Source: http://www.ats.ucla.edu/stat/mult_pkg/whatstat/nominal_ordinal_interval.htm
What is the difference between categorical, ordinal and interval variables?
In talking about variables, sometimes you hear variables being described as categorical (or sometimes nominal), or ordinal, or interval. Below we will define these terms and explain why they are important.
Categorical
A categorical variable (sometimes called a nominal variable) is one that has two or more categories, but there is no intrinsic ordering to the categories. For example, gender is a categorical variable having two categories (male and female) and there is no intrinsic ordering to the categories. Hair color is also a categorical variable having a number of categories (blonde, brown, brunette, red, etc.) and again, there is no agreed way to order these from highest to lowest. A purely categorical variable is one that simply allows you to assign categories but you cannot clearly order the variables. If the variable has a clear ordering, then that variable would be an ordinal variable, as described below.
Ordinal
An ordinal variable is similar to a categorical variable. The difference between the two is that there is a clear ordering of the variables. For example, suppose you have a variable, economic status, with three categories (low, medium and high). In addition to being able to classify people into these three categories, you can order the categories as low, medium and high. Now consider a variable like educational experience (with values such as elementary school graduate, high school graduate, some college and college graduate). These also can be ordered as elementary school, high school, some college, and college graduate. Even though we can order these from lowest to highest, the spacing between the values may not be the same across the levels of the variables. Say we assign scores 1, 2, 3 and 4 to these four levels of educational experience and we compare the difference in education between categories one and two with the difference in educational experience between categories two and three, or the difference between categories three and four. The difference between categories one and two (elementary and high school) is probably much bigger than the difference between categories two and three (high school and some college). In this example, we can order the people in level of educational experience but the size of the difference between categories is inconsistent (because the spacing between categories one and two is bigger than categories two and three). If these categories were equally spaced, then the variable would be an interval variable.
Interval
An interval variable is similar to an ordinal variable, except that the intervals between the values of the interval variable are equally spaced. For example, suppose you have a variable such as annual income that is measured in dollars, and we have three people who make $10,000, $15,000 and $20,000. The second person makes $5,000 more than the first person and $5,000 less than the third person, and the size of these intervals is the same. If there were two other people who make $90,000 and $95,000, the size of that interval between these two people is also the same ($5,000).
Why does it matter whether a variable is categorical, ordinal or interval?
Statistical computations and analyses assume that the variables have a specific levels of measurement. For example, it would not make sense to compute an average hair color. An average of a categorical variable does not make much sense because there is no intrinsic ordering of the levels of the categories. Moreover, if you tried to compute the average of educational experience as defined in the ordinal section above, you would also obtain a nonsensical result. Because the spacing between the four levels of educational experience is very uneven, the meaning of this average would be very questionable. In short, an average requires a variable to be interval. Sometimes you have variables that are “in between” ordinal and interval, for example, a fivepoint likert scale with values “strongly agree”, “agree”, “neutral”, “disagree” and “strongly disagree”. If we cannot be sure that the intervals between each of these five values are the same, then we would not be able to say that this is an interval variable, but we would say that it is an ordinal variable. However, in order to be able to use statistics that assume the variable is interval, we will assume that the intervals are equally spaced.
Does it matter if my dependent variable is normally distributed?
When you are doing a ttest or ANOVA, the assumption is that the distribution of the sample means are normally distributed. One way to guarantee this is for the distribution of the individual observations from the sample to be normal. However, even if the distribution of the individual observations is not normal, the distribution of the sample means will be normally distributed if your sample size is about 30 or larger. This is due to the “central limit theorem” that shows that even when a population is nonnormally distributed, the distribution of the “sample means” will be normally distributed when the sample size is 30 or more, for example seeCentral limit theorem demonstration .
If you are doing a regression analysis, then the assumption is that your residuals are normally distributed. One way to make it very likely to have normal residuals is to have a dependent variable that is normally distributed and predictors that are all normally distributed, however this is not necessary for your residuals to be normally distributed. You can see Regression with SAS: Chapter 2 – Regression Diagnostics, Regression with SAS: Chapter 2 – Regression Diagnostics, or Regression with SAS: Chapter 2 – Regression Diagnostics
Means difference
Chisquare (Categorical data)
[youtube]http://www.youtube.com/watch?v=DBsMPZqJjo[/youtube]
ttest independent samples
[youtube]http://www.youtube.com/watch?v=by4c3h3WXQc[/youtube]
Analysis questionnaire (likert) – non parametric test using Chisquare test
[youtube]http://www.youtube.com/watch?v=B3jjm_AyMI[/youtube]
Number of Dependent Variables 
Nature of Independent Variables 
Test(s) 

1 
0 IVs (1 population) 
interval & normal 
onesample ttest 
ordinal or interval 
onesample median 

categorical (2 categories) 
binomial test 

categorical 
Chisquare goodnessoffit 

1 IV with 2 levels (independent groups) 
interval & normal 
2 independent sample ttest 

ordinal or interval 

WilcoxonMann Whitney test 

categorical 
Chi square test 

Fisher’s exact test 

1 IV with 2 or more levels (independent groups) 
interval & normal 
oneway ANOVA 

ordinal or interval 
Kruskal Wallis 

categorical 
Chi square test 

1 IV with 2 levels (dependent/matched groups) 
interval & normal 
paired ttest 

ordinal or interval 
Wilcoxon signed ranks test 

categorical 
McNemar 

1 IV with 2 or more levels (dependent/matched groups) 
interval & normal 
oneway repeated measures ANOVA 

ordinal or interval 
Friedman test 

categorical 
repeated measures logistic regression 

2 or more IVs (independent groups) 
interval & normal 
factorial ANOVA 

ordinal or interval 
ordered logistic regression 

categorical 
factorial logistic regression 

1 interval IV 
interval & normal 
correlation 

simple linear regression 

ordinal or interval 
nonparametric correlation 

categorical 
simple logistic regression 

1 or more interval IVs and/or 1 or more categorical IVs 
interval & normal 
multiple regression 

