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|STAT Statistical Data Analysis
Free Data Analysis Programs for UNIX and DOS
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| by Gary Perlman |
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Home | | | Preface | | | Intro | | | Example | | | Conventions | | | Manipulation | | | Analysis | | | DM | | | Calc | | | Manuals | | | History |
Last updated: Accesses since 2001-12-20:
A concrete example with several |STAT programs is worked in detail. The example shows the style of analysis in |STAT. New users of |STAT should not try to understand all the details in the examples. Details about all the programs can be found in on-line manual entries and more examples of program use appear in following chapters. Explanations about features common to all |STAT programs can be found in the next chapter.
To show the |STAT style of interactive data analysis, I will work through a concrete example. The example is based on a familiar problem: grades in a course based on two midterm exams and a final exam. Scores on exams will be broken down by student gender (male or female) and by the lab section taught by one of two teaching assistants: John or Jane. Assume the following data are in the file exam.dat. Each line in the file includes a student identification number, the student's section's teaching assistant, the student's gender, and the scores (out of 100) on the two midterm exams and the final.
S-1 john male 56 42 58 S-2 john male 96 90 91 S-3 john male 70 59 65 S-4 john male 82 75 78 S-5 john male 85 90 92 S-6 john male 69 60 65 S-7 john female 82 78 60 S-8 john female 84 81 82 S-9 john female 89 80 68 S-10 john female 90 93 91 S-11 jane male 42 46 65 S-12 jane male 28 15 34 S-13 jane male 49 68 75 S-14 jane male 36 30 48 S-15 jane male 58 58 62 S-16 jane male 72 70 84 S-17 jane female 65 61 70 S-18 jane female 68 75 71 S-19 jane female 62 50 55 S-20 jane female 71 72 87We are interested in computing final grades based on the exam scores, and comparing the performances of males versus females, and of the different teaching assistants. The following analyses can be tried by typing in the above file and running the commands in the examples. Minor variations on the example commands will help show how the programs work.
Computing final scores is easy with the data manipulation program dm. Assume that the first midterm is worth 20 percent, the second 30 percent, and the final exam, 50 percent. The following command tells dm to repeat each input line with INPUT, and then print the weighted sum of columns 4, 5, and 6, treated as numbers. To print numbers, dm uses an x before the column number. The input to dm is read from the file exam.dat and the result is saved in the file scores.dat. Once all the original data and the final scores are in scores.dat, only that file will be used in following analyses.
dm INPUT ".2*x4 + .3*x5 + .5*x6" < exam.dat > scores.datThe standard input is redirected from the file exam.dat with the < on the command line. Similarly, the standard output, which would ordinarily go to the screen, is redirected to the file scores.dat with the > on the command line. The second expression for dm is in quotes. This allows the insertion of spaces to make the expression more readable, and to make sure that any special characters (e.g., * is special to UNIX shells) are hidden from the command line interpreter. The output from the above command, saved in the file scores.dat, would begin with the following.
S-1 john male 56 42 58 52.8 S-2 john male 96 90 91 91.7 S-3 john male 70 59 65 64.2 S-4 john male 82 75 78 77.9 S-5 john male 85 90 92 90 S S-6 john male 69 60 65 64.3 etc.This could be sorted by final grade by reversing the columns and sending the output to the standard UNIX or MSDOS sort utility program using the ``pipe'' symbol |.
reverse -f < scores.dat | sortThe above command would produce the following output.
