Investigating relationships between student marks and majors taken. A descriptive and inferential statistics using SAS

Contents

Aims

1 Problem Definition
1.1 Introduction
1.2 Problem Visualisation
1.3 Plan Overview
1.4 Risk Management

2 Data Collection and Preparation
2.1 Source of the Data
2.2 Data Dictionary
2.3 Issues with the Data
2.4 Data Cleaning
2.5 Data Transformation

3 Statistical Analysis
3.1 Descriptive Statistics
3.1.1 Central Tendency
3.1.2 Variation
3.1.3 Shape of the Distribution
3.2 Inferential Statistics
3.2.1 Two-sample t-tests
3.2.2 Chi-Square Tests

4 Delivery of Results
4.1 Conclusion

5 Appendix - Code Used

Aims

This project seeks to discover any correlation between students who did Mathemat­ics and Applied Mathematics and their chance of passing. This investigation will try to determine if for the double major students, they are more likely to pass without repeating a single time than the students only taking one major.

Chapter 1 Problem Definition

1.1 Introduction

This projectplan was created to detail the sections of the project and the specific steps that we planned to carry out during the course of the investigation. It helped to ensure the timeous completion of the project, as well as to meet all requirements and goals. Our project plan details the following:

- Outline of the environment of the business the project is being done for
- Outline of the deliverables that are needed from our project
- Investigation into risks and how to deal with them during the course of the project In addition, we hope to cover the following points:
- The importance of the work
- The specific steps we took to arrive at the end goals
- The questions that are answered by carrying out the data transformation, and subsequently, the statistical analysis

From examining the business requirements, as well as the specific project goals, we determined the direction to take to start with. We used the plan to understand the particular business environment (in this case, a University) that the problem is re­lated to and thus were able to take steps specific to that environment. This allowed us to understand which statistical methods were the most appropriate and thus we performed the appropriate tests and procedures to answer our questions.

1.2 Problem Visualisation

The dataset we are working with in this investigation is related to the personal infor­mation of students at the University of the Witwatersrand. This includes their details such as age and sex. In addition, the subject choices of students are given as well as their final marks and grades for each of their subjects, in each of their years of study. We are looking to determine whether there is a relationship between students who either majored in one subject (Applied Mathematics only) or in two subjects (both Mathematics and Applied Mathematics) and their chance of passing all 3 years with­out repeating. We will also investigate any other possible meaningful trends within the data.

The work being done is important because it will allow the University of the Witwa- tersrand to respond to common trends observed in the data. For example, if it is ob­served that a certain group of students performed worse than another, it will enable the faculty within the University to put measures in place to help increase the pass rate of the struggling students. In order to achieve this, we must follow the project plan detailed in the next section.

1.3 Plan Overview

In this investigation, we are required to:

- Clean the dataset, removing any erroneous records and other sources of prob­lematic data that may skew our results,
- Transform the dataset to conform to the needs of a statistical analysis, removing and modifying variables where necessary,
- Apply statistical methods over the dataset using the software SAS in order to ex­tract important information from it,
- Create summary and contingency tables to better understand the data,
- Detect patterns by plotting graphs to gain useful knowledge from the data,
- Use the information gained to answer questions raised in our project plan.

1.4 Risk Management

A project of this nature does incur significant risk. For example, the dataset we are working with may have an overwhelming amount of erroneous data. This would skew our results heavily and the outcome of the project would be unsuccessful. In addition, we must consider that the sample size of the data we are working with may turn out to be too small to gain any meaningful information from it.

Therefore, we must take steps to remove as much erroneous data as possible. Pro­cessing the data before running statistical analyses on it will prevent problems from occurring later on and having to backtrack. If there is simply too much erroneous data or if the dataset is too small, then there is not a simple solution to achieve meaningful results. The investigation may need to be repeated in the future, by taking a larger dataset from the source.

Chapter 2 Data Collection and Preparation

The data used for this investigation was not initially clean enough for meaningful analysis to be performed on it. We thus outline the various steps taken to prepare this data in order to make it ready for statistical tests.

2.1 Source of the Data

The educational dataset used for this project have been provided by the System Sup­port Consultant from the Academic Information System Unit at the University of the Witwatersrand. The dataset contains students who were registered for Applied Math­ematics or Mathematics between 2010 and 2015 and enrolled for a BSc degree of 3 years.

