Chapter 5 t-test

Chapter 5: Comparing two means using the t-test
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Upon completion of this chapter, you should be able to:
 explain what is the t-Test and its use in hypothesis testing
 demonstrate using the t-Test for INDEPENDENT MEANS
 identify the assumptions for using the t-test
 demonstrate the use of the t-Test for DEPENDENT MEANS
CHAPTER OVERVIEW
 What is the t-test?
 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
Chapter 6: Oneway Analysis of Variance
Chapter 7: Correlation
Chapter 8: Chi-Square
This chapter introduces you to the t-test which is statistical tool used to test the significant
differences between the means of two groups. The independent t-test is used when the
means of two groups when the sample is drawn from two different or independent
samples. The dependent or pairwise t-test is used when the sample is tested twice the
means are compared.

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What is the T-Test?
The t-test was developed by a statistician, W.S.
Gossett (1878-1937) who worked in a brewery in Dublin,
Ireland. His pen name was ‘student’ and hence the term
‘student’s t-test’ which was published in the scientific
journal, Biometrika in 1908. The t-test is a statistical tool
used to infer differences between small samples based on
the mean and standard deviation.
In many educational studies, the researcher is
interested in testing the differences between means on
some variable. The researcher is keen to determine
whether the differences observed between two samples
represents a real difference between the populations from
which the samples were drawn. In other words, did the observed difference just
happen by chance when, in reality, the two populations do not differ at all on the
variable studies.
or example, a teacher wanted to find out whether the Discovery method of
teaching science to primary school children was more effective than the Lecture
method. She conducted an experiment among 70 primary school children of which 35
pupils were taught using the Discovery method and 35 children were taught using the
Lecture method. The results of the study showed that subjects in the Discovery group
scored 43.0 marks while subjects in the Lecture method group score 38.0 marks on a
the science test. Yes, the Discovery group did better than the Lecture group. Does the
difference between the two groups represent a real difference or was it due to
chance? To answer this question, the t-test is often used by researchers.

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The Hypothesis Tested Using the T-Test
How do we go about establishing whether the differences in the two means are
statistically significant or due to chance? You begin by formulating a hypothesis
about the difference. This hypothesis states that the two means are equal or the
difference between the two means is zero and is called the null hypothesis.
Using the null hypothesis, you begin testing the significance by saying:
"There is no difference in the score obtained in science between subjects in the
Discovery group and the Lecture group".
More commonly the null hypothesis may be stated as follows:
a) Ho : U1 = U2 which translates into 43.0 = 38.0
b) Ho : U1 ─ U2 = 0 which translate into 43.0 ─ 38.0 = 0
 If you reject the null hypothesis, it means that the difference between the two
means have statistical significance
 If you do not reject the null hypothesis, it means that the difference between
the two means are NOT statistically significant and the difference is due to
chance.
Note:
For a null hypothesis to be accepted, the difference between the two means need not
be equal to zero since sampling may account for the departure from zero. Thus, you
can accept the null hypothesis even if the difference between the two means is not
zero provided the difference is likely to be due to chance. However, if the difference
between the two means appears too large to have been brought about by chance, you
reject the null hypothesis and conclude that a real difference exists.
LEARNING ACTIVITY
a) State TWO null hypothesis in your area of interest
that can be tested using the t-test.
b) What do you mean when you reject or do not reject
the null hypothesis?

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Using the T-Test for INDEPEDENT MEANS
The t-test is a powerful statistic that enables you to determine that the
differences obtained between two groups is statistically significant. When two groups
are INDEPENDENT of each other; it means that the sample drawn came from two
populations. Other words used to mean that the two groups are independent are
"unpaired" groups and "unpooled” groups.
a) What is meant by Independent Means or Unpaired Means?
Say for example you conduct a study to determine the spatial reasoning ability
of 70 ten-year old children in Malaysia. The sample consisted of 35 males and 35
females. See figure 5.1. The sample of 35 males was drawn from the population of ten
year old males in Malaysia and the sample of 35 females was drawn from the
population of ten year olds females in Malaysia.
Note that they are independent samples because they come from two completely
different populations.
Figure 5.1 Samples drawn from two independent populations
Population of ten year old
MALES in Malaysia
Population of ten year old
FEMALES in Malaysia
Sample of 35 MALES Sample of 35 FEMALES
Research Question:
"Is there a significant difference in spatial reasoning between male and female ten
year old children?"
Null Hypothesis or Ho:
"There is no significant difference in spatial reasoning between male and female
ten year old children"