27.1 34 15 28 male jane S-12 40.2 48 30 36 male jane S-14 52.8 58 42 56 male john S-1 54.7 65 46 42 male jane S-11 54.9 55 50 62 female jane S-19 ... 79.3 87 72 71 female jane S-20 82.1 82 81 84 female john S-8 90 92 90 85 male john S-5 91.4 91 93 90 female john S-10 91.7 91 90 96 male john S-2To restore the order of the fields, reverse could be called again. Another way, more efficient, would be to use the dsort filter to sort based on column 7:
dsort 7 < scores.dat
desc prints summary statistics, histograms, and frequency tables. The following command takes the final scores (the weighted average from the previous section).
dm s7 < scores.datSummary order statistics are printed with the -o option and the distribution is tested against the passing grade of 75 with the -t 75 option. desc makes a histogram (the -h option) with 10 point intervals (the -i 10 option) starting at a minimum value of 0 (the -m 0 option).
dm s7 < scores.dat | desc -o -t 75 -h -i 10 -m 0
------------------------------------------------------------
Under Range In Range Over Range Missing Sum
0 20 0 0 1359.200
------------------------------------------------------------
Mean Median Midpoint Geometric Harmonic
67.960 68.750 59.400 65.564 62.529
------------------------------------------------------------
SD Quart Dev Range SE mean
16.707 10.575 64.600 3.736
------------------------------------------------------------
Minimum Quartile 1 Quartile 2 Quartile 3 Maximum
27.100 57.450 68.750 78.600 91.700
------------------------------------------------------------
Skew SD Skew Kurtosis SD Kurt
-0.586 0.548 2.844 1.095
------------------------------------------------------------
Null Mean t prob (t) F prob (F)
75.000 -1.884 0.075 3.551 0.075
------------------------------------------------------------
Midpt Freq
5.000 0
15.000 0
25.000 1 *
35.000 0
45.000 1 *
55.000 4 ****
65.000 5 *****
75.000 5 *****
85.000 2 **
95.000 2 **
The next analysis predicts final exam scores with those of the two midterm exams. The regress program assumes its input has the predicted variable in column 1 and the predictors in following columns. dm can extract the columns in the correct order from the file scores.dat. The command for dm looks like this.
dm x6 x4 x5 < scores.datThe output from dm looks like this.
58 56 42 91 96 90 65 70 59 78 82 75 92 85 90 65 69 60 60 82 78 etc.This is the correct format for input for regress, which is given mnemonic names for the columns. The -e option tells regress to save the regression equation in the file regress.eqn for a later analysis.
dm x6 x4 x5 < scores.dat | regress -e final midterm1 midterm2The output from regress includes summary statistics for all the variables, a correlation matrix (e.g., the correlation of midterm1 and midterm2 is .9190), the regression equation relating the predicted variable, and the significance test of the multiple correlation coefficient. The squared multiple correlation coefficient of 0.7996 shows a strong relationship between midterm exams and the final.
Analysis for 20 cases of 3 variables:
Variable final midterm1 midterm2
Min 34.0000 28.0000 15.0000
Max 92.0000 96.0000 93.0000
Sum 1401.0000 1354.0000 1293.0000
Mean 70.0500 67.7000 64.6500
SD 15.3502 18.6720 20.4303
Correlation Matrix:
final 1.0000
midterm1 0.7586 1.0000
midterm2 0.8838 0.9190 1.0000
Variable final midterm1 midterm2
Regression Equation for final:
final = -0.2835 midterm1 + 0.9022 midterm2 + 30.9177
Significance test for prediction of final
Mult-R R-Squared SEest F(2,17) prob (F)
0.8942 0.7996 7.2640 33.9228 0.0000
Predicted Plot s1 (x2 * -0.283512...) + (x3 * 0.902182...) + 30.9177...Extra precision is used in regress.eqn, compared to the equation in the output from regress to allow more accurate calculations. These two expressions, one on each line, are the obtained and predicted final exam scores, respectively. To plot these against each other, we duplicate the input used to regress, and process regress's output with dm, reading its expressions from the expression file regress.eqn that follows the letter E. The result is passed through a pipe to the paired data analysis program pair with the plotting option -p, options to control the height and width of the plot, the -h and -w options, and -x and -y options to label the plot.
