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9Chapter 5: Comparing two means using the t-testconclusions reached. Then again what is a reasonable sample size? It hasbeen argued that as long as you have enough people in each group(typically greater or equal to 30 cases) and the groups are close to equal insize, you can be confident that the t-test will be a good, strong tool forgetting the correct conclusions. Statisticians say that the t-test is a "robust"test. Departure from normality is most serious when sample sizes aresmall. As sample sizes increase, the sampling distribution of the meanapproaches a normal distribution regardless of the shape of the originalpopulation. Homogeneity of Variance.It has often been suggested by some researchers that homogeneity ofvariance or equality of variance is actually more important than theassumption of normality. In other words, are the standard deviations of thetwo groups pretty close to equal? Most statistical software packagesprovide a "test of equality of variances" along with the results of the t-testand the most common being Levene's test of homogeneity of variance(see Table 5.2).Levene's Testof Equalityof med95% ConfidenceIntervalFSigtdSign.Two-tailMeanStd. Error UpperDifference .84816.70.0491.001.18-1.493.40Table 5.2 Levene’s Test of Equality of VariancesBegin by putting forward the null hypothesis that:"There are no significant differences between the variances of the twogroups" and you set the significant level at .05.

10Chapter 5: Comparing two means using the t-testIf the Levene statistic is significant, i.e. LESS than .05 level (p .05), then the nullhypothesis is: REJECTED and one accepts the alternative hypothesis and conclude thatthe VARIANCES ARE UNEQUAL. [The unequal variances in the SPSSoutput is used] If the Levene statistic is not significant, i.e. MORE than .05 level (p .05),then you DO NOT REJECT (or Accept) the null hypothesis and concludethat the VARIANCES ARE EQUAL. [The equal variances in the SPSSoutput is used]The Levene test is robust in the face of departures from normality. The Levene's testis based on deviations from the group mean. SPSS provides two options'; i.e. "homogeneity of variance assumed" and"homogeneity of variance not assumed" (see Table below). The Levene test is more robust in the face of non-normality than moretraditional tests like Bartlett's test.LEARNING ACTIVITYRefer to the table above. Based on the Levene’s Test ofHomogeneity of variance, what is your conclusion. Explain.Let’s examine an EXAMPLE:In the CoPs Project, an Inductive Reasoning scale consisting of 11 items wasadministered to 946 eighteen year. One of the research questions put forward is: "Is there a significant difference between in inductive reasoning betweenmale and female subjects"? To establish the statistical significance of the means of these two groups, the ttest was used. Using SPSS.

11Chapter 5: Comparing two means using the t-testTHE SPSS STEPS to answer the Research Question.SPSS PROCEDURES for the independent groups t-test:1. Select the Analyze menu.2. Click on Compare Means and then IndependentSamples T Test .to open the Independent SamplesT Test dialogue box.3.Select the test variable(s). [i.e. Inductive Reasoning] andthen click on the button to move the variables intothe Test Variables(s): box4.Select the grouping variables [i.e. gender] and click onthe button to move the variable into the Grouping Variable:box5. Click on the Define Groups .command pushbutton toopen the Define Groups sub-dialogue box.6. In the Group 1: box, type the lowest value for the variable[i.e. 1 for 'males'], then tab. Enter the second value for thevariables [i.e. 2 for 'females'] in the Group 2: box.7.Click on Continue and then OK.SPSS OUTPUTS:Output #1:The ‘Group statistics’ table above reports that the mean values on the variable(inductive reasoning) for the two different groups (males and females). Here, we seethat the 495 females in the sample scored 8.99 while the 451 males had a mean scoreof 7.95 on inductive reasoning. The standard deviation for the males is 3.46 whilethat for the females is 3.14. The scores for the females are less dispersed compared tothe males.