2.2 Data Dictionary

The dataset we utilised for the purposes of this project contained a number of varied attributes. These attributes had different data types, and for the purpose of the anal­ysis, different importance. The dataset is described in the data dictionary in Table 2.1 below:

Table 2.1: Data dictionary

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2.3 Issues with the Data

After completing a thorough examination of the dataset, a number of key issues were noted. The dataset, as is commonly the case, was not very clean and contained a number of issues that needed to be sorted out. The first thing to be noticed was that there were a number of observations with missing values in the "Final Mark" of the students. These missing or incomplete observations have the potential to skew the results of the statistical analysis if they were taken into account. Thus, they must be dealt with.

In addition, some final results of students were erroneously recorded. While the range of marks can only be between 0 and 100, some observations contained values outside this range, which were clearly incorrect. These incorrect marks likely came about as a result of human error whilst recording students’ achievements.

Next, some attributes in the dataset were insignificant. Attributes such as "Unit At­tempt Status" and "Program Title" did not provide any new information and thus were redundant.

Finally, there were an abnormally large number of students that had ’0’ recorded as their final mark, which was likely to skew the results of the analysis unfairly. We looked at factors that caused these types of results to occur.

2.4 Data Cleaning

Referring to the previous section, all the issues we outlined were tackled as follows in this section. Firstly, all data with missing observations was removed from the dataset. This eliminates any potential problems that these missing values could cause, such as not allowing sums or averages to be calculated. The data analysis will be focused strictly on only those observations that are complete.

Next, all observations that had a final mark outside the range of 0-100 were removed. As it is impossible to tell what mark these students actually achieved, assumptions cannot be made and it is more beneficial to just remove them completely from the dataset. Inserting any assumed values would just skew the dataset in an unnatural way.

The attributes "Calendar Instance Year", "Program Code", "Program Title", "Unit Ti­tle", and "Unit Attempt Status" were then removed from the dataset. All necessary information required to match students to a particular course can be found from the "Program Code" attribute. This one attribute tells us whether the course was a Math­ematics course (indicated by containing "MATH") or whether it was an Applied Math­ematics course (indicated by "APPM"). In addition, it also tells us which year or level each course was at, indicated by the numbers following the aforementioned prefixes. The "Year of Study" attribute tells us how many years the student had completed and thus if they managed to complete the required number of years for a major.

In terms of the large numbers of 0s in the data for final marks, we noticed that these were often associated with Final Grades such as "FAB" or "FNQ". FAB refers to failing due to absence [Wits 2016], thus students were given a 0 simply for not attending the final exam, and the 0 may not truly reflect their real result. FNQ stood for students that did not qualify to write the exam [Wits 2016], through not submitting assignments and writing tests. These students were also automatically given 0 as their final mark. Therefore, we removed all observations with these Final Grades to ensure our data was not skewed towards 0 too greatly.

2.5 Data Transformation

With the remaining useful attributes we selected, it was then necessary to transform the data further to prepare it for analysis. We sorted the data according to the Index Key attribute, meaning that each individual student would have all their observations and courses taken put together. The specifics of this investigation were as follows - we wanted to determine if there was a relationship between students that majored (did all 3 years) in either Applied Mathematics only or both Mathematics and Applied Math­ematics and whether they passed or had to repeat at least 1 year. Since the data was taken from the year 2010 to 2015, there is no data about students that took courses in years prior to 2010, and who are finishing their majors in 2010 or 2011. In addition, some students mighthave only started studying in 2014 or 2015, meaning they would not have finished their majors yet and there would thus be insufficient data to deter­mine whether they have passed or repeated. Finally, some students may have simply taken one year of mathematics courses as a requirement of another degree and not done any more mathematics courses since.

As a result of this, we removed the aforementioned data as it did not contribute to the analysis we intended to perform. This involved creating a new table and only inserting those students (marked by their Index Keys) that took courses in YOS (year of study) 1, 2 and 3. Next, it was necessary to separate the data further into different tables ac­cording to students that majored in both Mathematics and Applied Mathematics and students that only majored in Applied Mathematics. A student that majored in a field was defined as one that did all 3 years of courses in that field. This allows for analysis to be performed on these two different groups, in order to see how their chances of passing match up.

Next, in order to determine the relationships between students passing or repeating and their course choices, we grouped the data according to each index key (there is one key per student) and placed it in a summary table. This meant that we needed to calculate the average mark for each student for all courses they took. Any students that repeated any course at least once were recorded as ’Repeat’ students in the Final Grade column. Students that didn’t repeat a single time were recorded as ’Pass’ stu­dents. We also kept a column indicating whether each student was single major or double major.