analysis of covariance 

categorical 
multiple logistic regression 

discriminant analysis 

2 or more 
1 IV with 2 or more levels (independent groups) 
interval & normal 
oneway MANOVA 
2 or more 
2 or more 
interval & normal 
multivariate multiple linear regression 
2 sets of 2 or more 
0 
interval & normal 
canonical correlation 
2 or more 
0 
interval & normal 
factor analysis 
Number of Dependent Variables 
Nature of Independent Variables 
Test(s) 
Source: http://www.ats.ucla.edu/stat/stata/whatstat/whatstat.htm#wilc
What statistical analysis should I use?
Statistical analyses using Stata
Version info: Code for this page was tested in Stata 12.
Introduction
This page shows how to perform a number of statistical tests using Stata. Each section gives a brief description of the aim of the statistical test, when it is used, an example showing the Stata commands and Stata output with a brief interpretation of the output. You can see the page Choosing the Correct Statistical Test for a table that shows an overview of when each test is appropriate to use. In deciding which test is appropriate to use, it is important to consider the type of variables that you have (i.e., whether your variables are categorical, ordinal or interval and whether they are normally distributed), see What is the difference between categorical, ordinal and interval variables? for more information on this.
About the hsb data file
Most of the examples in this page will use a data file called hsb2, high school and beyond. This data file contains 200 observations from a sample of high school students with demographic information about the students, such as their gender (female), socioeconomic status (ses) and ethnic background (race). It also contains a number of scores on standardized tests, including tests of reading (read), writing (write), mathematics (math) and social studies (socst). You can get the hsb2data file from within Stata by typing:
use http://www.ats.ucla.edu/stat/stata/notes/hsb2
One sample ttest
A one sample ttest allows us to test whether a sample mean (of a normally distributed interval variable) significantly differs from a hypothesized value. For example, using the hsb2 data file, say we wish to test whether the average writing score (write) differs significantly from 50. We can do this as shown below.
ttest write=50Onesample t test  Variable  Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] + write  200 52.775 .6702372 9.478586 51.45332 54.09668  Degrees of freedom: 199 Ho: mean(write) = 50 Ha: mean < 50 Ha: mean ~= 50 Ha: mean > 50 t = 4.1403 t = 4.1403 t = 4.1403 P < t = 1.0000 P > t = 0.0001 P > t = 0.0000The mean of the variable write for this particular sample of students is 52.775, which is statistically significantly different from the test value of 50. We would conclude that this group of students has a significantly higher mean on the writing test than 50.
See also
One sample median test
A one sample median test allows us to test whether a sample median differs significantly from a hypothesized value. We will use the same variable, write, as we did in the one sample ttest example above, but we do not need to assume that it is interval and normally distributed (we only need to assume that write is an ordinal variable and that its distribution is symmetric). We will test whether the median writing score (write) differs significantly from 50.
signrank write=50Wilcoxon signedrank test sign  obs sum ranks expected + positive  126 13429 10048.5 negative  72 6668 10048.5 zero  2 3 3 + all  200 20100 20100 unadjusted variance 671675.00 adjustment for ties 1760.25 adjustment for zeros 1.25  adjusted variance 669913.50 Ho: write = 50 z = 4.130 Prob > z = 0.0000The results indicate that the median of the variable write for this group is statistically significantly different from 50.
See also
Binomial test
A one sample binomial test allows us to test whether the proportion of successes on a twolevel categorical dependent variable significantly differs from a hypothesized value. For example, using the hsb2 data file, say we wish to test whether the proportion of females (female) differs significantly from 50%, i.e., from .5. We can do this as shown below.
bitest female=.5Variable  N Observed k Expected k Assumed p Observed p + female  200 109 100 0.50000 0.54500 Pr(k >= 109) = 0.114623 (onesided test) Pr(k <= 109) = 0.910518 (onesided test) Pr(k <= 91 or k >= 109) = 0.229247 (twosided test)The results indicate that there is no statistically significant difference (p = .2292). In other words, the proportion of females does not significantly differ from the hypothesized value of 50%.
See also
Chisquare goodness of fit
A chisquare goodness of fit test allows us to test whether the observed proportions for a categorical variable differ from hypothesized proportions. For example, let’s suppose that we believe that the general population consists of 10% Hispanic, 10% Asian, 10% African American and 70% White folks. We want to test whether the observed proportions from our sample differ significantly from these hypothesized proportions. To conduct the chisquare goodness of fit test, you need to first download the csgof program that performs this test. You can download csgof from within Stata by typing findit csgof (seeHow can I used the findit command to search for programs and get additional help? for more information about using findit).
Now that the csgof program is installed, we can use it by typing:
csgof race, expperc(10 10 10 70) race expperc expfreq obsfreq hispanic 10 20 24 asian 10 20 11 africanamer 10 20 20 white 70 140 145 chisq(3) is 5.03, p = .1697These results show that racial composition in our sample does not differ significantly from the hypothesized values that we supplied (chisquare with three degrees of freedom = 5.03, p = .1697).
See also
Two independent samples ttest
An independent samples ttest is used when you want to compare the means of a normally distributed interval dependent variable for two independent groups. For example, using the hsb2 data file, say we wish to test whether the mean for write is the same for males and females.
ttest write, by(female) Twosample t test with equal variances  Group  Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] + male  91 50.12088 1.080274 10.30516 47.97473 52.26703 female  109 54.99083 .7790686 8.133715 53.44658 56.53507 + combined  200 52.775 .6702372 9.478586 51.45332 54.09668 + diff  4.869947 1.304191 7.441835 2.298059  Degrees of freedom: 198 Ho: mean(male)  mean(female) = diff = 0 Ha: diff < 0 Ha: diff ~= 0 Ha: diff > 0 t = 3.7341 t = 3.7341 t = 3.7341 P < t = 0.0001 P > t = 0.0002 P > t = 0.9999The results indicate that there is a statistically significant difference between the mean writing score for males and females (t = 3.7341, p = .0002). In other words, females have a statistically significantly higher mean score on writing (54.99) than males (50.12).
See also
WilcoxonMannWhitney test
The WilcoxonMannWhitney test is a nonparametric analog to the independent samples ttest and can be used when you do not assume that the dependent variable is a normally distributed interval variable (you only assume that the variable is at least ordinal). You will notice that the Stata syntax for the WilcoxonMannWhitney test is almost identical to that of the independent samples ttest. We will use the same data file (the hsb2 data file) and the same variables in this example as we did in theindependent ttest example above and will not assume that write, our dependent variable, is normally distributed.
