📄 mat_manova_example.hlp
字号:
{smcl}
{* 10mar2005}{...}
{* this hlp file called by testmanova.dlg}{...}
{title:Examples of using matrices to test linear expressions after help manova}
After
{cmd:. manova y1 y2 = a}
{pstd}
where {cmd:a} has four levels, you test if the coefficients for the first and
second levels of {cmd:a} are equal for dependent variable {cmd:y2} by defining
a matrix using the command
{cmd:. matrix c = (0,0,0,0,0,0,1,-1,0,0)}
{pstd}
and then supplying the matrix name, {cmd:c} in this example, as an argument to
the {cmd:test()} option of {cmd:test}.
{pstd}
Notice that the matrix has ten columns. The first five are related to
{cmd:y1} and the last five are related to {cmd:y2}. Each has five instead of
four columns because a constant is automatically included in the model for
each dependent variable. If in doubt of the number or order of columns, use
the {bind:{cmd:test , showorder}} command, obtainable by selecting
"{cmd:Show order of columns in the design matrix}" in the dialog box, to
examine the order and definitions of the columns.
The matrix
{cmd:. matrix x = (0,3,-1,-1,-1,0,3,-1,-1,-1)}
{pstd}
could be used to test if {cmd:a} at level 1 is equal to the average of {cmd:a}
at levels 2, 3, and 4 for {cmd:y1} and {cmd:y2} combined into one single
degree-of-freedom test.
The matrix
{phang2}
{cmd:. matrix z = (0,3,-1,-1,-1,0,0,0,0,0 \ 0,0,0,0,0,0,3,-1,-1,-1)}
{pstd}
could be used to test if {cmd:a} at level 1 is equal to the average of {cmd:a}
at levels 2, 3, and 4 for {cmd:y1} jointly with {cmd:a} at level 1 equal to
the average of {cmd:a} at levels 2, 3, and 4 for {cmd:y2} -- a two
degrees-of-freedom test.
{title:Also see}
{psee}
Online: {helpb manova}, {help manova postestimation}
{p_end}
⌨️ 快捷键说明
复制代码
Ctrl + C
搜索代码
Ctrl + F
全屏模式
F11
切换主题
Ctrl + Shift + D
显示快捷键
?
增大字号
Ctrl + =
减小字号
Ctrl + -