In addition to these tables, we also further broke down the tables with data split ac­cording to double major students that passed, double major students that repeated, single major students that passed, and single major students that repeated. This was done to perform descriptive statistical analysis on these sets of data. In addition, we opted to look at the relation between age and pass chance, sex and pass chance and field of study and pass chance, separating data into further tables in order to achieve this (for example, putting all Engineering students in one separate table, and putting male and female students in different tables). For the age analysis, age values were binned according to whether they were under 25 or 25 and over. This allows for the creation of contingency tables during chi-square analysis - the counts of the number of students that passed and failed according to whether they were male or female can be retrieved, for example.

Finally, we also created summary tables to perform t-tests on with two columns, one representing whether each student was a single major student or a double major stu­dent (marked as ’A’ or ’MA’) and the other showing what their final marks were. This allows for averages and standard deviations of single major marks and double major marks to be calculated, respectively, and used in a t-test. To examine age, gender and field using t-tests, we also created tables showing, for example, whether each student was male or female and their corresponding final marks, and similar tables for the other two tests.

Since the student marks already exhibited a roughly normal distribution, it was not necessary to perform any further normalisation steps on the dataset.

Chapter 3 Statistical Analysis

The overall shape of the data was first examined with a frequency distribution, as seen below:

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Figure 3.1: Frequency distribution of final marks for all students

As we can see in Figure 3.1, the data follows a relatively normal distribution in terms of final marks for students.

3.1 Descriptive Statistics

In this section, we examined the central tendency, variation and shape of distribution of the data. We utilised built-in functions in SAS [SAS 2010a] in order to perform the calculations (Proc Means), as seen in the code in the appendix.

3.1.1 Central Tendency

The formula used by SAS to calculate the mean in the following section was as follows (from class notes):

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As we can see in Table 3.1 above, the average value of the final mark for all major stu­dents is 57.542. This means that, on average, a student was more likely to pass overall than have to repeat. The middle observation, or median, was 58. The most often oc­curring mark was 50. We also examined the values for the mean, median and mode of the tables after being split from the original data. For students that majored in both mathematics and applied mathematics, the average final mark across all subjects was 57.667, with a median of 57 and mode of 50. In contrast, for students who majored only in applied mathematics, the average final mark was found to be 59.071, which is slightly higher than that of the double major students.

We then looked at a further broken-down set of tables, broken into students that passed and students that repeated, for both single major and double major students. As described in the previous section, we grouped data according to index key, mean­ing averages of each student’s marks for all their subjects were taken. Thus, statistical measures were taken on these averages rather than the full dataset in the following segment. The central tendency statistics for these tables are displayed below:

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In Table 3.2 above, we see the mean, median and mode calculated for each portion of the dataset. An interesting point to note is the percentage of students that passed and failed for the sets of students that did a single major and those that did a double major. For double major students, only 35.465% of students passed their courses on the first try, while 64.535% of students repeated a subject at least once. In contrast, 58.371% of single students passed their courses in the first try, while 41.629% repeated at least once. Therefore single major students had a higher pass rate.

Table 3.3: Central Tendency Statistics Split According to Faculty, Sex, Age

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The mean final marks for engineering students and students under 25 years old were higher than their counterparts, as seen in Table 3.3. Other averages were relatively similar to each other.

3.1.2 Variation

The formulae used by SAS to calculate the variance and standard deviation in the fol­lowing section were as follows (from class notes):

Table 3.4: Variation Statistics for all Major Students

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As seen in Table 3.4, variances for all 3 groups were relatively high, indicating the final marks of students varied an average of around 14 points from each other.

We then grouped students together by their index keys, causing the marks for each of their subjects to need to be averaged.

Table 3.5: Variation Statistics for Different Major Combinations

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The result is smaller variances and ranges, as seen in Table 3.5. The average marks of first-time passing students generally had a greater variance.

Table 3.6: Variation Statistics for Different Faculties, Ages and Genders

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Variances and standard deviations were evenly spread when looking at different fac­ulties, ages and genders and their relationships with marks, as seen in Table 3.6.

3.1.3 Shape of the Distribution

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Table 3.7: Shape Statistics for All Majors

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For all 3 distributions in Table 3.7, the data was skewed slightly to the right (as given by the skewness values being negative) and had a relatively flat peak (kurtosis values being close to 0).

Table 3.8: Shape Statistics for Different Major Combinations

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For the double major students, for both first time passers and repeaters, the data was skewed heavily to the left (positive skewness values), with repeaters having a more pronounced peak than the first time passers (higher kurtosis value). Single major ap­plied student data (first time passers) was also skewed heavily to the left but had a much flatter peak, while single major repeaters had a more symmetrical distribution and a flat peak.

