** Student's t-Test for Independent Samples**

**© 1998 by Dr. Thomas W. MacFarland -- All Rights Reserved**

************* student_t.doc ************* Background: Student's t-test is a very common (and possibly overused) test for determining differences between two groups. The t-test (developed in 1915 by Gosset for the Guinness Breweries of Dublin) is the appropriate test for small samples, as opposed to samples with greater than 30 or more observations. And recall: -- Student's t-test is still the appropriate test with greater than 30 observations. -- With n => 30 observations, t approximates z. When using Student's t-test to determine if the difference between two groups is indeed a true difference, or if the difference between the two groups is due only to chance: -- Both groups should approximate normal distribution. -- Random selection should ideally be used for all members of each of the two groups. Scenario: This study examines if there are differences in final examination test scores between two groups of students in a data structures programming course: -- One group of students was taught by Computer Based Training. -- The other group of students was taught by traditional lecture. Students were all from a university freshman- level data structures course (using C++ as the programming language platform) who were assigned, through random selection, to placement into one of two groups: instruction by CBT (Computer Based Training) vs. instruction by traditional lecture. Because the teacher was confident that final examination scores represented interval data ( i.e., the data are parametric, with the difference between "89" and "90" equal to the difference between "75" and "76"), Student's t-Test for Independent Samples was correctly judged to be the appropriate test for this analysis of differences between two groups. Test scores for both groups of students are summarized in Table 1. Table 1 Final Examination Test Scores in a Data Structures Course by Student Group: Students Taught by Computer Based Training and Students Taught by Traditional Lecture ==================================================== Teaching Method =============== 1 = CBT Student Number 2 = Traditional Final Score ---------------------------------------------------- 01 1 089 02 1 081 03 1 092 04 1 094 05 1 074 06 1 056 07 1 077 08 1 085 09 1 078 10 1 069 11 1 089 12 1 088 13 1 045 14 1 083 15 1 095 16 2 091 17 2 057 18 2 089 19 2 083 20 2 080 21 2 083 22 2 091 23 2 084 24 2 084 25 2 094 26 2 096 27 2 088 28 2 097 29 2 091 30 2 094 ---------------------------------------------------- Ho: Null Hypothesis: There is no difference in final examination test scores between students in a data structures course taught by Computer Based Training and their counterparts who were taught by the use of traditional lecture (p <= .05). Files: 1. studen_t.doc 2. studen_t.dat 3. studen_t.r01 4. studen_t.o01 5. studen_t.con 6. studen_t.lis Command: At the Unix prompt (%), key: %spss -m < studen_t.r01 > studen_t.o01 ************ studen_t.dat ************ 01 1 089 02 1 081 03 1 092 04 1 094 05 1 074 06 1 056 07 1 077 08 1 085 09 1 078 10 1 069 11 1 089 12 1 088 13 1 045 14 1 083 15 1 095 16 2 091 17 2 057 18 2 089 19 2 083 20 2 080 21 2 083 22 2 091 23 2 084 24 2 084 25 2 094 26 2 096 27 2 088 28 2 097 29 2 091 30 2 094 ************ studen_t.r01 ************ SET WIDTH = 80 SET LENGTH = NONE SET CASE = UPLOW SET HEADER = NO TITLE = Student's t-Test for Independent Samples COMMENT = This file examines if there are differences in final examination test scores between two groups of students in a data structures programming course: one group of students was taught by Computer Based Training and the other group of students was taught by traditional lecture. Students were all from a university freshman- level data structures course (using C++ as the