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b) Formula for the Independent T-Test
Note that the formula for the t-test shown below is a ratio. It is Group 1 mean (i.e.
males) minus Group 2 mean (i.e. females) divided by the Standard Error multiplied by
Group 1 mean minus Group 2 mean.
Computation of the Standard Error
Use the formula below. To compute the standard error (SE), you take the variance
(i.e. standard deviation squared) for Group 1 and divide it by the number of subjects
in that group minus "1". Do the same for Group 2. Than add these two values and take
the square root.
The top part of the equation is the
difference between the two means
The bottom part of the equation is
the Standard Error (SE) which is a
measure of the variability of
dispersion of the scores.
This is the formula for the
Standard Error:
Combine the two
formulas and you get
this version of the t-test
formula:

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b) Example:
The results of the study are as follow:
Let's try using the formula:
t =
2
= ------------- = 4.124
0.485
Note:
The t-value will be positive if the mean for Group I is larger or more than (>) the
mean of Group 2 and negative if it is smaller or less than (<).
c) What do you do after computing the t-value?
Once you compute the t-value (which is 4.124) you look up the t-value in The
Student's t-test Probabilities or The Table of Critical Values for Student’s T-Test
which tells us whether the ratio is large enough to say that the difference between the
groups is significant. In other words the difference observed is not likely due to
chance or sampling error.
 Alpha Level: As with any test of significance, you need to set the alpha level.
In most educational and social research, the "rule of thumb" is to set the alpha
level at .05. This means that 5% of the time (five times out of a hundred) you
would find a statistically significant difference between the means even if
there is none ("chance").
 Degrees of Freedom: The t-test also requires that we determine the degrees of
freedom (df) for the test. In the t-test, the degrees of freedom is the sum of the
subjects or persons in both groups minus 2. Given the alpha level, the df, and
the t-value, you look up in the Table (available as an appendix in the back of
12 -10
4.0 1
0.1177 + 0.1177
4.0 2
(35-1) (35-1)
2
=
+

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most statistics texts) to determine whether the t-value is large enough to be
significant.
d) Look up in the Table of Critical Values for Student's t-test shown on the right:
The df is 70 minus 2 = 68. You take the nearest df which is 70 and read the
column for the two-tailed alpha of 0.050. See Table 5.1.
The t-value you obtained is 4.124. The critical value shown is 1.677. Since,
the t-values is greater than the critical value of 1.677, you Reject Ho and conclude
that the difference between the means for the two groups is different. In other words,
males scored significantly higher than females on the spatial reasoning test.
However, you do not have to go through this tedious process, as statistical
computer programs such as SPSS, provides the significance test results, saving you
from looking them up in the Table of Critical Values.
Table 5. 1: Table of Critical Values for Student's t-test
Tailed
Two 0.250 0.100 0.050 0.025 0.010 0.005
One 0.500 0.200 0.100 0.050 0.020 0.010
df
30 0.683 1.310 1.697 2.042 2.457 2.750
40 0.681 1.303 1.684 2.021 2.423 2.704
50 0.679 1.299 1.676 2.009 2.403 2.678
60 0.679 1.296 1.671 2.000 2.390 2.660
70 0.678 1.294 1.667 1.994 2.381 2.648
80 0.678 1.292 1.664 1.990 2.374 2.639
90 0.677 1.291 1.662 1.987 2.368 2.632
100 0.677 1.290 1.660 1.984 2.364 2.626
100 0.674 1.282 1.645 1.960 2.326 2.576