Table 3.9: Shape Statistics for Different Faculties, Ages and Genders

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All sets of data in Table 3.9 displayed heavy skewing to the left, as a result of positive skewness values. Peaks were mostly pronounced too, especially for female students, due to the high skewness values. Only engineering student data had a particularly flat peak.

3.2 Inferential Statistics

Once again, SAS functions (Proc Ttest [SAS 2010c] and Proc Freq with chisq option [SAS 2010b]) were used to do the calculations for our data.

3.2.1 Two-sample t-tests

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We performed two-sample t-tests on the data to determine whether there was a dif­ference in the average final marks for students in different situations. The general hypotheses were as follows:

- H0: There is no significant difference between the student mark group means.
- Ha: There is a significant difference between the student mark group means.

We started off by comparing the means of final marks between students that did a single major and students that did a double major. We took random samples, ap­proximately equal in size, from each group of data. The results are seen in Figure 3.2 below:

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Figure 3.2: Results of t-test for comparing single major and double major student marks

We selected the ’pooled’ method of t-value calculation, since the group variances are approximately equal. We discovered a t-value of -1.41. At a = 0.05, the critical values for a two-tail test are -1.96 and +1.96. Since our calculated value falls between these two values, we fail to reject the null hypothesis. There is no significant difference be­tween the average student marks between single major (indicated as ’A’ in the table) and double major students (indicated as ’MA’). However, even though the means are not different, this does not necessarily mean that the amounts of passes and repeats will be the same too, as we will see later.

We then looked to see if the mean marks between students in the science faculty and students from other faculties were different.

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Figure 3.3: Results of t-test for comparing marks between different faculties

As we can see, the discovered t-value is 3.27. This value falls outside the range of -1.96 to +1.96. As a result of this, we must reject the null hypothesis. This means that there is a relationship between the field or area the student is in and their average final mark. Students in other fields (indicated as ’O’ in the table) displayed a higher average mark compared to students in the science field (indicated as ’S’).

Next, we looked to see if the mean marks between students of different genders were different.

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Figure 3.4: Results of t-test for comparing marks between different genders

We found a t-value of 1.70. This falls inside the required range, leading us to fail to re­ject the null hypothesis. There is no significant difference between the average marks of students with different genders, and as such, gender has no meaningful effect on performance.

Lastly, we looked to see if the mean marks between students of different ages were different.

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Figure 3.5: Results of t-test for comparing marks between different ages

We found a t-value of 0.27. This falls within the required critical value range, meaning we fail to reject the null hypothesis. Students below 25 years of age (indicated by ’U’ in the table) did not have a significantly higher or lower average final mark compared to students 25 years and older (indicated by ’O’).

3.2.2 Chi-Square Tests

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The formula used by SAS to calculate the chi-square value in the following section was as follows (from class notes):

We first performed a chi square test to examine whether there was a relationship be­tween the combination of majors a student took and their final result. The hypotheses we intended to test were as follows:

- H0: There is no relationship between the majors a student takes and their final result.
- Ha: There is a relationship between the majors a student takes and their final result.

We created a contingency table from the data as seen in Table 3.10 below. We then ran the SAS chi-square test function on this contingency table. The results were as follows:

Table 3.10: Contingency Table for Math + Applied Math vs Applied Math Only

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Degrees of freedom: 1; Chi-squared value: 20.322

As we can see, the chi-square value was calculated to be 20.322 at 1 degree of freedom. At a degree of freedom 1 and a = 0.05, the critical value of chi-square is 3.841. Since the calculated value is greater than the critical value, we must reject the null hypothesis. This means that there is indeed a relationship between the majors a student takes and their result. When students only major in applied mathematics, they are significantly more likely to pass than if they are doing a double major. This difference is not due to chance.

We then wanted to investigate whether gender had any effect on chances of passing. Our hypotheses were as follows:

- H0: There is no relationship between the gender of a student and their final re­sult.
- Ha: There is a relationship between the gender of a student and their final result.

We created another contingency table from the data as seen in Table 3.11 below. We then ran the SAS chi-square test function on this contingency table once more. The results were as follows:

Table 3.11: Contingency Table for Male vs Female

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Degrees of freedom: 1; Chi-squared value: 0.132

The chi-square value was calculated to be 0.132 at 1 degree of freedom. At a degree of freedom 1 and a = 0.05, the critical value of chi-square is 3.841. Since the calculated value is smaller than the critical value, we fail to reject the null hypothesis. This means that there is no relationship between the gender of a student and their result. Students are statistically equally likely to pass or repeat whether they are male or female. Any differences are due to chance.

