Lecture slides to increase understanding of statistics by all category of researchers

T-Tests and ANOVA
Olaniyi Taiwo

Disclaimer
• This presentation contains materials adapted or referenced from
various online sources. These materials are included solely for
educational and informational purposes. I do not claim ownership or
original authorship of the content sourced from external references,
and full credit is due to the respective authors and creators of these
works.

Research
Title and
Question
Literature
Review
Research
Hypo
Approach
and
Designs
Sample
size
Sampling
Variables
and
Measure-
ment
Data
collection
Data
Analysis
Report /
Presentati
ons
RESEARCH
PROCESS

Variables
• A variable is a characteristic that may assume more than one set of
values to which a numerical measure can be assigned
• Variable is a characteristic that changes.
• This differs from a constant, which does not change but remains the
same.
Quantitative
Continuous
Discrete
Qualitative
Ordinal
Nominal

Measurement Scales
• Measurement scales are used to quantify numeric values of variables
or observations
There are 4 types of measurement scales:
• Nominal Scale
• Ordinal Scale
• Interval Scale
• Ratio Scale

The Hierarchy of Measurement Scales
Nominal
Interval
Ratio
Attributes are only named; weakest
Attributes can be ordered
Distance is meaningful
Absolute zero
Ordinal

Types of Data
Data
Quantitative
or Numeric
Continuous Data Discrete Data
Qualitative or
Categorical
Nominal Data Ordinal Data

Inferential Statistics
▪ Inferential statistics allows researchers to make decisions or
inferences by interpreting data patterns
▪ Inferential statistics can be classified as either Estimation or
Hypothesis testing.

Lecture slides to increase understanding of statistics by all category of researchers

Correlation
• A correlation typically evaluates three aspects of the relationship:
• the direction
• the form
• the degree
• How to study correlation
• Scatter Diagram Method
• Karl Pearson’s Correlation
• Spearman’s Rank correlation coefficient
• Correlation Coefficient (r) and Coefficient of determination (r2)

Chi-Square
• Chi-square analysis is primarily used to deal with categorical
(frequency) data
• Steps in tests of Hypothesis
1. Determine the appropriate test
2. Establish the level of significance: α
3. Formulate the statistical hypothesis
4. Calculate the test statistic
5. Determine the degree of freedom
6. Compare computed test statistic against a tabled/critical value
2
2 ( )
O E
E

−
= 

Today
• Understanding and interpreting T-tests
• Understanding and interpreting ANOVA

Objectives
• Determine when the t-test is appropriate to use
• Discuss the assumptions for t-tests
• Select the appropriate form of t-test for use in a given situation
• Understand the role of the F-statistic in determining which
independent sample t-test to use in SPSS
• Interpret t-test analyses

• Parametric test used to compare means of two group
• Data must be continuous, and the variable(s) of interest must be
normally distributed
• Compares the means of two groups; it cannot compare more than
two
• So when your research question requires the comparison of more
than two groups the t-test is not the correct test to use

• Please note that the t-test can also be used with only one group
• Used with small samples (≤ 30) - theoretical.
• For large samples (N>100) can use z to test hypotheses about means
• The null hypothesis can be expressed as
H0 = x1 = x2, or that the means of the two groups are equal

• The t-test yields a t-statistic, which can help you determine the
confidence intervals associated with a given statistical hypothesis
about the mean of a population (or about the means of two
populations)

Assumptions
• The distribution must be close to normal (so the data are continuous)
• The grouping variable must be categorical and dichotomous

• t statistic is used to estimate the magnitude of difference in the
means of two groups; the bigger the t statistic, the greater the
difference and the more significant the difference
• To have a mean, the variable must be at least approximately
normal; therefore, categorical variables, whether nominal or
ordinal, cannot be compared using the t-test

• t statistic helps determine confidence intervals associated with a
statistical hypothesis about the mean of a population or the means
of two populations
• significance is based on the confidence interval
• sample size influences the interpretation of the t statistic, but not
the interpretation of the 95% confidence interval