ranksum write, by(female)Twosample Wilcoxon ranksum (MannWhitney) test female  obs rank sum expected + male  91 7792 9145.5 female  109 12308 10954.5 + combined  200 20100 20100 unadjusted variance 166143.25 adjustment for ties 852.96  adjusted variance 165290.29 Ho: write(female==male) = write(female==female) z = 3.329 Prob > z = 0.0009The results suggest that there is a statistically significant difference between the underlying distributions of the write scores of males and the write scores of females (z = 3.329, p = 0.0009). You can determine which group has the higher rank by looking at the how the actual rank sums compare to the expected rank sums under the null hypothesis. The sum of the female ranks was higher while the sum of the male ranks was lower. Thus the female group had higher rank.
See also
Chisquare test
A chisquare test is used when you want to see if there is a relationship between two categorical variables. In Stata, the chi2option is used with the tabulate command to obtain the test statistic and its associated pvalue. Using the hsb2 data file, let’s see if there is a relationship between the type of school attended (schtyp) and students’ gender (female). Remember that the chisquare test assumes the expected value of each cell is five or higher. This assumption is easily met in the examples below. However, if this assumption is not met in your data, please see the section on Fisher’s exact test below.
tabulate schtyp female, chi2 type of  female school  male female  Total ++ public  77 91  168 private  14 18  32 ++ Total  91 109  200 Pearson chi2(1) = 0.0470 Pr = 0.828These results indicate that there is no statistically significant relationship between the type of school attended and gender (chisquare with one degree of freedom = 0.0470, p = 0.828).
Let’s look at another example, this time looking at the relationship between gender (female) and socioeconomic status (ses). The point of this example is that one (or both) variables may have more than two levels, and that the variables do not have to have the same number of levels. In this example, female has two levels (male and female) and ses has three levels (low, medium and high).
tabulate female ses, chi2  ses female  low middle high  Total ++ male  15 47 29  91 female  32 48 29  109 ++ Total  47 95 58  200 Pearson chi2(2) = 4.5765 Pr = 0.101Again we find that there is no statistically significant relationship between the variables (chisquare with two degrees of freedom = 4.5765, p = 0.101).
See also
Fisher’s exact test
The Fisher’s exact test is used when you want to conduct a chisquare test, but one or more of your cells has an expected frequency of five or less. Remember that the chisquare test assumes that each cell has an expected frequency of five or more, but the Fisher’s exact test has no such assumption and can be used regardless of how small the expected frequency is. In the example below, we have cells with observed frequencies of two and one, which may indicate expected frequencies that could be below five, so we will use Fisher’s exact test with the exact option on the tabulate command.
tabulate schtyp race, exact type of  race school  hispanic asian africana white  Total ++ public  22 10 18 118  168 private  2 1 2 27  32 ++ Total  24 11 20 145  200 Fisher's exact = 0.597These results suggest that there is not a statistically significant relationship between race and type of school (p = 0.597). Note that the Fisher’s exact test does not have a “test statistic”, but computes the pvalue directly.
See also
Oneway ANOVA
A oneway analysis of variance (ANOVA) is used when you have a categorical independent variable (with two or more categories) and a normally distributed interval dependent variable and you wish to test for differences in the means of the dependent variable broken down by the levels of the independent variable. For example, using the hsb2 data file, say we wish to test whether the mean of write differs between the three program types (prog). The command for this test would be:
anova write prog Number of obs = 200 Rsquared = 0.1776 Root MSE = 8.63918 Adj Rsquared = 0.1693 Source  Partial SS df MS F Prob > F + Model  3175.69786 2 1587.84893 21.27 0.0000  prog  3175.69786 2 1587.84893 21.27 0.0000  Residual  14703.1771 197 74.635417 + Total  17878.875 199 89.843593The mean of the dependent variable differs significantly among the levels of program type. However, we do not know if the difference is between only two of the levels or all three of the levels. (The F test for the Model is the same as the F test forprog because prog was the only variable entered into the model. If other variables had also been entered, the F test for theModel would have been different from prog.) To see the mean of write for each level of program type, you can use thetabulate command with the summarize option, as illustrated below.
tabulate prog, summarize(write) type of  Summary of writing score program  Mean Std. Dev. Freq. + general  51.333333 9.3977754 45 academic  56.257143 7.9433433 105 vocation  46.76 9.3187544 50 + Total  52.775 9.478586 200From this we can see that the students in the academic program have the highest mean writing score, while students in the vocational program have the lowest.
See also
Kruskal Wallis test
The Kruskal Wallis test is used when you have one independent variable with two or more levels and an ordinal dependent variable. In other words, it is the nonparametric version of ANOVA and a generalized form of the MannWhitney test method since it permits 2 or more groups. We will use the same data file as the one way ANOVA example above (the hsb2 data file) and the same variables as in the example above, but we will not assume that write is a normally distributed interval variable.
kwallis write, by(prog)Test: Equality of populations (KruskalWallis test) prog _Obs _RankSum general < 45 4079.00 academic 105 12764.00 vocation 50 3257.00 chisquared = 33.870 with 2 d.f. probability = 0.0001 chisquared with ties = 34.045 with 2 d.f. probability = 0.0001If some of the scores receive tied ranks, then a correction factor is used, yielding a slightly different value of chisquared. With or without ties, the results indicate that there is a statistically significant difference among the three type of programs.
Paired ttest
A paired (samples) ttest is used when you have two related observations (i.e. two observations per subject) and you want to see if the means on these two normally distributed interval variables differ from one another. For example, using the hsb2 data file we will test whether the mean of read is equal to the mean of write.
ttest read = writePaired t test  Variable  Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] + read  200 52.23 .7249921 10.25294 50.80035 53.65965 write  200 52.775 .6702372 9.478586 51.45332 54.09668 + diff  200 .545 .6283822 8.886666 1.784142 .6941424  Ho: mean(read  write) = mean(diff) = 0 Ha: mean(diff) < 0 Ha: mean(diff) ~= 0 Ha: mean(diff) > 0 t = 0.8673 t = 0.8673 t = 0.8673 P < t = 0.1934 P > t = 0.3868 P > t = 0.8066These results indicate that the mean of read is not statistically significantly different from the mean of write (t = 0.8673, p = 0.3868).
See also
Wilcoxon signed rank sum test
The Wilcoxon signed rank sum test is the nonparametric version of a paired samples ttest. You use the Wilcoxon signed rank sum test when you do not wish to assume that the difference between the two variables is interval and normally distributed (but you do assume the difference is ordinal). We will use the same example as above, but we will not assume that the difference between read and write is interval and normally distributed.