dm x6 x4 x5 < scores.dat | dm Eregress.eqn | pair -p -h 10 -w 30 -x final -y predicted
|------------------------------|89.3045
| 3|
| 1 1 |
| 1 1 11 1 1 |
| |
| 1 2 1 |predicted
| 1 1 |
| 1 |
| 1 |
| |
|1 |
|------------------------------|36.5121
34.000 92.000
final r= 0.894
Residual Plot dm x6 x4 x5 < scores.dat | dm Eregress.eqn | dm x2 x1-x2 | pair -p -h 10 -w 30 -x predicted -y residuals
|------------------------------|11.2546
| 11 |
| 1 |
| 1 1 1 1 1|
| 1 1 1 1|
|1 1 1 |residuals
| 1 1 |
| 1 |
| 1 |
| |
| 1 |
|------------------------------|-18.0399
36.512 89.304
predicted r= 0.000
Now suppose the passing grade in the course is 75. To see how many people of each sex in the two sections passed, we can use the contab program to print contingency tables. First dm extracts the columns containing teaching assistant, gender, and the final grade (the weighted average computed earlier). Rather than include the final grade, a label indicating pass or fail is added, as appropriate.
dm s2 s3 "if x7 >= 75 then 'pass' else 'fail'" 1 < scores.datThe huge third expression says ``if the final grade is greater than or equal to 75, then insert the string pass, else insert the string fail.'' Such expressions can be placed in files rather than be typed on the command line, and usually dm is used for simpler expressions. The fourth expression is the constant 1 used to tell contab that there was one replication for each combination of factor levels. Part of the output from dm follows.
john male fail 1
john male pass 1
john male fail 1
...
jane female fail 1
jane female fail 1
jane female pass 1
This is used as input to contab, which is given mnemonic
factor names.
dm s2 s3 "if x7 >= 75 then 'pass' else 'fail'" 1 < scores.dat | contab assistant gender success countParts of the output from this command follow. First, there is a summary of the input, which contained three factors, each with 2 levels, and a sum of observation counts.
FACTOR: assistant gender success count LEVELS: 2 2 2 20The first contingency table does not provide new information. It shows that both Jane's section and John's section had 6 male and 4 female students.
SOURCE: assistant gender
male female Totals
john 6 4 10
jane 6 4 10
Totals 12 8 20
The second contingency table
tells us that 12 of 20 students failed the course--4 in John's section and 8 in
Jane's. A significance test follows, and the warning about small expected
frequencies suggests that the chi-square test for independence might be
invalid. No matter, the Fisher exact test applies because we are dealing with
a 2x2 table and total frequencies less than 100. It does not show a
significant association of factors (ie. Jane's section did not do significantly
better than John's).
SOURCE: assistant success
fail pass Totals
john 4 6 10
jane 8 2 10
Totals 12 8 20
Analysis for assistant x success: NOTE: Yates' correction for continuity applied WARNING: 2 of 4 cells had expected frequencies < 5 chisq 1.875000 df 1 p 0.170904 Fisher Exact One-Tailed Probability 0.084901 Fisher Exact Two-Tailed Probability 0.169802 phi Coefficient == Cramer's V 0.306186 Contingency Coefficient 0.292770The third contingency table shows that 8 male students and 4 female students failed the course.
SOURCE: gender success
fail pass Totals
male 8 4 12
female 4 4 8
Totals 12 8 20
The final table, the three-way interaction,
shows all the effects listed above, but no significance test
is computed by contab. Some hints about the reason for the poorer
performance of Jane's section follow from the next section's analysis of
variance.