Variations of the t-test
• One Sample t-test
• Paired Samples t-test or Dependent t-test
• Independent Samples t-test

One Sample t-test
• Used to evaluate whether the population mean of a test variable
is different from a constant
• A major decision in using the one sample t-test is choosing the
test value (or constant)
• The test value usually represents a neutral point
• Those above the constant are given a label, those below are also
given a label while those on the neutral point are without a label

• The test value or constant may be:
• Midpoint on the test variable
• Average value of test variable based on past research
• Chance level of performance of the test variable

Formula for one sample t-test
• (x̄) = The sample mean.
• (μ) = The population mean.
• (s) = The sample standard
deviation
• (n) = Number of
observations

Class example
• Prof. Morenike wants to improve the number of scientific publications
by members of MWH WhatsApp platform. In the past among MWH
members, data indicate that the average paper publication was 100
per year. After a training on research methodology and data analysis,
the recent publication data (taken from a sample of 25 participants)
indicates an average publication of 130, with a standard deviation of
15. Did the training work? Test your hypothesis at a 5% alpha level.

• Step 1: Write your null hypothesis statement
• Step 2: Write your alternate hypothesis. This is the one you’re
testing. You think there is a difference (that the mean publication
increased), so: H1: x > 100. (one-tail) or H1: x ≠ 100 (two-tail)
• Step 3: Identify the following information you’ll need to calculate the
test statistic. The question should give you these items:
• The sample mean(x̄). This is given in the question as 130.
• The population mean(μ). Given as 100 (from past data).
• The sample standard deviation(s) = 15.
• Number of observations(n) = 25.

• Step 4: Insert the items from above into the t score formula.
• Step 5: Find the t-table value. You need two values to find this:
• The alpha level: given as 5% in the question.
• The degrees of freedom, which is the number of items in the sample (n - 1)
: 25 – 1 = 24.
• Step 6: Find the t-table value for one and two tails
• Decision rule (If calculated t is greater than Table t then reject Ho)
• Interpret

The sample mean(x̄). This is given in the question as 130.
The population mean(μ). Given as 100 (from past data).
The sample standard deviation(s) = 15.
Number of observations(n) = 25.

SPSS example
• Dr Lawal wants to test a research question about the knowledge of
mathematics in a group of students in a class in a high-brow
environment. He hypothesizes that the average math score of these
students is higher than the average for their age and class in the
population. He samples a 100 students in the school at the high brow
area, gives a standard mathematics exam and scores them
• So there are 100 students and 100 scores
• He then conducts a one-sample t-test. (using a test value of 50)

SPSS demonstration with SPSS exam data set
One-Sample Statistics
N Mean Std. Deviation Std. Error
Mean
Scores of Mathematics
exam
100 59.765 21.6848 2.1685
One-Sample Test
Test Value = 50
t df Sig. (2-tailed) Mean
Difference
95% Confidence Interval of
the Difference
Lower Upper
Scores of
Mathematics Exam
4.503 99 .000 9.7650 5.462 14.068

Paired (correlated/dependent) Sample t-Test
• Each case must have scores on two variables for a paired sample t-test
• The paired sample t-test evaluates whether the mean of the difference
between these two variables is different from the population
• It is applicable to two types of studies
• Repeated measures
• With intervention
• Without intervention
• Matched subjects designs
• With intervention
• Without intervention

• For repeated measure design, a participant is assessed on two
occasions or under two conditions on a single measure
• Repeated tests of the same person: This can be weights measured
before and after 6 months for people on a diet (intervention)
• Knowledge before and after for people data analysis class
(intervention).
• A group of people are asked to compare their preference – (on a
measured scale) for healthcare if they prefer Doctor A or B OR Buhari
and Tinubu (no intervention)
• Most of the characteristics between measurements are constant,
such as gender and ethnicity

