023 0.0290 0.0563

The coefficients on the interaction terms (in the last two rows
of `b`

) do not appear significant. You can use the
value of the loglikelihood objective function for this fit, `loglikF`

,
to compare this model to one without the interaction terms using a
likelihood ratio test.

Plot the fitted lines for females and males.

Yhatf = X1*b; Yhatm = X2*b; figure() plot(t,Y) hold on plot(t,Y(1:8,:),'^',t,Y(9:16,:),'o') hf = plot(t,Yhatf,'k--','LineWidth',3); hm = plot(t,Yhatm,'k','LineWidth',3); legend([hf,hm],'Females','Males','Location','NorthEast') title('Longitudinal Response') ylabel('Blood Drug Level') xlabel('Time') hold off

Fit the model without interaction terms,

$${y}_{ij}={\beta}_{0}+{\beta}_{1}{G}_{i}+{\beta}_{2}{t}_{ij}+{\beta}_{3}{t}_{ij}^{2}+{\epsilon}_{ij},$$

This model has four coefficients, which correspond to the first
four columns of the design matrices `X1`

and `X2`

(for
females and males, respectively).

X1R = X1(:,1:4); X2R = X2(:,1:4); XR = cell(8,1); XR(1:8) = {X1R}; XR(9:16) = {X2R};

Fit this model using maximum likelihood estimation. Display the estimated coefficients and their standard errors.

[bR,sigR,ER,VR,loglikR] = mvregress(XR,Y); [bR,sqrt(diag(VR))]

ans = 19.3765 0.6898 12.0936 0.8591 2.2919 0.2139 -0.3623 0.0283

Compare the two models using a likelihood ratio test. The null hypothesis is that the reduced model is sufficient. The alternative is that the reduced model is inadequate (compared to the full model with the interaction terms).

The likelihood ratio test statistic is compared to a chi-squared distribution with two degrees of freedom (for the two coefficients being dropped).

LR = 2*(loglikF-loglikR); pval = 1 - chi2cdf(LR,2)

pval = 0.0803

`0.0803`

indicates
that the null hypothesis is not rejected at the 5% significance level.
Therefore, there is insufficient evidence that the extra terms improve
the fit.- Set Up Multivariate Regression Problems
- Multivariate General Linear Model
- Fixed Effects Panel Model with Concurrent Correlation

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