We then examined whether there was a relationship between the field the student is studying in and their result. We formulated the following hypotheses:

- H0: There is no relationship between the field of study of a student and their final result.
- Ha: There is a relationship between the field of study of a student and their final result.

We created a third contingency table from the data as seen in Table 3.12 below. We then ran the SAS chi-square test function on this contingency table again. The results were as follows:

Table 3.12: Contingency Table for Different Fields

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Degrees of freedom: 2; Chi-squared value: 30.419

The chi-square value was calculated to be 30.419 at 2 degrees of freedom. At a degree of freedom 2 and a = 0.05, the critical value of chi-square is 5.991. Since the calcu­lated value is greater than the critical value, we must reject the null hypothesis. This means that there is a relationship between the faculty or field the student is studying in and their result. Students in the engineering field were shown to be statistically more likely to pass without repeating than those in the commerce and science fields. This difference is not due to chance.

We then examined whether there was a relationship between the age of a student and their result. We formulated the following hypotheses:

- H0: There is no relationship between the age of a student and their final result.
- Ha: There is a relationship between the age of a student and their final result.

We created a final contingency table from the data as seen in Table 3.13 below. We then ran the SAS chi-square test function on this contingency table again. The results were as follows:

Table 3.13: Contingency Table for Different Ages of Students

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Degrees of freedom: 1; Chi-squared value: 1.217

The chi-square value was calculated to be 1.217 at 1 degree of freedom. At a degree of freedom 1 and a = 0.05, the critical value of chi-square is 3.841. Since the calculated value is less than the critical value, we fail to reject the null hypothesis. This means that there is no signficant relationship between how old the student is and their result.

Chapter 4 Delivery of Results

From our results, we determined that students taking only one major were more likely to perform better than those doing a double major. This may be due to the increased workload, which would not allow students to devote an appropriate amount of time to each subject. For this reason, it may be worthwhile to look at avenues that may help improve the chance of a double major student’s success. Extra tutorials, for ex­ample, could be introduced to assist any students who feel they are struggling and likely to repeat a course. Another possible option would be to reduce the number of assignments given out in each course, allowing students enough time to focus on everything.

We also noticed that students from different faculties had different chances of pass­ing. This may be due to students taking on other field-specific subjects and courses in addition to their mathematics courses, increasing their workload and giving them less time to prepare for everything. Students may need to be warned before they com­mence their studies that the amount of subjects they are taking on is too great, based on past students’ results.

4.1 Conclusion

Our investigation set out to determine if there was a relationship between the number of majors a student took and their likely achievement. We cleaned the data, remov­ing missing values and unnecessary attributes. We then transformed the data and performed statistical tests to acquire both descriptive and inferential statistics. Our investigation discovered the association between majors taken and results obtained. Students majoring only in applied maths were found to be more likely to pass than students doing a double major. In addition, students in different fields had different chances of succeeding. The University can now determine appropriate actions to take to increase its students’ performance.

Bibliography

[SAS 2010a] SAS. Base SAS(R) 9.2 Procedures Guide: Statistical Procedures, Third Edi­tion, 2010. Retrieved May 2016, from http://support.sas.com/documentation/ cdl/en/proc/61895/HTML/default/viewer.htm#a000146729.htm

[SAS 2010b] SAS. Base SAS(R) 9.2 Procedures Guide: Statistical Procedures, Third Edi­tion, 2010. Retrieved May 2016, from http://support.sas.com/documentation/ cdl/en/procstat/63104/HTML/default/viewer.htm#procstat_freq_sect027. htm

[SAS 2010c] SAS. SAS/STAT(R) 9.2 User’s Guide, Second Edition, 2010. Retrieved May 2016, from https://support.sas.com/documentation/cdl/en/statug/63033/ HTML/default/viewer.htm#ttest_toc.htm

[Wits 2016] Wits. University of the witwatersrand - explanation of result codes. Wits Examinations and Graduations Office, 2016.