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Assumptions that Must be Observed when Using the T-Test
While the t-test has been described as a robust statistical tool, it is based on a
model that makes several assumptions about the data that must be met prior to
analysis. Unfortunately, students conducting research tend not to report whether their
data meet the assumptions of the t-test. These assumptions need are be observed,
because the accuracy of your interpretation of the data depends on whether
assumptions are violated. The following are three main assumptions that are generic
to all t-tests.
 Instrumentation (Scale of Measurement)
The data that you collect for the dependent variable should be based on an
instrument or scale that is continuous or ordinal. For example, scores that
you obtain from a 5-point Likert scale; 1,2,3,4,5 or marks obtained in a
mathematics test, the score obtained on an IQ test or the score obtained on
an aptitude test.
 Random Sampling
The sample of subjects should be randomly sampled from the population
of interest.
 Normality
The data come from a distribution that has one of those nice bell-shaped
curves known as a normal distribution. Refer to Chapter 3: The Normal
Distribution which provides both graphical and statistical methods for
assessing normality of a sample or samples.
 Sample Size
Fortunately, it has been shown that if the sample size is reasonably large,
quite severe departures from normality do not seem to affect the
LEARNING ACTIVITY
a) Would you reject Ho if you had set the alpha at 0.01 for a
two-tailed test?
b) When do you use the one-tailed test and two-tailed t-test?

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conclusions reached. Then again what is a reasonable sample size? It has
been 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 in
size, you can be confident that the t-test will be a good, strong tool for
getting the correct conclusions. Statisticians say that the t-test is a "robust"
test. Departure from normality is most serious when sample sizes are
small. As sample sizes increase, the sampling distribution of the mean
approaches a normal distribution regardless of the shape of the original
population.
 Homogeneity of Variance.
It has often been suggested by some researchers that homogeneity of
variance or equality of variance is actually more important than the
assumption of normality. In other words, are the standard deviations of the
two groups pretty close to equal? Most statistical software packages
provide a "test of equality of variances" along with the results of the t-test
and the most common being Levene's test of homogeneity of variance
(see Table 5.2).
Levene's Test 95% Confidence
of Equality Interval
of Variances
F Sig t d Sign. Mean Std. Error Upper Lower
Two-tail Difference Difference
Equal
Variances 3.39 .080 .848 20 .047 1.00 1.18 -1.46 3.46
Assumed
Unequal
Variances .848 16.70 .049 1.00 1.18 -1.49 3.40
Assumed
Table 5.2 Levene’s Test of Equality of Variances
Begin by putting forward the null hypothesis that:
"There are no significant differences between the variances of the two
groups" and you set the significant level at .05.

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If the Levene statistic is significant, i.e. LESS than .05 level (p < .05), then the null
hypothesis is:
 REJECTED and one accepts the alternative hypothesis and conclude that
the VARIANCES ARE UNEQUAL. [The unequal variances in the SPSS
output 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 conclude
that the VARIANCES ARE EQUAL. [The equal variances in the SPSS
output is used]
The Levene test is robust in the face of departures from normality. The Levene's test
is 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 more
traditional tests like Bartlett's test.
Let’s examine an EXAMPLE:
In the CoPs Project, an Inductive Reasoning scale consisting of 11 items was
administered to 946 eighteen year. One of the research questions put forward is:
 "Is there a significant difference between in inductive reasoning between
male and female subjects"?
 To establish the statistical significance of the means of these two groups, the t-
test was used. Using SPSS.
LEARNING ACTIVITY
Refer to the table above. Based on the Levene’s Test of
Homogeneity of variance, what is your conclusion. Explain.

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THE SPSS STEPS to answer the Research Question.
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 see
that the 495 females in the sample scored 8.99 while the 451 males had a mean score
of 7.95 on inductive reasoning. The standard deviation for the males is 3.46 while
that for the females is 3.14. The scores for the females are less dispersed compared to
the males.
SPSS PROCEDURES for the independent groups t-test:
1. Select the Analyze menu.
2. Click on Compare Means and then Independent-
Samples T Test ....to open the Independent Samples
T Test dialogue box.
3. Select the test variable(s). [i.e. Inductive Reasoning] and
then click on the button to move the variables into
the Test Variables(s): box
4. Select the grouping variables [i.e. gender] and click on
the button to move the variable into the Grouping Variable:
box
5. Click on the Define Groups ....command pushbutton to
open 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 the
variables [i.e. 2 for 'females'] in the Group 2: box.
7. Click on Continue and then OK.