signrank read = writeWilcoxon signedrank test sign  obs sum ranks expected + positive  88 9264 9990 negative  97 10716 9990 zero  15 120 120 + all  200 20100 20100 unadjusted variance 671675.00 adjustment for ties 715.25 adjustment for zeros 310.00  adjusted variance 670649.75 Ho: read = write z = 0.887 Prob > z = 0.3753The results suggest that there is not a statistically significant difference between read and write.
If you believe the differences between read and write were not ordinal but could merely be classified as positive and negative, then you may want to consider a sign test in lieu of sign rank test. Again, we will use the same variables in this example and assume that this difference is not ordinal.
signtest read = writeSign test sign  observed expected + positive  88 92.5 negative  97 92.5 zero  15 15 + all  200 200 Onesided tests: Ho: median of read  write = 0 vs. Ha: median of read  write > 0 Pr(#positive >= 88) = Binomial(n = 185, x >= 88, p = 0.5) = 0.7688 Ho: median of read  write = 0 vs. Ha: median of read  write < 0 Pr(#negative >= 97) = Binomial(n = 185, x >= 97, p = 0.5) = 0.2783 Twosided test: Ho: median of read  write = 0 vs. Ha: median of read  write ~= 0 Pr(#positive >= 97 or #negative >= 97) = min(1, 2*Binomial(n = 185, x >= 97, p = 0.5)) = 0.5565This output gives both of the onesided tests as well as the twosided test. Assuming that we were looking for any difference, we would use the twosided test and conclude that no statistically significant difference was found (p=.5565).
See also
McNemar test
You would perform McNemar’s test if you were interested in the marginal frequencies of two binary outcomes. These binary outcomes may be the same outcome variable on matched pairs (like a casecontrol study) or two outcome variables from a single group. For example, let us consider two questions, Q1 and Q2, from a test taken by 200 students. Suppose 172 students answered both questions correctly, 15 students answered both questions incorrectly, 7 answered Q1 correctly and Q2 incorrectly, and 6 answered Q2 correctly and Q1 incorrectly. These counts can be considered in a twoway contingency table. The null hypothesis is that the two questions are answered correctly or incorrectly at the same rate (or that the contingency table is symmetric). We can enter these counts into Stata using mcci, a command from Stata’s epidemiology tables. The outcome is labeled according to casecontrol study conventions.
mcci 172 6 7 15 Controls  Cases  Exposed Unexposed  Total ++ Exposed  172 6  178 Unexposed  7 15  22 ++ Total  179 21  200 McNemar's chi2(1) = 0.08 Prob > chi2 = 0.7815 Exact McNemar significance probability = 1.0000 Proportion with factor Cases .89 Controls .895 [95% Conf. Interval]   difference .005 .045327 .035327 ratio .9944134 .9558139 1.034572 rel. diff. .047619 .39205 .2968119 odds ratio .8571429 .2379799 2.978588 (exact)McNemar’s chisquare statistic suggests that there is not a statistically significant difference in the proportions of correct/incorrect answers to these two questions.
Oneway repeated measures ANOVA
You would perform a oneway repeated measures analysis of variance if you had one categorical independent variable and a normally distributed interval dependent variable that was repeated at least twice for each subject. This is the equivalent of the paired samples ttest, but allows for two or more levels of the categorical variable. This tests whether the mean of the dependent variable differs by the categorical variable. We have an example data set called rb4, which is used in Kirk’s book Experimental Design. In this data set, y is the dependent variable, a is the repeated measure and s is the variable that indicates the subject number.
use http://www.ats.ucla.edu/stat/stata/examples/kirk/rb4 anova y a s, repeated(a)Number of obs = 32 Rsquared = 0.7318 Root MSE = 1.18523 Adj Rsquared = 0.6041 Source  Partial SS df MS F Prob > F + Model  80.50 10 8.05 5.73 0.0004  a  49.00 3 16.3333333 11.63 0.0001 s  31.50 7 4.50 3.20 0.0180  Residual  29.50 21 1.4047619 + Total  110.00 31 3.5483871 Betweensubjects error term: s Levels: 8 (7 df) Lowest b.s.e. variable: s Repeated variable: a HuynhFeldt epsilon = 0.8343 GreenhouseGeisser epsilon = 0.6195 Box's conservative epsilon = 0.3333  Prob > F  Source  df F Regular HF GG Box + a  3 11.63 0.0001 0.0003 0.0015 0.0113 Residual  21 +You will notice that this output gives four different pvalues. The “regular” (0.0001) is the pvalue that you would get if you assumed compound symmetry in the variancecovariance matrix. Because that assumption is often not valid, the three other pvalues offer various corrections (the HuynhFeldt, HF, GreenhouseGeisser, GG and Box’s conservative, Box). No matter which pvalue you use, our results indicate that we have a statistically significant effect of a at the .05 level.
See also
Repeated measures logistic regression
If you have a binary outcome measured repeatedly for each subject and you wish to run a logistic regression that accounts for the effect of these multiple measures from each subjects, you can perform a repeated measures logistic regression. In Stata, this can be done using the xtgee command and indicating binomial as the probability distribution and logit as the link function to be used in the model. The exercise data file contains 3 pulse measurements of 30 people assigned to 2 different diet regiments and 3 different exercise regiments. If we define a “high” pulse as being over 100, we can then predict the probability of a high pulse using diet regiment.
First, we use xtset to define which variable defines the repetitions. In this dataset, there are three measurements taken for each id, so we will use id as our panel variable. Then we can use i: before diet so that we can create indicator variables as needed.
use http://www.ats.ucla.edu/stat/stata/whatstat/exercise, clear xtset id xtgee highpulse i.diet, family(binomial) link(logit)Iteration 1: tolerance = 1.753e08 GEE populationaveraged model Number of obs = 90 Group variable: id Number of groups = 30 Link: logit Obs per group: min = 3 Family: binomial avg = 3.0 Correlation: exchangeable max = 3 Wald chi2(1) = 1.53 Scale parameter: 1 Prob > chi2 = 0.2157  highpulse  Coef. Std. Err. z P>z [95% Conf. Interval] + 2.diet  .7537718 .6088196 1.24 0.216 .4394927 1.947036 _cons  1.252763 .4621704 2.71 0.007 2.1586 .3469257 These results indicate that diet is not statistically significant (Z = 1.24, p = 0.216).
Factorial ANOVA
A factorial ANOVA has two or more categorical independent variables (either with or without the interactions) and a single normally distributed interval dependent variable. For example, using the hsb2 data file we will look at writing scores (write) as the dependent variable and gender (female) and socioeconomic status (ses) as independent variables, and we will include an interaction of female by ses. Note that in Stata, you do not need to have the interaction term(s) in your data set. Rather, you can have Stata create it/them temporarily by placing an asterisk between the variables that will make up the interaction term(s).