Chapter 5 Appendix - Code Used

/* Nicholas Telford (671271) and Mike Nkongolo (1171635)

SAS code for Data Analysis Project Submitted: 16 June 2016 */

FILENAME REFFILE ’/folders/myfolders/Project_Data_2016 . xls ’;

PROC IMPORT DATAFILE=REFFILE DBMS=XLS

OUT=WORK. PROJDATA;

GETNAMES=YES;

RUN;

PROC CONTENTS DATA=WORK. PROJDATA; RUN;

/* Clean the dataset */

Data PData (Drop=Program_Code Drop= Program_Title Drop=Unit_Title Drop=Unit_Attempt_Status); Set Projdata;

If cmiss(of _all_) Then Delete;

If Final_Mark > 100 Then Delete;

If Final_Grade = 'FAB' Then Delete;

If Final_Grade = ’FNQ’ Then Delete;

Run ;

Proc Sort Data = PData; By Index_Key;

Run;

/* Create empty tables */

Data AllYears;

Set PData ;

If index (Unit_Code , ’APPM’) Then Delete; If index (Unit_Code , ’MATH’) Then Delete;

Run ;

Data AppmOnly;

Set PData;

If index (Unit_Code, ’APPM’) Then Delete; If index (Unit_Code, ’MATH’) Then Delete;

Run;

D ata MathAppm ;

Set PData ;

If index (Unit_Code, ’APPM’) Then Delete; If index (Unit_Code, ’MATH’) Then Delete;

Run ;

Data AllStudents;

Set PData;

If index(Unit_Code, ’APPM’) Then Delete; If index(Unit_Code, ’MATH’) Then Delete;

Run;

Abbildung in dieser Leseprobe nicht enthalten

If index(Unit_Code, ’APPM’) Then Delete;

If index(Unit_Code, ’MATH’) Then Delete;

Data AppmOnlyPass (Drop= Calendar_Instance_Year Drop= Year_Of_Study Drop= Unit_Code); Set PData ;

If index(Unit_Code, ’APPM’) Then Delete;

If index(Unit_Code, ’MATH’) Then Delete;

Run ;

Data AppmOnlyRepeat (Drop= Calendar_Instance_Year Drop= Year_Of_Study Drop= Unit_Code); Set PData ;

If index(Unit_Code, ’APPM’) Then Delete;

If index(Unit_Code, ’MATH’) Then Delete;

Run ;

Data MathAppmPass (Drop= Calendar_Instance_Year Drop= Year_Of_Study Drop= Unit_Code); Set PData ;

If index(Unit_Code, ’APPM’) Then Delete;

If index(Unit_Code, ’MATH’) Then Delete;

Run ;

Data MathAppmRepeat (Drop= Calendar_Instance_Year Drop= Year_Of_Study Drop= Unit_Code); Set PData ;

If index(Unit_Code, ’APPM’) Then Delete;

If index(Unit_Code, ’MATH’) Then Delete;

Run ;

/* Perform data transformations and table creation */ Proc SQL;

Insert into AllYears

Select * from PData as pi

Where pi.Index_Key in (

Select p2.Index_Key from PData as p2 Where p2. Year_Of_Study = ’YOS 1’

)

and pi.Index_Key in (

Select p3.Index_Key from PData as p3 Where p3. Year_Of_Study = ’YOS 2’

)

and pi.Index_Key in (

Select p4.Index_Key from PData as p4 Where p4. Year_Of_Study = ’YOS 3’

);

Insert into MathAppm Select * From AllYears as ai Where ai . Index_Key in (

Select a2.Index_Key from AllYears as a2 Where a2.Unit_Code like ’MATH3%’

)

and a1.Index_Key in (

Select a3.Index_Key from AllYears as a3 Where a3. Unit_Code like ’APPM3%’

);

Insert into AppmOnly Select * From AllYears as a1 Where a1. Index_Key not in (

Select a2.Index_Key from AllYears as a2 Where a2.Unit_Code like ’MATH3%’

)

and a1.Index_Key in (

Select a3.Index_Key from AllYears as a3 Where a3. Unit_Code like ’APPM3%’

)；

Insert into AllStudents Select * From MathAppm Union

Select * From AppmOnly;

/* Group students by their index keys */

Insert into MathAppmGrouped

Select Distinct Area, Index_Key, Age, Sex, Avg(Final_Mark) , ’ Pass ’ From MathAppm as a1 Where a1. Index_Key not in (

Select a2.Index_Key from MathAppm as a2 Where a2. Final_Grade like ’F%’

)

Group By a1. Index_Key;

Insert into MathAppmGrouped

Select Distinct Area, Index_Key, Age, Sex, Avg(Final_Mark) , ’Repeat’ From MathAppm as a1 Where a1. Index_Key in (

Select a2.Index_Key from MathAppm as a2 Where a2. Final_Grade like ’F%’

)