programming language platform) who were assigned, through random selection, to placement into one of two groups: instruction by CBT (Computer Based Training) vs. instruction by traditional lecture. Because the teacher was confident that final examination scores represented interval data ( i.e., the data are parametric, with the difference between "89" and "90" equal to the difference between "75" and "76"), Student's t-Test for Independent Samples was correctly judged to be the appropriate test for this analysis of differences between two groups. DATA LIST FILE = 'studen_t.dat' FIXED / Stu_Code 20-21 Method 35 Score 50-52 Variable Labels Stu_Code "Student Code" / Method "Method: CBT vs. Lecture" / Score "Final Examination Score" Value Labels Method 1 'CBT: Computer Based Training' 2 'Traditional Lecture' T-TEST GROUPS = Method(1,2) / VARIABLES = Score ************ studen_t.o01 ************ 1 SET WIDTH = 80 2 SET LENGTH = NONE 3 SET CASE = UPLOW 4 SET HEADER = NO 5 TITLE = Student's t-Test for Independent Samples 6 COMMENT = This file examines if there are differences 7 in final examination test scores between two 8 groups of students in a data structures 9 programming course: one group of students 10 was taught by Computer Based Training and the 11 other group of students was taught by 12 traditional lecture. 13 14 Students were all from a university freshman- 15 level data structures course (using C++ as the 16 programming language platform) who were assigned, 17 through random selection, to placement into one 18 of two groups: instruction by CBT (Computer 19 Based Training) vs. instruction by traditional 20 lecture. Because the teacher was confident that 21 final examination scores represented interval 22 data ( i.e., the data are parametric, with the 23 difference between "89" and "90" equal to the 24 difference between "75" and "76"), Student's 25 t-Test for Independent Samples was correctly 26 judged to be the appropriate test for this 27 analysis of differences between two groups. 28 DATA LIST FILE = 'studen_t.dat' FIXED 29 / Stu_Code 20-21 30 Method 35 31 Score 50-52 32 This command will read 1 records from studen_t.dat Variable Rec Start End Format STU_CODE 1 20 21 F2.0 METHOD 1 35 35 F1.0 SCORE 1 50 52 F3.0 33 Variable Labels 34 Stu_Code "Student Code" 35 / Method "Method: CBT vs. Lecture" 36 / Score "Final Examination Score" 37 38 Value Labels 39 Method 1 'CBT: Computer Based Training' 40 2 'Traditional Lecture' 41 42 T-TEST GROUPS = Method(1,2) 43 / VARIABLES = Score T-TEST requires 72 bytes of workspace for execution. t-tests for Independent Samples of METHOD Method: CBT vs. Lecture Number Variable of Cases Mean SD SE of Mean ----------------------------------------------------------------------- SCORE Final Examination Score CBT: Computer Base 15 79.6667 14.130 3.648 Traditional Lecture 15 86.8000 9.748 2.517 ----------------------------------------------------------------------- Mean Difference = -7.1333 Levene's Test for Equality of Variances: F= 1.768 P= .194 t-test for Equality of Means 95% Variances t-value df 2-Tail Sig SE of Diff CI for Diff ------------------------------------------------------------------------------- Equal -1.61 28 .119 4.432 (-16.213,1.946) Unequal -1.61 24.87 .120 4.432 (-16.265,1.998) ------------------------------------------------------------------------------- ************ studen_t.con ************ Outcome: Computed t = | - 1.61 | Criterion t = + or - 2.05 (alpha = .05, df = 28) Computed t |-1.61| < Criterion t |-2.05| Note. The | and | characters are used to indicate absolute value. Therefore, the null hypothesis is accepted and it can be claimed that there is no difference (p <= .05) in final examination test scores between students in a data structures course taught by Computer Based Training and their counterparts who were taught by the use of traditional