anova write female ses female##ses Number of obs = 200 Rsquared = 0.1274 Root MSE = 8.96748 Adj Rsquared = 0.1049 Source  Partial SS df MS F Prob > F + Model  2278.24419 5 455.648837 5.67 0.0001  female  1334.49331 1 1334.49331 16.59 0.0001 ses  1063.2527 2 531.626349 6.61 0.0017 female#ses  21.4309044 2 10.7154522 0.13 0.8753  Residual  15600.6308 194 80.4156228 + Total  17878.875 199 89.843593These results indicate that the overall model is statistically significant (F = 5.67, p = 0.001). The variables female and ses are also statistically significant (F = 16.59, p = 0.0001 and F = 6.61, p = 0.0017, respectively). However, that interaction betweenfemale and ses is not statistically significant (F = 0.13, p = 0.8753).
See also
Friedman test
You perform a Friedman test when you have one withinsubjects independent variable with two or more levels and a dependent variable that is not interval and normally distributed (but at least ordinal). We will use this test to determine if there is a difference in the reading, writing and math scores. The null hypothesis in this test is that the distribution of the ranks of each type of score (i.e., reading, writing and math) are the same. To conduct the Friedman test in Stata, you need to first download the friedman program that performs this test. You can download friedman from within Stata by typing findit friedman (see How can I used the findit command to search for programs and get additional help? for more information about using findit). Also, your data will need to be transposed such that subjects are the columns and the variables are the rows. We will use the xpose command to arrange our data this way.
use http://www.ats.ucla.edu/stat/stata/notes/hsb2 keep read write math xpose, clear friedman v1v200Friedman = 0.6175 Kendall = 0.0015 Pvalue = 0.7344Friedman’s chisquare has a value of 0.6175 and a pvalue of 0.7344 and is not statistically significant. Hence, there is no evidence that the distributions of the three types of scores are different.
Ordered logistic regression
Ordered logistic regression is used when the dependent variable is ordered, but not continuous. For example, using the hsb2 data file we will create an ordered variable called write3. This variable will have the values 1, 2 and 3, indicating a low, medium or high writing score. We do not generally recommend categorizing a continuous variable in this way; we are simply creating a variable to use for this example. We will use gender (female), reading score (read) and social studies score (socst) as predictor variables in this model.
use http://www.ats.ucla.edu/stat/stata/notes/hsb2 generate write3 = 1 replace write3 = 2 if write >= 49 & write <= 57 replace write3 = 3 if write >= 58 & write <= 70ologit write3 female read socst Iteration 0: log likelihood = 218.31357 Iteration 1: log likelihood = 157.692 Iteration 2: log likelihood = 156.28133 Iteration 3: log likelihood = 156.27632 Iteration 4: log likelihood = 156.27632 Ordered logistic regression Number of obs = 200 LR chi2(3) = 124.07 Prob > chi2 = 0.0000 Log likelihood = 156.27632 Pseudo R2 = 0.2842  write3  Coef. Std. Err. z P>z [95% Conf. Interval] + female  1.285435 .3244567 3.96 0.000 .6495115 1.921359 read  .1177202 .0213565 5.51 0.000 .0758623 .1595781 socst  .0801873 .0194432 4.12 0.000 .0420794 .1182952 + /cut1  9.703706 1.197002 7.357626 12.04979 /cut2  11.8001 1.304306 9.243705 14.35649 The results indicate that the overall model is statistically significant (p < .0000), as are each of the predictor variables (p < .000). There are two cutpoints for this model because there are three levels of the outcome variable.
One of the assumptions underlying ordinal logistic (and ordinal probit) regression is that the relationship between each pair of outcome groups is the same. In other words, ordinal logistic regression assumes that the coefficients that describe the relationship between, say, the lowest versus all higher categories of the response variable are the same as those that describe the relationship between the next lowest category and all higher categories, etc. This is called the proportional odds assumption or the parallel regression assumption. Because the relationship between all pairs of groups is the same, there is only one set of coefficients (only one model). If this was not the case, we would need different models (such as a generalized ordered logit model) to describe the relationship between each pair of outcome groups. To test this assumption, we can use either the omodel command (findit omodel, see How can I used the findit command to search for programs and get additional help? for more information about using findit) or the brant command. We will show both below.
omodel logit write3 female read socst Iteration 0: log likelihood = 218.31357 Iteration 1: log likelihood = 158.87444 Iteration 2: log likelihood = 156.35529 Iteration 3: log likelihood = 156.27644 Iteration 4: log likelihood = 156.27632 Ordered logit estimates Number of obs = 200 LR chi2(3) = 124.07 Prob > chi2 = 0.0000 Log likelihood = 156.27632 Pseudo R2 = 0.2842  write3  Coef. Std. Err. z P>z [95% Conf. Interval] + female  1.285435 .3244565 3.96 0.000 .649512 1.921358 read  .1177202 .0213564 5.51 0.000 .0758623 .159578 socst  .0801873 .0194432 4.12 0.000 .0420794 .1182952 + _cut1  9.703706 1.197 (Ancillary parameters) _cut2  11.8001 1.304304  Approximate likelihoodratio test of proportionality of odds across response categories: chi2(3) = 2.03 Prob > chi2 = 0.5658 brant, detail Estimated coefficients from j1 binary regressions y>1 y>2 female 1.5673604 1.0629714 read .11712422 .13401723 socst .0842684 .06429241 _cons 10.001584 11.671854 Brant Test of Parallel Regression Assumption Variable  chi2 p>chi2 df + All  2.07 0.558 3 + female  1.08 0.300 1 read  0.26 0.608 1 socst  0.52 0.470 1  A significant test statistic provides evidence that the parallel regression assumption has been violated.Both of these tests indicate that the proportional odds assumption has not been violated.
See also
Factorial logistic regression
A factorial logistic regression is used when you have two or more categorical independent variables but a dichotomous dependent variable. For example, using the hsb2 data file we will use female as our dependent variable, because it is the only dichotomous (0/1) variable in our data set; certainly not because it common practice to use gender as an outcome variable. We will use type of program (prog) and school type (schtyp) as our predictor variables. Because prog is a categorical variable (it has three levels), we need to create dummy codes for it. The use of i.prog does this. You can use the logitcommand if you want to see the regression coefficients or the logistic command if you want to see the odds ratios.
logit female i.prog##schtyp Iteration 0: log likelihood = 137.81834 Iteration 1: log likelihood = 136.25886 Iteration 2: log likelihood = 136.24502 Iteration 3: log likelihood = 136.24501 Logistic regression Number of obs = 200 LR chi2(5) = 3.15 Prob > chi2 = 0.6774 Log likelihood = 136.24501 Pseudo R2 = 0.0114  female  Coef. Std. Err. z P>z [95% Conf. Interval] + prog  2  .3245866 .3910782 0.83 0.407 .4419125 1.091086 3  .2183474 .4319116 0.51 0.613 .6281839 1.064879  2.schtyp  1.660724 1.141326 1.46 0.146 .5762344 3.897683  prog#schtyp  2 2  1.934018 1.232722 1.57 0.117 4.350108 .4820729 3 2  1.827778 1.840256 0.99 0.321 5.434614 1.779057  _cons  .0512933 .3203616 0.16 0.873 .6791906 .576604 The results indicate that the overall model is not statistically significant (LR chi2 = 3.15, p = 0.6774). Furthermore, none of the coefficients are statistically significant either. We can use the test command to get the test of the overall effect of prog as shown below. This shows that the overall effect of prog is not statistically significant.