12Chapter 5: Comparing two means using the t-testGROUP 128.9980Std. Deviation Std. Error Mean3.46183.14272.3453.879The question remains: Is this sample difference in inductive reasoning largeenough to convince us that there it is a real significant difference in inductivereasoning ability between the population 18 year old females and the population of 18year-old males?Output #2:Let’s examine this output in two parts:First is to determine that the data meet the "Homogeneity of Variance" assumptionyou can use the Levene's Test and set the alpha at 0.05. The alpha obtained is 0.054which is greater ( ) than 0.05 and you do not Reject the Ho: and conclude that thevariances are equal. Hence, you have not violated the "Homogeneity of Variance"assumption.Levene's Testof Equalityof umedSig4.720 .03095% ConfidenceIntervaltdSign.Two-tail-4.875944-4.853 911.4MeanStd. Error UpperDifference 1.0468-2.146-1.4701-.6234

13Chapter 5: Comparing two means using the t-testSECOND is to examine the following: The SPSS output below displays the results of the t-test to test whether or notthe difference between the two sample means is significantly different fromzero. Remember the null hypothesis states that there is no real difference betweenthe means (Ho: X1 X2). Any observed difference just occurred by chance.Interpretation:t-valueThis "t" value tells you how far away from 0, in terms of the number of standarderrors, the observed difference between the two sample means falls. The "t" value isobtained by dividing the difference in the Means ( - 1.0468) by the Std. Error (-.2147)which is equal to - 4.875p-valueIf the p-value as shown in the "sig (2 tailed) column is smaller than your chosen alphalevel you do not reject the null hypothesis and argue that there is a real differencebetween the populations. In other words, we can conclude, that the observeddifference between the samples is statistically significant.Mean DifferenceThis is the difference between the means (labelled "Mean Difference"); i.e. 7.9512 –8.9980 – 1.0468.

14Chapter 5: Comparing two means using the t-testUsing the T-Test for Dependent MeansThe Dependent means t-test or the Paired t-test or the Repeated measurest-test is used when you have data from only one group of subjects. i.e. each subjectobtains two scores under different conditions. For example, when you give a pre-testand after a particular treatment or intervention you give the same subjects a post-test.In this form of design, the same subjects obtain a score on the pretest and, after someintervention or manipulation obtain a score on the posttest. Your objective is todetermine whether the difference between means for the two sets of scores is the sameor different.You want to find answers to the following:Research Questions: Is there a significant difference in pretest and posttest scores in mathematicsfor subjects taught using visualisation techniques?Null Hypotheses: There is no significant difference between the pretest and the posttest scores inmathematics for subjects taught using visualisation techniques.Treatment: Students taught usingVisualisation techniquesPRETESTNote: The pretest and posttest should be similar or equivalentPOSTTEST

15Chapter 5: Comparing two means using the t-testFORMULA OF THE DEPENDENT t-TESTThe top part of the equation is the sum of thedifference between the two means divided by ‘n’or the number of subjectsdt sdThe bottom part of the equation is theStandard Deviation (sd) which is a measure ofthe variability of dispersion of the scores dividedby the square root of ‘n’ or the number ofsubjects.nLet’s look at an EXAMPLE where the formula is applied:A researcher wanted to determine if teaching 12 year children memory techniquesimproved their performance in science. Randomly selected 12 year olds were trainedin memory techniques for two weeks and the results of the study is shown in the 4567891012101591113141110918141915141716131612 Standard deviation 0.443 Ʃ d 38 Ʃ d ² 162Paireddifferenced6446342243Ʃd 38d²3616163691644169Ʃ d ² 162The 4th column in the table above shows the difference, d, between the science pretestand the science posttest scores for each of the 10 students sampled. You refer to each

16Chapter 5: Comparing two means using the t-testdifference as a paired difference because it is the difference of a pair of observations.For example, student #1 got 12 on the pretest and 18 on the posttest, giving a paireddifference of d 18 – 12 6 marks, an increase in 6 marks as a result of the memorytechniques training. If the null hypothesis is true, the paired differences between the pretest andthe posttest for the 10 students sampled should average about 0 (zero).If the paired differences is greater than zero, the null hypothesis is false.STEPS IN THE COMPUTATION OF THE T-VALUEStep 1:You begin by computing the dd Ʃ (posttest score – pretest score ) 38number of students 3.8010Step 2:Next is to compute the value of sd.Ʃd² sd (Ʃd)² n162 n–1 n–11.399Step 3:Applying the t-test for Dependent Means formula:dt 3.80 sd n(38)² n 1.399 108.589