• For matched subject design, or matched-pairs design uses two
treatment conditions and pairs subjects based on common variables
• Example include parents following their ward or child to a clinic and
being accessed on their experience – like fear scale (no intervention)
• Investigating whether husbands and wives with infertility problems
feel equally anxious using the Infertility Anxiety Measure (no
intervention)

• You want to check the effects of a new BP drug A. You match 40
selected people with similar BP readings into pairs (20 pairs). A
member of the pair is randomized to used drug A, the other uses
existing drug B over 3 months (intervention)
• Each participant pair is a case and has scores on two variables – the
score obtained by one participant under one condition and the score
obtained by the other participant under the other condition

• Assumptions:
• The distribution is close to normal (so the data are continuous)
• Grouping variable is categorical and dichotomous
• Groups are not independent
• Examples of groups that are not independent:
• Repeated tests of the same individual
• Different parts of the same individual (two legs, two eyes)
• Identical twins with identical genes

• Different parts of the same person: This could be comparing drops
using 2 eyes, or lotion using 2 legs, or left side to the right side of
the mouth.
• Since those 2 eyes or 2 legs and the teeth belong to one person,
there is very little difference between groups other than the
treatment that was assigned

Example of how to calculate paired t-test
manually
• Eleven (11) students were told to do push-ups. The number was
recorded for each person. After one hour and a big bottle of
Lucozade Boost, they were told to repeat the push-ups, and the
number of push-ups after the drink was noted for each student.
• The researcher wants to see if Lucozade boost has any statistically
significant effect on the number of push-ups the students can do
after consumption.
• Table of his findings is on the next slide

Sample Research question
• What is the effect of a big bottle of Lucozade boost on the mean
(average) number of push-ups students can do before and after
consumption?

Statistical Hypotheses
• H0: A = B versus H1: A  B
• There is no statistically significant difference in the mean push-ups of
students before and after drinking a large bottle of Lucozade boost
• There is a statistically significant difference in the mean push-ups of
students before and after drinking a large bottle of Lucozade boost

Decision rule
228
.
2
10
1
11
1
scores
difference
of
number
)
1
(
05
.

=
=
−
=
−
=
−
=
=
crit
t
N
df


Calculate test statistic
Formula for calculating Paired sample t-score where:
D = difference between matched scores
N = number of pairs of scores

• Step 1: Subtract each Y score
from each X score (Note that
X - Y = D)
• Step 2: Add up all of the
values from Step 1. Set this
number aside for a moment

• Step 3: Square the
differences from Step1
[i.e (X – Y)2 = D2]
• Step 4: Add up all of the
squared differences from
Step 3

• Step 5: Use the formula to
calculate t-score
• Step 6: Subtract 1 from the
sample size to get the
degrees of freedom. We
have 11 items, so 11-1 = 10.

• Step 7: Find the p-value in the t-
table, using the degrees of
freedom in Step 6. If you don’t
have a specified alpha level, use
0.05 (5%). For this sample
problem, with df=10, the t-value
is 2.228. (next slide)
• Step 8: Compare your t-table
value from Step 7 (2.228) to
your calculated t-value (-2.74).
The calculated t-value is greater
than the table value at an alpha
level of .05. The p-value is less
than the alpha level: p <.05.
• We can reject the null
hypothesis that there is no
difference between means.

• Note: You can ignore the minus sign
when comparing the two t-values, as
± indicates the direction; the p-value
remains the same for both directions.
• It would be useful to calculate a
confidence interval for the mean
difference to tell us within what limits
the true difference is likely to lie.
95% confidence interval for the true mean difference is:
d ± t∗ sd
√n or, equivalently d ± (t∗ × SE( d))
where t∗ is the 2.5% point of the t-distribution on n − 1 degrees
of freedom.

t Test for Dependent Samples: CLASS WORK

Hypothesis Testing
1. State the research question.
2. State the statistical hypothesis.
3. Set decision rule.
4. Calculate the test statistic.
5. Decide if result is significant.
6. Interpret result as it relates to your research question.