test 2.prog 3.prog ( 1) [female]2.prog = 0 ( 2) [female]3.prog = 0 chi2( 2) = 0.69 Prob > chi2 = 0.7086Likewise, we can use the testparm command to get the test of the overall effect of the prog by schtyp interaction, as shown below. This shows that the overall effect of this interaction is not statistically significant.
testparm prog#schtyp ( 1) [female]2.prog#2.schtyp = 0 ( 2) [female]3.prog#2.schtyp = 0 chi2( 2) = 2.47 Prob > chi2 = 0.2902If you prefer, you could use the logistic command to see the results as odds ratios, as shown below.
logistic female i.prog##schtyp Logistic regression Number of obs = 200 LR chi2(5) = 3.15 Prob > chi2 = 0.6774 Log likelihood = 136.24501 Pseudo R2 = 0.0114  female  Odds Ratio Std. Err. z P>z [95% Conf. Interval] + prog  2  1.383459 .5410405 0.83 0.407 .6428059 2.977505 3  1.244019 .5373063 0.51 0.613 .5335599 2.900487  2.schtyp  5.263121 6.006939 1.46 0.146 .5620107 49.28811  prog#schtyp  2 2  .1445662 .1782099 1.57 0.117 .0129054 1.619428 3 2  .1607704 .2958586 0.99 0.321 .0043629 5.924268 
Correlation
A correlation is useful when you want to see the linear relationship between two (or more) normally distributed interval variables. For example, using the hsb2 data file we can run a correlation between two continuous variables, read and write.
corr read write(obs=200)  read write + read  1.0000 write  0.5968 1.0000In the second example, we will run a correlation between a dichotomous variable, female, and a continuous variable, write. Although it is assumed that the variables are interval and normally distributed, we can include dummy variables when performing correlations.
corr female write(obs=200)  female write + female  1.0000 write  0.2565 1.0000In the first example above, we see that the correlation between read and write is 0.5968. By squaring the correlation and then multiplying by 100, you can determine what percentage of the variability is shared. Let’s round 0.5968 to be 0.6, which when squared would be .36, multiplied by 100 would be 36%. Hence read shares about 36% of its variability with write. In the output for the second example, we can see the correlation between write and female is 0.2565. Squaring this number yields .06579225, meaning that female shares approximately 6.5% of its variability with write.
See also
Simple linear regression
Simple linear regression allows us to look at the linear relationship between one normally distributed interval predictor and one normally distributed interval outcome variable. For example, using the hsb2 data file, say we wish to look at the relationship between writing scores (write) and reading scores (read); in other words, predicting write from read.
regress write read  write  Coef. Std. Err. t P>t [95% Conf. Interval] + read  .5517051 .0527178 10.47 0.000 .4477446 .6556656 _cons  23.95944 2.805744 8.54 0.000 18.42647 29.49242 We see that the relationship between write and read is positive (.5517051) and based on the tvalue (10.47) and pvalue (0.000), we would conclude this relationship is statistically significant. Hence, we would say there is a statistically significant positive linear relationship between reading and writing.
See also
 Regression With Stata: Chapter 1 – Simple and Multiple Regression
 Stata Annotated Output: Regression
 Stata Frequently Asked Questions
 Stata Topics: Regression
 Stata Textbook Example: Introduction to the Practice of Statistics, Chapter 10
 Stata Textbook Examples: Regression with Graphics, Chapter 2
 Stata Textbook Examples: Applied Regression Analysis, Chapter 5
Nonparametric correlation
A Spearman correlation is used when one or both of the variables are not assumed to be normally distributed and interval (but are assumed to be ordinal). The values of the variables are converted in ranks and then correlated. In our example, we will look for a relationship between read and write. We will not assume that both of these variables are normal and interval .
spearman read writeNumber of obs = 200 Spearman's rho = 0.6167 Test of Ho: read and write are independent Prob > t = 0.0000The results suggest that the relationship between read and write (rho = 0.6167, p = 0.000) is statistically significant.
Simple logistic regression
Logistic regression assumes that the outcome variable is binary (i.e., coded as 0 and 1). We have only one variable in thehsb2 data file that is coded 0 and 1, and that is female. We understand that female is a silly outcome variable (it would make more sense to use it as a predictor variable), but we can use female as the outcome variable to illustrate how the code for this command is structured and how to interpret the output. The first variable listed after the logistic (or logit) command is the outcome (or dependent) variable, and all of the rest of the variables are predictor (or independent) variables. You can use thelogit command if you want to see the regression coefficients or the logistic command if you want to see the odds ratios. In our example, female will be the outcome variable, and read will be the predictor variable. As with OLS regression, the predictor variables must be either dichotomous or continuous; they cannot be categorical.
logistic female read Logit estimates Number of obs = 200 LR chi2(1) = 0.56 Prob > chi2 = 0.4527 Log likelihood = 137.53641 Pseudo R2 = 0.0020  female  Odds Ratio Std. Err. z P>z [95% Conf. Interval] + read  .9896176 .0137732 0.75 0.453 .9629875 1.016984  logit female readIteration 0: log likelihood = 137.81834 Iteration 1: log likelihood = 137.53642 Iteration 2: log likelihood = 137.53641 Logit estimates Number of obs = 200 LR chi2(1) = 0.56 Prob > chi2 = 0.4527 Log likelihood = 137.53641 Pseudo R2 = 0.0020  female  Coef. Std. Err. z P>z [95% Conf. Interval] + read  .0104367 .0139177 0.75 0.453 .0377148 .0168415 _cons  .7260875 .7419612 0.98 0.328 .7281297 2.180305 The results indicate that reading score (read) is not a statistically significant predictor of gender (i.e., being female), z = 0.75, p = 0.453. Likewise, the test of the overall model is not statistically significant, LR chisquared 0.56, p = 0.4527.
See also
 Stata Textbook Examples: Applied Logistic Regression (2nd Ed) Chapter 1
 Stata Web Books: Logistic Regression in Stata
 Stata Topics: Logistic Regression
 Stata Data Analysis Example: Logistic Regression
 Annotated Stata Output: Logistic Regression Analysis
 Stata FAQ: How do I interpret odds ratios in logistic regression?
 Stata Library
 Teaching Tools: Graph Logistic Regression Curve
Multiple regression
Multiple regression is very similar to simple regression, except that in multiple regression you have more than one predictor variable in the equation. For example, using the hsb2 data file we will predict writing score from gender (female), reading, math, science and social studies (socst) scores.
regress write female read math science socstSource  SS df MS Number of obs = 200 + F( 5, 194) = 58.60 Model  10756.9244 5 2151.38488 Prob > F = 0.0000 Residual  7121.9506 194 36.7110855 Rsquared = 0.6017 + Adj Rsquared = 0.5914 Total  17878.875 199 89.843593 Root MSE = 6.059  write  Coef. Std. Err. t P>t [95% Conf. Interval] + female  5.492502 .8754227 6.27 0.000 3.765935 7.21907 read  .1254123 .0649598 1.93 0.055 .0027059 .2535304 math  .2380748 .0671266 3.55 0.000 .1056832 .3704665 science  .2419382 .0606997 3.99 0.000 .1222221 .3616542 socst  .2292644 .0528361 4.34 0.000 .1250575 .3334713 _cons  6.138759 2.808423 2.19 0.030 .599798 11.67772 The results indicate that the overall model is statistically significant (F = 58.60, p = 0.0000). Furthermore, all of the predictor variables are statistically significant except for read.