17Chapter 5: Comparing two means using the t-testExcerpt of the Table of Critical Values for Student's t-testTailedTwo 0.100One .055Step 4:Having computed the t-value (which is 8.589) you look up the t-value in The Tableof Critical Values for Student's t-test or The Table of Significance which tells uswhether the ratio is large enough to say that the difference between the groups issignificant. In other words the difference observed is not likely due to chance orsampling error.Alpha Level:The researcher set the alpha level at 0.05. This means that 5% of the time (five out ofa hundred) you would find a statistically significant difference between the meanseven if there is none ("chance").Degrees of Freedom:The t-test also requires that we determine the degrees of freedom (df) for the test. Inthe t-test, the degrees of freedom is the sum of the subjects or persons which is 10minus 1 9. Given the alpha level, the df, and the t-value, you look up in the Table(available as an appendix in the back of most statistics texts) to determine whether thet-value is large enough to be significant.Step 5:The t-value obtained is 8.589 which is greater than the critical value shown which is1.833 (one tailed). Hence, the null hypothesis [Ho:] is Rejected and Ha: is accepted

18Chapter 5: Comparing two means using the t-testwhich states Mean 1 than Mean 2. It can be concluded that the difference betweenthe means is different. In other words, there is overwhelming evidence that a "gain"has taken place on the science posttest as a result of training students on memorytechniques.Again, you do not have to go through this tedious process, as statisticalcomputer programs such as SPSS, provides the significance testresults, saving you from looking them up in a table.Note: Misapplication of the Formula A common error made by some research students is the misapplication of theformula. Researchers who have Dependent Samples fail to recognise this fact,and inappropriately apply the t-test for Independent Groups to test thehypothesis that X¹ X² 0. If an inappropriate Independent Groups t-test isperformed with Dependent Groups the standard error will be greatlyoverestimated and significant differences between the two means may beconsidered "non-significant" (Type 1 Error). The opposite error, mistaking non-significant differences for significant ones(Type 2 Error), may be made if the Independent Groups t-test is applied toDependent Groups t-test. Thus, when using the t-test, you need to recogniseand distinguish Independent and Dependent samples.Using SPSS: T-Test for Dependent MeansEXAMPLE:In a study, a researcher was keen to determine if teaching note-taking techniquesimproved achievement in history. A sample of 22 students selected for the study andtaught note-taking techniques for a period of 4 weeks. The research questions putforward is:"Is there a significant difference in performance in history before and after thetreatment?" i.e. You wish to determine whether the difference between the meansfor the two sets of score is the same or different.To establish the statistical significance of the means obtained on the pretestand posttest, the dependent-samples or paired t-test was used.

19Chapter 5: Comparing two means using the t-testData was collected from the same group of subjects on both conditions andeach subject obtains a score on the pretest, and after the treatment (or intervention ormanipulation), a score on the posttest.Ho: U1 U2or Ha: U1 U2SPSS PROCEDURES for the dependent groups t-test:1. Select the Analyze menu.2. Click on Compare Means and then Paired-Samples T Test.to open the Paired-Sample T Test dialogue box.3.Select the test variable(s). [i.e. History Test] andthen press thebutton to move the variables intothe Paired Variables: box4.Click on Continue and then OK.You will notice that the syntax for the Independent Groups t-test is different from thatof the Dependent groups t-test. In the case of the Independent Groups t-test you havea grouping variable so you can distinguish between Group 1 and Group 2 whereasthis is not found with the Dependent groups t -test.The following are the SPSS OUTPUTS:Paired Sample StatisticsHISTORY TESTPairPretestPosttestN4040Mean Std. Deviation Std. Error Mean43.1512.972.0563.9813.162.08The ‘Paired sample statistics’ table above reports that the mean values on the variable(history test) for the pretest and posttest. The posttest mean is higher (63.98) than theposttest mean (43.15) indicating improved performance in the history test after the