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GROUP STATISTICS
The question remains: Is this sample difference in inductive reasoning large
enough to convince us that there it is a real significant difference in inductive
reasoning ability between the population 18 year old females and the population of 18
year-old males?
Output #2:
Let’s examine this output in two parts:
First is to determine that the data meet the "Homogeneity of Variance" assumption
you can use the Levene's Test and set the alpha at 0.05. The alpha obtained is 0.054
which is greater (>) than 0.05 and you do not Reject the Ho: and conclude that the
variances are equal. Hence, you have not violated the "Homogeneity of Variance"
assumption.
Levene's Test 95% Confidence
of Equality Interval
of Variances
F Sig t d Sign. Mean Std. Error Upper Lower
Two-tail Difference Difference
Equal
Variances 4.720 .030 -4.875 944 .000 -1.0468 -2.147 -1.4682 -.6254
Assumed
Unequal
Variances -4.853 911.4 .049 -1.0468 -2.146 -1.4701 -.6234
Assumed
INDUCTIVE N Mean Std. Deviation Std. Error Mean
GENDER Male 451 7.9512 3.4618 2.345
Female 495 8.9980 3.1427 3.879

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SECOND is to examine the following:
 The SPSS output below displays the results of the t-test to test whether or not
the difference between the two sample means is significantly different from
zero.
 Remember the null hypothesis states that there is no real difference between
the means (Ho: X1 = X2).
 Any observed difference just occurred by chance.
Interpretation:
t-value
This "t" value tells you how far away from 0, in terms of the number of standard
errors, the observed difference between the two sample means falls. The "t" value is
obtained by dividing the difference in the Means ( - 1.0468) by the Std. Error (-.2147)
which is equal to - 4.875
p-value
If the p-value as shown in the "sig (2 tailed) column is smaller than your chosen alpha
level you do not reject the null hypothesis and argue that there is a real difference
between the populations. In other words, we can conclude, that the observed
difference between the samples is statistically significant.
Mean Difference
This is the difference between the means (labelled "Mean Difference"); i.e. 7.9512 –
8.9980 = – 1.0468.

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Using the T-Test for Dependent Means
The Dependent means t-test or the Paired t-test or the Repeated measures
t-test is used when you have data from only one group of subjects. i.e. each subject
obtains two scores under different conditions. For example, when you give a pre-test
and 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 some
intervention or manipulation obtain a score on the posttest. Your objective is to
determine whether the difference between means for the two sets of scores is the same
or different.
You want to find answers to the following:
Research Questions:
 Is there a significant difference in pretest and posttest scores in mathematics
for subjects taught using visualisation techniques?
Null Hypotheses:
 There is no significant difference between the pretest and the posttest scores in
mathematics for subjects taught using visualisation techniques.
Treatment: Students taught using
Visualisation techniques
Note: The pretest and posttest should be similar or equivalent
PRETEST
POSTTEST

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The top part of the equation is the sum of the
difference between the two means divided by ‘n’
or the number of subjects
FORMULA OF THE DEPENDENT t-TEST
d
t =
sd
n
Let’s look at an EXAMPLE where the formula is applied:
A researcher wanted to determine if teaching 12 year children memory techniques
improved their performance in science. Randomly selected 12 year olds were trained
in memory techniques for two weeks and the results of the study is shown in the table
below:
Student Science
Pretest
Science
Posttest
Paired
difference
d d ²
1 12 18 6 36
2 10 14 4 16
3 15 19 4 16
4 9 15 6 36
5 11 14 3 9
6 13 17 4 16
7 14 16 2 4
8 11 13 2 4
9 10 16 4 16
10 9 12 3 9
Ʃd = 38 Ʃ d ² = 162
 Standard deviation = 0.443
 Ʃ d = 38
 Ʃ d ² = 162
The 4th
column in the table above shows the difference, d, between the science pretest
and the science posttest scores for each of the 10 students sampled. You refer to each
The bottom part of the equation is the
Standard Deviation (sd) which is a measure of
the variability of dispersion of the scores divided
by the square root of ‘n’ or the number of
subjects.