See also
 Regression with Stata: Lesson 1 – Simple and Multiple Regression
 Annotated Output: Multiple Linear Regression
 Stata Annotated Output: Regression
 Stata Teaching Tools
 Stata Textbook Examples: Applied Linear Statistical Models
 Stata Textbook Examples: Introduction to the Practice of Statistics, Chapter 11
 Stata Textbook Examples: Regression Analysis by Example, Chapter 3
Analysis of covariance
Analysis of covariance is like ANOVA, except in addition to the categorical predictors you also have continuous predictors as well. For example, the one way ANOVA example used write as the dependent variable and prog as the independent variable. Let’s add read as a continuous variable to this model, as shown below.
anova write prog c.read Number of obs = 200 Rsquared = 0.3925 Root MSE = 7.44408 Adj Rsquared = 0.3832 Source  Partial SS df MS F Prob > F + Model  7017.68123 3 2339.22708 42.21 0.0000  prog  650.259965 2 325.129983 5.87 0.0034 read  3841.98338 1 3841.98338 69.33 0.0000  Residual  10861.1938 196 55.4142539 + Total  17878.875 199 89.843593The results indicate that even after adjusting for reading score (read), writing scores still significantly differ by program type (prog) F = 5.87, p = 0.0034.
See also
Multiple logistic regression
Multiple logistic regression is like simple logistic regression, except that there are two or more predictors. The predictors can be interval variables or dummy variables, but cannot be categorical variables. If you have categorical predictors, they should be coded into one or more dummy variables. We have only one variable in our data set that is coded 0 and 1, and that isfemale. We understand that female is a silly outcome variable (it would make more sense to use it as a predictor variable), but we can use female as the outcome variable to illustrate how the code for this command is structured and how to interpret the output. The first variable listed after the logistic (or logit) command is the outcome (or dependent) variable, and all of the rest of the variables are predictor (or independent) variables. You can use the logit command if you want to see the regression coefficients or the logistic command if you want to see the odds ratios. In our example, female will be the outcome variable, and read and write will be the predictor variables.
logistic female read write Logit estimates Number of obs = 200 LR chi2(2) = 27.82 Prob > chi2 = 0.0000 Log likelihood = 123.90902 Pseudo R2 = 0.1009  female  Odds Ratio Std. Err. z P>z [95% Conf. Interval] + read  .9314488 .0182578 3.62 0.000 .8963428 .9679298 write  1.112231 .0246282 4.80 0.000 1.064993 1.161564 These results show that both read and write are significant predictors of female.
See also
 Stata Annotated Output: Logistic Regression
 Stata Library
 Stata Web Books: Logistic Regression with Stata
 Stata Topics: Logistic Regression
 Stata Textbook Examples: Applied Logistic Regression, Chapter 2
 Stata Textbook Examples: Applied Regression Analysis, Chapter 8
 Stata Textbook Examples: Introduction to Categorical Analysis, Chapter 5
 Stata Textbook Examples: Regression Analysis by Example, Chapter 12
Discriminant analysis
Discriminant analysis is used when you have one or more normally distributed interval independent variables and a categorical dependent variable. It is a multivariate technique that considers the latent dimensions in the independent variables for predicting group membership in the categorical dependent variable. For example, using the hsb2 data file, say we wish to useread, write and math scores to predict the type of program a student belongs to (prog). For this analysis, you need to first download the daoneway program that performs this test. You can download daoneway from within Stata by typing findit daoneway (see How can I used the findit command to search for programs and get additional help? for more information about using findit).
You can then perform the discriminant function analysis like this.
daoneway read write math, by(prog)Oneway Disciminant Function Analysis Observations = 200 Variables = 3 Groups = 3 Pct of Cum Canonical After Wilks' Fcn Eigenvalue Variance Pct Corr Fcn Lambda Chisquare df Pvalue  0 0.73398 60.619 6 0.0000 1 0.3563 98.74 98.74 0.5125  1 0.99548 0.888 2 0.6414 2 0.0045 1.26 100.00 0.0672  Unstandardized canonical discriminant function coefficients func1 func2 read 0.0292 0.0439 write 0.0383 0.1370 math 0.0703 0.0793 _cons 7.2509 0.7635 Standardized canonical discriminant function coefficients func1 func2 read 0.2729 0.4098 write 0.3311 1.1834 math 0.5816 0.6557 Canonical discriminant structure matrix func1 func2 read 0.7785 0.1841 write 0.7753 0.6303 math 0.9129 0.2725 Group means on canonical discriminant functions func1 func2 prog1 0.3120 0.1190 prog2 0.5359 0.0197 prog3 0.8445 0.0658Clearly, the Stata output for this procedure is lengthy, and it is beyond the scope of this page to explain all of it. However, the main point is that two canonical variables are identified by the analysis, the first of which seems to be more related to program type than the second. For more information, see this page on discriminant function analysis.
See also
Oneway MANOVA
MANOVA (multivariate analysis of variance) is like ANOVA, except that there are two or more dependent variables. In a oneway MANOVA, there is one categorical independent variable and two or more dependent variables. For example, using thehsb2 data file, say we wish to examine the differences in read, write and math broken down by program type (prog). For this analysis, you can use the manova command and then perform the analysis like this.
manova read write math = prog, category(prog)Number of obs = 200 W = Wilks' lambda L = LawleyHotelling trace P = Pillai's trace R = Roy's largest root Source  Statistic df F(df1, df2) = F Prob>F + prog  W 0.7340 2 6.0 390.0 10.87 0.0000 e  P 0.2672 6.0 392.0 10.08 0.0000 a  L 0.3608 6.0 388.0 11.67 0.0000 a  R 0.3563 3.0 196.0 23.28 0.0000 u  Residual  197 + Total  199  e = exact, a = approximate, u = upper bound on FThis command produces three different test statistics that are used to evaluate the statistical significance of the relationship between the independent variable and the outcome variables. According to all three criteria, the students in the different programs differ in their joint distribution of read, write and math.
See also
 Stata Data Analysis Examples: Oneway MANOVA
 Stata Annotated Output: Oneway MANOVA
 Stata FAQ: How can I do multivariate repeated measures in Stata?
Multivariate multiple regression
Multivariate multiple regression is used when you have two or more dependent variables that are to be predicted from two or more predictor variables. In our example, we will predict write and read from female, math, science and social studies (socst) scores.
mvreg write read = female math science socstEquation Obs Parms RMSE "Rsq" F P  write 200 5 6.101191 0.5940 71.32457 0.0000 read 200 5 6.679383 0.5841 68.4741 0.0000   Coef. Std. Err. t P>t [95% Conf. Interval] + write  female  5.428215 .8808853 6.16 0.000 3.69093 7.165501 math  .2801611 .0639308 4.38 0.000 .1540766 .4062456 science  .2786543 .0580452 4.80 0.000 .1641773 .3931313 socst  .2681117 .049195 5.45 0.000 .1710892 .3651343 _cons  6.568924 2.819079 2.33 0.021 1.009124 12.12872 + read  female  .512606 .9643644 0.53 0.596 2.414529 1.389317 math  .3355829 .0699893 4.79 0.000 .1975497 .4736161 science  .2927632 .063546 4.61 0.000 .1674376 .4180889 socst  .3097572 .0538571 5.75 0.000 .2035401 .4159744 _cons  3.430005 3.086236 1.11 0.268 2.656682 9.516691 Many researchers familiar with traditional multivariate analysis may not recognize the tests above. They do not see Wilks’ Lambda, Pillai’s Trace or the HotellingLawley Trace statistics, the statistics with which they are familiar. It is possible to obtain these statistics using the mvtest command written by David E. Moore of the University of Cincinnati. UCLA updated this command to work with Stata 6 and above. You can download mvtest from within Stata by typing findit mvtest (see How can I used the findit command to search for programs and get additional help? for more information about using findit).