Group By a1. Index_Key;

Insert into AppmOnlyGrouped

Select Distinct Area, Index_Key, Age, Sex, Avg(Final_Mark) , ’ Pass ’ From AppmOnly as a1 Where a1. Index_Key not in (

Select a2.Index_Key from AppmOnly as a2 Where a2 . Final_Grade like ’F%’

)

Group By a1. Index_Key;

Insert into AppmOnlyGrouped

Select Distinct Area, Index_Key, Age, Sex, Avg(Final_Mark) , ’Repeat’ From AppmOnly as a1 Where a1. Index_Key in (

Select a2.Index_Key from AppmOnly as a2 Where a2 . Final_Grade like ’F%’

)

Group By a1. Index_Key;

Run ;

Proc Sort Data = MathAppmGrouped; By Index_Key;

Run ;

Proc Sort Data = AppmOnlyGrouped; By Index_Key;

Run ;

Proc SQL;

/‘Create and insert into summary table where students are grouped by their index keys, marked as ’Pass’ or ’Repeat’ and as ’MathAppm’ or ’AppmOnly’ */

Create Table SummaryTbl (

Index_Key int ,

Area varchar ,

Age varchar ,

Sex varchar ,

MajorType varchar,

FinalMark num,

Result varchar )；

Insert into MathAppmPass

Select * From MathAppmGrouped as a1

Where a1. Index_Key in (

Select a2.Index_Key from MathAppmGrouped as a2 Where a2. Final_Grade like ’P%’

)；

Insert into MathAppmRepeat Select * From MathAppmGrouped as a1 Where a1. Index_Key in (

Select a2.Index_Key from MathAppmGrouped as a2 Where a2. Final_Grade like ’R%’

)；

Insert into AppmOnlyPass

Select * From AppmOnlyGrouped as a1

Where a1. Index_Key in (

Select a2.Index_Key from AppmOnlyGrouped as a2 Where a2. Final_Grade like ’P%’

)；

Insert into AppmOnlyRepeat Select * From AppmOnlyGrouped as a1 Where a1. Index_Key in (

Select a2.Index_Key from AppmOnlyGrouped as a2 Where a2. Final_Grade like ’R%’

)；

Abbildung in dieser Leseprobe nicht enthalten

If index(Result, ’Repeat’) Then Delete;

);

Insert into FemaleStudents Select * From SummaryTbl as a1 Where a1. Index_Key in (

Select a2.Index_Key from SummaryTbl as a2 Where a2. Sex = ’F’

);

Insert into AgeUnder25 Select * From SummaryTbl as a1 Where a1 . Index_Key in (

Select a2.Index_Key from SummaryTbl as a2

Where a2.Age like ’1%’ or a2.Age = ’20’ or a2.Age = ’21’ or a2.Age = ’22’ or a2.Age = ’23’ or a2.Age = ’24’

)；

Insert into Age25Up

Select * From SummaryTbl as a1

Where a1 . Index_Key in (

Select a2.Index_Key from SummaryTbl as a2

Where a2.Age = ’25’ or a2.Age = ’26’ or a2.Age = ’27’ or a2.Age = ’28’ or a2.Age = ’29’ or a2.Age like ’3%’ or a2.Age like ’4%’ or a2.Age like ’5%’ )；

Create Table SummaryTblAgeTests (

Index_Key int ,

Area varchar ,

Age varchar ,

Sex varchar ,

MajorType varchar,

FinalMark num,

Result varchar )；

Insert Into SummaryTblAgeTests

Select s1 . Index_Key, s1.Area, ’Under25’, s1.Sex, s1 . MajorType, s1 .FinalMark, s1 . Result From SummaryTbl as s1 Where s1 . Index_Key in (

Select s2. Index_Key From AgeUnder25 as s2 )；

Insert Into SummaryTblAgeTests

Select s1 . Index_Key , s1.Area, ’25Up’, s1.Sex, s1 . MajorType , s1 .FinalMark, s1 . Result From SummaryTbl as s1 Where s1 . Index_Key in (

Select s2 . Index_Key From Age25Up as s2 )；

Run;

Proc Sort Data = SummaryTblAgeTests; By Index_Key;

Run ;

ods noproctitle ;

ods graphics / imagemap=on;

/* produce frequency distribution */ proc univariate data=WORK.ALLSTUDENTS; ods select Histogram; var Final_Mark; histogram Final_Mark;

run ;

/* Calculate descriptive statistics */

Proc Means Data = AllStudents Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis;

Var Final_Mark;