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difference 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 paired
difference of d = 18 – 12 = 6 marks, an increase in 6 marks as a result of the memory
techniques training.
 If the null hypothesis is true, the paired differences between the pretest and
the 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-VALUE
Step 1:
You begin by computing the d
d = Ʃ (posttest score – pretest score ) = 38 = 3.80
number of students 10
Step 2:
Next is to compute the value of sd.
sd =
= 1.399
Step 3:
Applying the t-test for Dependent Means formula:
d 3.80
t = = = 8.589
sd 1.399 √ 10
√ n
Ʃd² ─ (Ʃd)² ∕ n
n – 1
162 ─ (38)² ∕ n
n – 1

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Excerpt of the Table of Critical Values for Student's t-test
Tailed
Two 0.100 0.050 0.025 0.010 0.005
One 0.200 0.100 0.050 0.020 0.010
df
9 1.383 1.833 2.262 2.821 3.250
10 1.372 1.812 2.228 2.764 3.169
11 1.363 1.796 2.201 2.718 3.106
12 1.356 1.782 2.179 2.681 3.055
Step 4:
Having computed the t-value (which is 8.589) you look up the t-value in The Table
of Critical Values for Student's t-test or The Table of Significance which tells us
whether the ratio is large enough to say that the difference between the groups is
significant. In other words the difference observed is not likely due to chance or
sampling error.
Alpha Level:
The researcher set the alpha level at 0.05. This means that 5% of the time (five out of
a hundred) you would find a statistically significant difference between the means
even if there is none ("chance").
Degrees of Freedom:
The t-test also requires that we determine the degrees of freedom (df) for the test. In
the t-test, the degrees of freedom is the sum of the subjects or persons which is 10
minus 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 the
t-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 is
1.833 (one tailed). Hence, the null hypothesis [Ho:] is Rejected and Ha: is accepted

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which states Mean 1 > than Mean 2. It can be concluded that the difference between
the 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 memory
techniques.
Again, you do not have to go through this tedious process, as statistical
computer programs such as SPSS, provides the significance test
results, 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 the
formula. Researchers who have Dependent Samples fail to recognise this fact,
and inappropriately apply the t-test for Independent Groups to test the
hypothesis that X¹ = X² = 0. If an inappropriate Independent Groups t-test is
performed with Dependent Groups the standard error will be greatly
overestimated and significant differences between the two means may be
considered "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 to
Dependent Groups t-test. Thus, when using the t-test, you need to recognise
and distinguish Independent and Dependent samples.
Using SPSS: T-Test for Dependent Means
EXAMPLE:
In a study, a researcher was keen to determine if teaching note-taking techniques
improved achievement in history. A sample of 22 students selected for the study and
taught note-taking techniques for a period of 4 weeks. The research questions put
forward is:
"Is there a significant difference in performance in history before and after the
treatment?" i.e. You wish to determine whether the difference between the means
for the two sets of score is the same or different.
To establish the statistical significance of the means obtained on the pretest
and posttest, the dependent-samples or paired t-test was used.

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Data was collected from the same group of subjects on both conditions and
each subject obtains a score on the pretest, and after the treatment (or intervention or
manipulation), a score on the posttest.
Ho: U1 = U2 or Ha: U1 = U2
You will notice that the syntax for the Independent Groups t-test is different from that
of the Dependent groups t-test. In the case of the Independent Groups t-test you have
a grouping variable so you can distinguish between Group 1 and Group 2 whereas
this is not found with the Dependent groups t -test.
The following are the SPSS OUTPUTS:
Paired Sample Statistics
HISTORY TEST N Mean Std. Deviation Std. Error Mean
Pair Pretest 40 43.15 12.97 2.05
Posttest 40 63.98 13.16 2.08
The ‘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 the
posttest mean (43.15) indicating improved performance in the history test after the
SPSS 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] and
then press the button to move the variables into
the Paired Variables: box
4. Click on Continue and then OK.