Now that we have downloaded it, we can use the command shown below.
mvtest femaleMULTIVARIATE TESTS OF SIGNIFICANCE Multivariate Test Criteria and Exact F Statistics for the Hypothesis of no Overall "female" Effect(s) S=1 M=0 N=96 Test Value F Num DF Den DF Pr > F Wilks' Lambda 0.83011470 19.8513 2 194.0000 0.0000 Pillai's Trace 0.16988530 19.8513 2 194.0000 0.0000 HotellingLawley Trace 0.20465280 19.8513 2 194.0000 0.0000These results show that female has a significant relationship with the joint distribution of write and read. The mvtestcommand could then be repeated for each of the other predictor variables.
See also
Canonical correlation
Canonical correlation is a multivariate technique used to examine the relationship between two groups of variables. For each set of variables, it creates latent variables and looks at the relationships among the latent variables. It assumes that all variables in the model are interval and normally distributed. Stata requires that each of the two groups of variables be enclosed in parentheses. There need not be an equal number of variables in the two groups.
canon (read write) (math science) Linear combinations for canonical correlation 1 Number of obs = 200   Coef. Std. Err. t P>t [95% Conf. Interval] + u  read  .0632613 .007111 8.90 0.000 .0492386 .077284 write  .0492492 .007692 6.40 0.000 .0340809 .0644174 + v  math  .0669827 .0080473 8.32 0.000 .0511138 .0828515 science  .0482406 .0076145 6.34 0.000 .0332252 .0632561  (Std. Errors estimated conditionally) Canonical correlations: 0.7728 0.0235The output above shows the linear combinations corresponding to the first canonical correlation. At the bottom of the output are the two canonical correlations. These results indicate that the first canonical correlation is .7728. You will note that Stata is brief and may not provide you with all of the information that you may want. Several programs have been developed to provide more information regarding the analysis. You can download this family of programs by typing findit cancor (see How can I used the findit command to search for programs and get additional help? for more information about using findit).
Because the output from the cancor command is lengthy, we will use the cantest command to obtain the eigenvalues, Ftests and associated pvalues that we want. Note that you do not have to specify a model with either the cancor or the cantestcommands if they are issued after the canon command.
cantestCanon Can Corr Likelihood Approx Corr Squared Ratio F df1 df2 Pr > F 7728 .59728 0.4025 56.4706 4 392.000 0.0000 0235 .00055 0.9994 0.1087 1 197.000 0.7420 Eigenvalue Proportion Cumulative 1.4831 0.9996 0.9996 0.0006 0.0004 1.0000The Ftest in this output tests the hypothesis that the first canonical correlation is equal to zero. Clearly, F = 56.4706 is statistically significant. However, the second canonical correlation of .0235 is not statistically significantly different from zero (F = 0.1087, p = 0.7420).
See also
Factor analysis
Factor analysis is a form of exploratory multivariate analysis that is used to either reduce the number of variables in a model or to detect relationships among variables. All variables involved in the factor analysis need to be continuous and are assumed to be normally distributed. The goal of the analysis is to try to identify factors which underlie the variables. There may be fewer factors than variables, but there may not be more factors than variables. For our example, let’s suppose that we think that there are some common factors underlying the various test scores. We will first use the principal components method of extraction (by using the pc option) and then the principal components factor method of extraction (by using the pcf option). This parallels the output produced by SAS and SPSS.
factor read write math science socst, pc (obs=200) (principal components; 5 components retained) Component Eigenvalue Difference Proportion Cumulative  1 3.38082 2.82344 0.6762 0.6762 2 0.55738 0.15059 0.1115 0.7876 3 0.40679 0.05062 0.0814 0.8690 4 0.35617 0.05733 0.0712 0.9402 5 0.29884 . 0.0598 1.0000 Eigenvectors Variable  1 2 3 4 5 + read  0.46642 0.02728 0.53127 0.02058 0.70642 write  0.44839 0.20755 0.80642 0.05575 0.32007 math  0.45878 0.26090 0.00060 0.78004 0.33615 science  0.43558 0.61089 0.00695 0.58948 0.29924 socst  0.42567 0.71758 0.25958 0.20132 0.44269Now let’s rerun the factor analysis with a principal component factors extraction method and retain factors with eigenvalues of .5 or greater. Then we will use a varimax rotation on the solution.
factor read write math science socst, pcf mineigen(.5) (obs=200) (principal component factors; 2 factors retained) Factor Eigenvalue Difference Proportion Cumulative  1 3.38082 2.82344 0.6762 0.6762 2 0.55738 0.15059 0.1115 0.7876 3 0.40679 0.05062 0.0814 0.8690 4 0.35617 0.05733 0.0712 0.9402 5 0.29884 . 0.0598 1.0000 Factor Loadings Variable  1 2 Uniqueness + read  0.85760 0.02037 0.26410 write  0.82445 0.15495 0.29627 math  0.84355 0.19478 0.25048 science  0.80091 0.45608 0.15054 socst  0.78268 0.53573 0.10041rotate, varimax (varimax rotation) Rotated Factor Loadings Variable  1 2 Uniqueness + read  0.64808 0.56204 0.26410 write  0.50558 0.66942 0.29627 math  0.75506 0.42357 0.25048 science  0.89934 0.20159 0.15054 socst  0.21844 0.92297 0.10041Note that by default, Stata will retain all factors with positive eigenvalues; hence the use of the mineigen option or thefactors(#) option. The factors(#) option does not specify the number of solutions to retain, but rather the largest number of solutions to retain. From the table of factor loadings, we can see that all five of the test scores load onto the first factor, while all five tend to load not so heavily on the second factor. Uniqueness (which is the opposite of commonality) is the proportion of variance of the variable (i.e., read) that is not accounted for by all of the factors taken together, and a very high uniqueness can indicate that a variable may not belong with any of the factors. Factor loadings are often rotated in an attempt to make them more interpretable. Stata performs both varimax and promax rotations.
rotate, varimax(varimax rotation) Rotated Factor Loadings Variable  1 2 Uniqueness + read  0.62238 0.51992 0.34233 write  0.53933 0.54228 0.41505 math  0.65110 0.45408 0.36988 science  0.64835 0.37324 0.44033 socst  0.44265 0.58091 0.46660The purpose of rotating the factors is to get the variables to load either very high or very low on each factor. In this example, because all of the variables loaded onto factor 1 and not on factor 2, the rotation did not aid in the interpretation. Instead, it made the results even more difficult to interpret.
To obtain a scree plot of the eigenvalues, you can use the greigen command. We have included a reference line on the yaxis at one to aid in determining how many factors should be retained.
greigen, yline(1)See also