lecture. Any difference between the two groups is due only to chance. The p value is another way to view differences in the three graded activities: -- The calculated p value is .119. -- The delcared p value is .05. The calculated p value exceeds the declared p value and there is, accordingly, no difference between the two groups in terms of scores on the final examination. At p <= .05 any differences in test scores that exist are due only to chance. ************ studen_t.lis ************ % minitab MTB > outfile 'studen_t.lis' Collecting Minitab session in file: studen_t.lis MTB > # MINITAB addendum to studen_t.dat MTB > read 'studen_t.dat' c1 c2 c3 Entering data from file: studen_t.dat 30 rows read. MTB > name c1 'Stu_Code' c2 'Method' c3 'Score' MTB > print 'Stu_Code' 'Method' 'Score' ROW Stu_Code Method Score 1 1 1 89 2 2 1 81 3 3 1 92 4 4 1 94 5 5 1 74 6 6 1 56 7 7 1 77 8 8 1 85 9 9 1 78 10 10 1 69 11 11 1 89 12 12 1 88 13 13 1 45 14 14 1 83 15 15 1 95 16 16 2 91 17 17 2 57 18 18 2 89 Continue? y 19 19 2 83 20 20 2 80 21 21 2 83 22 22 2 91 23 23 2 84 24 24 2 84 25 25 2 94 26 26 2 96 27 27 2 88 28 28 2 97 29 29 2 91 30 30 2 94 MTB > # With MINITAB, it is possible to conduct Student's t-Test MTB > # with stacked and unstacked data. MTB > # MTB > # I will unstack the data in c3 and then conduct the MTB > # t-Test using both methods. MTB > # MTB > unstack (c2-c3) into (c5-c6) (c7-c8); SUBC> subscripts c2. MTB > print c1-c8 ROW Stu_Code Method Score C5 C6 C7 C8 1 1 1 89 1 89 2 91 2 2 1 81 1 81 2 57 3 3 1 92 1 92 2 89 4 4 1 94 1 94 2 83 5 5 1 74 1 74 2 80 6 6 1 56 1 56 2 83 7 7 1 77 1 77 2 91 8 8 1 85 1 85 2 84 9 9 1 78 1 78 2 84 10 10 1 69 1 69 2 94 11 11 1 89 1 89 2 96 12 12 1 88 1 88 2 88 13 13 1 45 1 45 2 97 14 14 1 83 1 83 2 91 15 15 1 95 1 95 2 94 16 16 2 91 17 17 2 57 18 18 2 89 Continue? y 19 19 2 83 20 20 2 80 21 21 2 83 22 22 2 91 23 23 2 84 24 24 2 84 25 25 2 94 26 26 2 96 27 27 2 88 28 28 2 97 29 29 2 91 30 30 2 94 * NOTE * One or more variables are undefined. MTB > histogram c6 Histogram of C6 N = 15 Midpoint Count 45 1 * 50 0 55 1 * 60 0 65 0 70 1 * 75 2 ** 80 2 ** 85 2 ** 90 4 **** 95 2 ** MTB > histogram c8 Histogram of C8 N = 15 Midpoint Count 55 1 * 60 0 65 0 70 0 75 0 80 1 * 85 4 **** 90 5 ***** 95 4 **** MTB > describe c6 c8 N MEAN MEDIAN TRMEAN STDEV SEMEAN C6 15 79.67 83.00 81.15 14.13 3.65 C8 15 86.80 89.00 88.31 9.75 2.52 MIN MAX Q1 Q3 C6 45.00 95.00 74.00 89.00 C8 57.00 97.00 83.00 94.00 MTB > # MTB > # And now notice how I conduct the t-Test on stacked data. MTB > # MTB > twot data in c3 groups in c2 TWOSAMPLE T FOR Score Method N MEAN STDEV SE MEAN 1 15 79.7 14.1 3.6 2 15 86.80 9.75 2.5 95 PCT CI FOR MU 1 - MU 2: ( -16.3, 2.0) TTEST MU 1 = MU 2 (VS NE): T= -1.61 P=0.12 DF= 24 MTB > # MTB > # And now notice how I conduct the t-Test on unstacked data. MTB > # MTB > twosamplet c6 c8 TWOSAMPLE T FOR C6 VS C8 N MEAN STDEV SE MEAN C6 15 79.7 14.1 3.6 C8 15 86.80 9.75 2.5 95 PCT CI FOR MU C6 - MU C8: ( -16.3, 2.0) TTEST MU C6 = MU C8 (VS NE): T= -1.61 P=0.12 DF= 24 MTB > stop -------------------------- Disclaimer: All care was used to prepare the information in this tutorial. Even so, the author does not and cannot guarantee the accuracy of this information. The author disclaims any and all injury that may come about from the use of this tutorial. As always, students and all others should check with their advisor(s) and/or other appropriate professionals for any and all assistance on research design, analysis, selected levels of significance, and interpretation of output file(s). The author is entitled to exclusive distribution of this tutorial. Readers have permission to print this tutorial for individual use, provided that the copyright statement appears and that there is no redistribution of this tutorial without permission. Prepared 980316 Revised 980914 end-of-file 'studen_t.ssi'