20Chapter 5: Comparing two means using the t-testtreatment. The standard deviation for the pretest 2.05 and is very close to the standarddeviation for the posttest which is 2.08.The question remains: Is this mean difference large enough to convince us that thereit is a real significant difference in performance in history a consequence of teachingnote taking techniques)?Paired DifferencesMeanStd .Std. ErrorDifference DeviationMeanPair PretestPosttest-20.8315.652.47tLower-25.83dfSig. (2 tailed)Upper-15.82 -8.4339.000t-ValueThis "t" value tells you how far away from 0, in terms of the number of standarderrors, the observed difference between the two sample means falls. The "t" value isobtained by dividing the Mean difference ( - 20.83) by the Std. Error (2.47) which isequal to – 8.43.p-valueThe p-value shown in the "sig (2 tailed) column is smaller than your chosen alphalevel (0.05) and so you Reject the null hypothesis and argue that there is a realdifference between the pretest and posttest.In other words, we can conclude, that the observedmeans is statistically significant.difference between the twoMean DifferenceThis is the difference between the means 43.15 – 63.98 – 20.83 which students didsignificantly better on the posttest.

21Chapter 5: Comparing two means using the t-testLEARNING ACTIVITYt-Test for Dependent Means or GroupsT-testPage 3 of 5CASE STUDY 1:In a study, a researcher was interested in finding outwhether attitude towards science would be enhanced whenstudents are taught science using the Inquiry Method. Asample of 22 students were administered an attitude towardscience scale before the experiment. The treatment wasconducted for one semester and after which the sameattitude scale was administered to the same group 013.86Std. Deviation Std. Error Mean3.332.75.71.59Paired DifferencesMeanPair Pretest -5.361 PosttestStd .Deviation2.90Std. ErrorMeanLower.62-6.65tdfUpper-4.08-8.6621Sig.(2 tailed).000

22Chapter 5: Comparing two means using the t-testANSWER THE FOLLOWING QUESTIONS:1. State a null hypothesis for the above study.2. State an alternative hypothesis for the above study.3. Briefly describe the 'Paired Sample Statistics' table with regards tothe means and variability of scores.4. What is the conclusion of the null hypothesis stated in (1).5. What is the conclusion of the alternative hypothesis stated in (2).

23Chapter 5: Comparing two means using the t-testLEARNING ACTIVITYt-Test for Independent Means or GroupsT-testCASE STUDY 2:A researcher was interested in finding out about thecreative thinking skills of secondary school students. Headministered a 10 item creative thinking to a sample of4400 sixteen year old students drawn from all overMalaysiae 3 of 5GENDERNStd. DeviationStd. Error .48625.035E-02Levene's Testfor Equalityof VariancesEqualEqualMeant-test forEquality ailedMeanDifference.145.142.1059.1059Std. ErrorDifference7.271E-027.206E-02

24Chapter 5: Comparing two means using the t-testANSWER THE FOLLOWING QUESTIONSE:1. State a null hypothesis for the above study.2. State an alternative hypothesis for the above study.3. Briefly describe the 'Group Statistics' table with regards tothe means and variability of scores.4. Is there evidence for homogeneity of variance? Explain.5. What would you do if the significance level is 0.053?6. What is the conclusion of the null hypothesis stated in (1).7. What is the conclusion of the alternative hypothesis stated in (2).SUMMARY The t-test was developed by a statistician, W.S. Gossett (1878-1937) whoworked in a brewery in Dublin, Ireland. Researchers are keen to determine whether the differences observed betweentwo samples represents a real difference between the populations from whichthe samples were drawn. The t-test is a powerful statistic that enables you to determine that thedifferences obtained between two groups is statistically significant. When two groups are INDEPENDENT of each other; it means that the sampledrawn came from two populations. Other words used to mean that the twogroups are independent are "unpaired" groups and "unpooled” groups. In most educational and social research, the "rule of thumb" is to set the alphalevel at .05. This means that 5% of the time (five times out of a hundred) youwould find a statistically significant difference between the means even ifthere is none ("chance").

25Chapter 5: Comparing two means using the t-test This "t" value tells you how far away fr

The hypothesis tested using the t-test Using the t-test for independent means Assumptions that must be observed when using the t-test Summary Key Terms References Chapter 1: Introduction Chapter 2: Descriptive Statistics Chapter 3: The Normal Distribution Chapter 4: Hypothesis Testing Chapter 5: T-test