The t Test for Dependent Samples: An
Example
• State the research hypothesis/question.
• Does listening to a pro-socialized medicine lecture change
an individual’s attitude toward socialized medicine?
• State the statistical hypotheses.
0
:
0
:

=
D
A
D
O
H
H



The t Test for Dependent Samples:
An Example
• Set the decision rule.
365
.
2
7
1
8
1
scores
difference
of
number
05
.

=
=
−
=
−
=
=
crit
t
df


The t Test for Dependent Samples
• Calculate the test statistic.

The t Test for Dependent Samples
• Decide if your results are significant.
• Reject H0, -4.73 > -2.365
• Ignore the “-” sign
• Interpret your results.
• After the pro-socialized medicine lecture, individuals’ attitudes
toward socialized medicine were significantly more positive than
before the lecture.

Above example using spss (Class work)
Paired Samples Statistics
Mean N Std.
Deviation
Std. Error
Mean
Pair 1
Before
Speech
3.6250 8 1.06066 .37500
After Speech
5.6250 8 1.40789 .49776
Paired Samples Correlations
N Correlation Sig.
Pair 1
Before Speech & After
Speech
8 .562 .147
Paired Samples Test
Paired Differences t df Sig. (2-
tailed)
Mean Std.
Deviation
Std. Error
Mean
95% Confidence Interval
of the Difference
Lower Upper
Pair 1
Before Speech -
After Speech
-2.00000 1.19523 .42258 -2.99924 -1.00076 -4.733 7 .002

Sample SPSS Output and Interpretation:
1. SPSS reports the mean and standard deviation of the difference scores for each pair of variables. The
mean is the difference between the sample means. It should be close to zero if the populations means
are equal.
2. The mean difference between exams 1 and 2 is not statistically significant at α = 0.05. This is because
‘Sig. (2-tailed)’ or p > 0.05.
3. The 95% confidence interval includes zero: a zero mean difference is well within the range of likely
population outcomes.

• The interpretation of this is based on the 95% confidence interval.
• SPSS subtracts the first measurement from the second and
calculates the 95% confidence interval of the difference between
the two paired groups.
• If the two groups were significantly different you would not expect
the lower and upper values to have different signs (negative and
positive). The different signs mean that there is a 0 in the
confidence interval.
• That means that in at least some of the samples there was no
difference between the before and after measurements.

• There are many different names for a t test for related samples.
• Dependent t test,
• Repeated samples t test,
• Paired samples t test,
• Correlated samples t test,
• Within subjects t test, and
• Within groups t test

• There are many advantages of a related samples t test.
1. First, you need fewer participants than you do for a t test for
independent samples.
2. You are able to study changes over time.
3. It also reduces the impact of individual differences.
4. And lastly, it is more powerful, or more sensitive, than a t test for
independent samples

• There are also some disadvantages.
1. The first disadvantage is called carryover effects. And these can
occur when the participant's response in the second treatment is
altered by the lingering after effects of the first treatment
2. The second disadvantage of a related samples t test is called
progressive error. And this can occur when the participant's
performance changes consistently over time due to fatigue or
practice

Independent Sample t-test
• This test evaluates the difference between the mean of two
independent or unrelated groups
• With an independent sample t-test, each case must have scores on
two variables: the grouping variable and the test variable
• The grouping variable divides the cases in two mutually exclusive
groups or categories
• The test variable describes each case on some quantitative
measurement

• This test is also known as:
• Independent t Test
• Independent Measures t Test
• Independent Two-sample t Test
• Student t Test
• Two-Sample t Test
• Uncorrelated Scores t Test
• Unpaired t Test
• Unrelated t Test