Title Overall;

Run ;

Proc Means Data = MathAppm Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis;

Var Final_Mark;

T i t l e MathAppm ;

Run ;

Proc Means Data = AppmOnly Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis;

Var Final_Mark;

Title AppmOnly;

Run ;

Proc Means Data = MathAppmPass Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis; Var Final_Mark;

Title MathAppmPass ;

Run ;

Proc Means Data = MathAppmRepeat Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis ; Var Final_Mark;

Title MathAppmRepeat;

Run ;

Proc Means Data = AppmOnlyPass Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis; Var Final_Mark;

T i t l e AppmOnlyPass ;

Run ;

Proc Means Data = AppmOnlyRepeat Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis ; Var Final_Mark;

Title AppmOnlyRepeat;

Run ;

Proc Means Data = ScienceFac Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis;

Var FinalMark;

Title ScienceFac;

Run ;

Proc Means Data = CommerceFac Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis;

Var FinalMark;

T i t l e CommerceFac ;

Run ;

Proc Means Data = EngineeringFac Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis ; Var FinalMark;

Title EngineeringFac;

Run ;

Proc Means Data = MaleStudents Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis; Var FinalMark;

Title MaleStudents;

Run ;

Proc Means Data = FemaleStudents Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis ; Var FinalMark;

Title FemaleStudents;

Run ;

Proc Means Data = AgeUnder25 Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis;

Var FinalMark;

Title AgeUnder25 ;

Run ;

Proc Means Data = Age25Up Mean Stddev Var Q1 Q3 Median Mode Max Min Range Skewness Kurtosis;

Var FinalMark;

T i t l e Age25Up ;

Run;

/* Perform chi—squared tests */

Proc Freq Data = SummaryTbl;

Tables MajorType * Result / chisq;

Title ’Chi—squared test for MathAppm vs AppmOnly’;

Run ;

Proc Freq Data = SummaryTbl;

Tables Sex * Result / chisq;

Title ’Chi—squared test for Male vs Female’;

Run ;

Proc Freq Data = SummaryTbl;

Tables Area * Result / chisq;

Title ’Chi—squared test for Different Fields’;

Run ;

Proc Freq Data = SummaryTblAgeTests;

Tables Age * Result / chisq;

Title ’Chi—squared test for Different Age’;

Run ;

/* Create sample data from population for two—tailed t—tests */ Proc SQL;

Create Table TwoPops (Major char (5),

Final_Mark num);

Insert into TwoPops

Select ’A’ , Final_Mark from AppmOnly Where Index_Key < 7202;

Insert into TwoPops

Select ’MA’, Final_Mark from MathAppm Where Index_Key < 6246;

Create Table FacTest (Area char (5) ,

Final_Mark num);

Insert into FacTest

Select ’S’, FinalMark from ScienceFac Where Index_Key < 8232;

Insert into FacTest

Select ’O’ , FinalMark from CommerceFac Where Index_Key < 15064;

Insert into FacTest

Select ’O’ , FinalMark from EngineeringFac Where Index_Key < 10308;

Create Table SexTest (Sex char (5) ,

Final_Mark num);

Insert into SexTest

Select ’M’ , FinalMark from MaleStudents Where Index_Key < 7313;

Insert into SexTest

Select ’F’ , FinalMark from FemaleStudents Where Index_Key < 9838;

Create Table AgeTest (Age char (5) ,

Final_Mark num);

Insert into AgeTest

Select ’U’ , FinalMark from AgeUnder25 Where Index_Key < 9472;

Insert into AgeTest

Select ’O’ , FinalMark from Age25Up

Where Index_Key < 11389;

Run;

/* Perform two-tailed t—tests */

Proc TTest Data=TwoPops Sides=2 Alpha=0.05 h0 = 0;

Title "Two sample t—test for student marks";

Class Major;

Var Final_Mark;

Run ;

Proc TTest Data=FacTest Sides=2 Alpha=0.05 h0 = 0;

Title "Two sample t—test for student marks and faculties"; Class Area;

Var Final_Mark;

Run ;

Proc TTest Data=SexTest Sides=2 Alpha=0.05 h0 = 0;

Title "Two sample t—test for student marks and genders"; Class Sex;

Var Final_Mark;

Run ;

Proc TTest Data=AgeTest Sides=2 Alpha=0.05 h0 = 0;

Title "Two sample t—test for student marks and ages";

Class Age ;

Var Final_Mark;

Run ;

/* Proc Corr Data = SummaryTbl fisher; */ /* Var Age FinalMark; */

/* Run; */