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treatment. The standard deviation for the pretest 2.05 and is very close to the standard
deviation for the posttest which is 2.08.
The question remains: Is this mean difference large enough to convince us that there
it is a real significant difference in performance in history a consequence of teaching
note taking techniques)?
Paired Differences
Mean Std . Std. Error t df Sig. (2 tailed)
Difference Deviation Mean Lower Upper
Pair Pretest -20.83 15.65 2.47 -25.83 -15.82 -8.43 39 .000
Posttest
t-Value
This "t" value tells you how far away from 0, in terms of the number of standard
errors, the observed difference between the two sample means falls. The "t" value is
obtained by dividing the Mean difference ( - 20.83) by the Std. Error (2.47) which is
equal to – 8.43.
p-value
The p-value shown in the "sig (2 tailed) column is smaller than your chosen alpha
level (0.05) and so you Reject the null hypothesis and argue that there is a real
difference between the pretest and posttest.
In other words, we can conclude, that the observed difference between the two
means is statistically significant.
Mean Difference
This is the difference between the means 43.15 – 63.98 = – 20.83 which students did
significantly better on the posttest.

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ATTITUDE N Mean Std. Deviation Std. Error Mean
Pair Pretest 22 8.50 3.33 .71
Posttest 22 13.86 2.75 .59
Paired Differences
Mean Std . Std. Error t df Sig.
Deviation Mean Lower Upper (2 tailed)
Pair Pretest -5.36 2.90 .62 -6.65 -4.08 -8.66 21 .000
1 Posttest
LEARNING ACTIVITY
t-Test for Dependent Means or Groups
T-test
Page 3 of 5
CASE STUDY 1:
In a study, a researcher was interested in finding out
whether attitude towards science would be enhanced when
students are taught science using the Inquiry Method. A
sample of 22 students were administered an attitude toward
science scale before the experiment. The treatment was
conducted for one semester and after which the same
attitude scale was administered to the same group of
students.

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ANSWER 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 to
the 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).

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GENDER
N Mean Std. Deviation Std. Error Mean
Male 1966 6.9410 2.2858 5.155E-02
Female 2438 6.8351 2.4862 5.035E-02
Levene's Test t-test for
for Equality Equality of
of Variances Means
F Sig. t df Sig. Mean Std. Error
2-tailed Difference Difference
Equal 19.408 .000 1.456 4402 .145 .1059 7.271E-02
Equal 1.469 4327 .142 .1059 7.206E-02
LEARNING ACTIVITY
t-Test for Independent Means or Groups
T-test
CASE STUDY 2:
A researcher was interested in finding out about the
creative thinking skills of secondary school students. He
administered a 10 item creative thinking to a sample of
4400 sixteen year old students drawn from all over
Malaysia
e 3 of 5

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ANSWER 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 to
the 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) who
worked in a brewery in Dublin, Ireland.
 Researchers are keen to determine whether the differences observed between
two samples represents a real difference between the populations from which
the samples were drawn.
 The t-test is a powerful statistic that enables you to determine that the
differences obtained between two groups is statistically significant.
 When two groups are INDEPENDENT of each other; it means that the sample
drawn came from two populations. Other words used to mean that the two
groups are independent are "unpaired" groups and "unpooled” groups..
 In most educational and social research, the "rule of thumb" is to set the alpha
level at .05. This means that 5% of the time (five times out of a hundred) you
would find a statistically significant difference between the means even if
there is none ("chance").

25
 This "t" value tells you how far away from 0, in terms of the number of
standard errors, the observed difference between the two sample means falls.
 The Dependent means t-test or the Paired t-test or the Repeated measures t-test
is used when you have data from only one group of subjects. i.e. each subject
obtains two scores under different conditions.
KEY WORDS:
 T-test
 Independent groups
 Dependent groups
 Paired groups
 t-value
 Levene’s test
 Critical values
 Alpha level
 Degress of freedom
 One tailed
 Two tailed
 Null hypothesis
 Alternative hypothesis

Chapter 5 t-test

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