SOURCE: assistant gender success
assistan gender success
john male fail 3
john male pass 3
john female fail 1
john female pass 3
jane male fail 5
jane male pass 1
jane female fail 3
jane female pass 1
We now want to compare the performance of the two teaching assistants and of male versus female students. We are interested to see how an assistant's students progress through the term. anova, the analysis of variance program, is the program to analyze these data, but we have to get the data into the correct format for input to anova. anova assumes that there is only one datum per line, preceded by the levels of factors under which it was obtained. This is unlike the format of scores.dat, which has the three exam scores after the student number, teaching assistant name, and gender. Several transformations are needed to get the data in the correct format. As an example, the data for student 1:
S-1 john male 56 42 58must be transformed to:
S-1 john male m1 56 S-1 john male m2 42 S-1 john male final 58This is made up of three replications of the labels with new labels, m1, m2, and final, for the exams inserted. First, dm extracts and inserts the desired information. The result is a 15 column output, one for each expression. Note that on UNIX, it is necessary to quote the quotes of the labels for the exam names. To insert the newlines, so that each datum is on one line, the program maketrix reformats the input to anova into 5 columns. Finally, mnemonic labels for factor names are given to anova.
dm s1 s2 s3 "'m1'" s4 ...
s1 s2 s3 "'m2'" s5 ...
s1 s2 s3 "'final'" s6 < scores.dat |
maketrix 5 | anova student assistant gender exam score
Only parts of the
output are shown below. First, John's students did better than Jane's students
(F(1,16)=8.311, p=.011).
john 76.7000 jane 58.2333Female students scored better than males, although the effect is not statistically significant (F(1,16)=3.102, p=.097).
male 62.8611 female 74.3750There was no interaction between these two factors (F(1,16)=.289), but there were some interactions between section assistant and gender and the different exam grades. If we look at the interaction of section assistant and exam, we get a better picture of the performances of John and Jane.
SOURCE: assistant exam assista exam N MEAN SD SE john m1 10 80.3000 11.9355 3.7743 john m2 10 74.8000 16.3761 5.1786 john final 10 75.0000 13.4247 4.2453 jane m1 10 55.1000 15.5167 4.9068 jane m2 10 54.5000 19.5973 6.1972 jane final 10 65.1000 16.2101 5.1261This is the first full cell-means table shown. It contains the names of factors and levels, cell counts, means, standard deviations, and standard errors. The results show that John's students started higher than Jane's (80.3 versus 55.1), and that over the term, Jane's students improved while John's got worse. The significance test for the interaction looks like this.
SOURCE SS df MS F p ================================================= ae 610.4333 2 305.2167 9.502 0.001 *** es/ag 1027.8889 32 32.1215A Scheffé confidence interval around the difference between two means of this interaction can be found with the following formula.
sqrt (df1 * critf * MSerror * 2 / N)df1 is the degrees of freedom numerator, critf is the critical F-ratio given the degrees of freedom and confidence level desired, MSerror is the mean-square error for the overall F-test, and N is the number of scores going into each cell. The critical F ratio for a 95% confidence interval based on 2 and 32 degrees of freedom can be found with the probdist program.
probdist crit F 2 32 .05 3.294537Then, the calculator program calc can be used interactively to substitute the values.
CALC: sqrt (2 * 3.294537 * 32.1215 * 2 / 10) sqrt(((((2 * 3.29454) * 32.1215) * 2) / 10)) = 6.50617Any difference of two means in this interaction greater than 6.5 is significant at the .05 level.
There was a similar pattern of males versus females on the three exams. Males started out lower than females, and males improved slightly while females stayed about the same.
SOURCE: gender exam gender exam N MEAN SD SE male m1 12 61.9167 20.7822 5.9993 male m2 12 58.5833 22.5931 6.5221 male final 12 68.0833 17.1329 4.9459 female m1 8 76.3750 11.1475 3.9413 female m2 8 73.7500 13.1557 4.6512 female final 8 73.0000 12.7167 4.4960After the cell means in the output from anova is a summary of the design, followed by an F-table, parts of which were seen above.
FACTOR: student assistant gender exam score LEVELS: 20 2 2 3 60 TYPE : RANDOM BETWEEN BETWEEN WITHIN DATA
The results
of the analysis show that John's section did better than Jane's. That must be
qualified because it seems that Jane's students may not have been as good as
John's. To Jane's credit, her students improved more than John's during the
term.
| © 1986 Gary Perlman |