Sample questions for independent sample T-tests
• Used extensively in experiments when you want to check a measure
in one group against the other (active agent Vs placebo)
• To compare the average score of the knowledge of research
methodology among Class A and Class B
• To compare fear level (measured in a scale) among those seen by a
particular dentist compared to another dentist
• Or how long it takes to have remission from a skin blister among
some patients using ointment A compared to those using ointment B

• The t-test evaluates whether the population mean of the test variable in
one group differs from the population mean of the test variable in the
second group
• Assumptions
• Test variable is normally distributed in each of the two population
• There are equal variances of the test variable (homogeneity)
• The cases represent random sample from population
• Scores of the test variables are independent of each other

Variation of independent sample t-test
There are two types of t-test that can be used on independent
groups. The decision on which to use is based on whether the
variances are equal or significantly different.
• Pooled t-test: meets all of the assumptions listed on previous slide,
and in addition, the variances of the two independent groups are
essentially equal.
• Separate t-test: meets all of the same assumptions listed on
previous slide and, in addition, the variances of the two
independent groups are unequal or unknown.

• Unfortunately, just like you cannot judge whether two means are
different by just looking at the data set, you cannot judge whether
two variances are equal or different without performing a statistical
test.
• The test statistic you use to compare variances is the F-statistic
(Levene’s Test for Equality of Variances)

• Recall that the Independent Samples t Test requires the assumption
of homogeneity of variance -- i.e., both groups have the same
variance.
• SPSS conveniently includes a test for the homogeneity of variance,
called Levene's Test, whenever you run an independent samples T
test.

• The hypotheses for Levene’s test are:
• H0: σ1
2 - σ2
2 = 0 ("the population variances of group 1 and 2 are
equal")
H1: σ1
2 - σ2
2 ≠ 0 ("the population variances of group 1 and 2 are not
equal")
• This implies that if we reject the null hypothesis of Levene's Test, it
suggests that the variances of the two groups are not equal; i.e., that
the homogeneity of variances assumption is violated.

F-statistic
• The F-statistic is similar to the t-statistic in that it compares groups
• Whereas the t-statistic is used to evaluate the differences in the
mean of 2 groups, the F-statistic evaluates differences in the
variance
• The interpretation of the F-statistic is similar to the interpretation of
the t-statistic, but is usually done based on the p-value rather than
the confidence interval

• The null hypothesis associated with the F-statistic is that there is no
difference between the variances of the two groups
• If the p-value or significance level is 0.05 or smaller you can reject
the null hypothesis and assume that the groups have significantly
different variances
• If the p-value is greater than 0.05 you cannot reject the null
hypothesis and you assume the variances are equal

Sample SPSS output
Independent Samples Test
Levene's Test
for Equality of
Variances t-test for Equality of Means
F Sig. t df Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence Interval of the
Difference
Lower Upper
histidin Equal
variances
assumed
5.743 .096 5.096 3 .015 83.500 16.385 31.355 135.645
Equal
variances
not assumed
6.418 2.296 .016 83.500 13.010 33.907 133.093

There are two steps to the interpretation of this output:
• First you look at the columns labeled Levene’s Test for Equality of
Variances.
• In those columns you will see the F-statistic and the significance of
that statistic.
• Since the significance is greater than 0.05 you cannot reject the null
hypothesis and you can assume that the variances are equal.

• Then, using that information you can use the row Equal variances
assumed, which gives you the result of the paired t-test.
• Again, the interpretation of this is based on the 95% confidence
interval.
• Here you see that upper and lower values have the same positive
value so the confidence interval does not cross 0 and you can reject
the null hypothesis of no difference in the means..

How to report t-tests
• Values to report: means (M) and standard deviations (SD) for each
group, t value (t), degrees of freedom (in parentheses next to t),
and significance level (p).
• Examples
• Women (M = 3.66, SD = .40) reported significantly higher levels of
happiness than men (M = 3.20, SD = .32), t(1) = 5.44, p < .05.
• Men (M = 4.05, SD = .50) and women (M = 4.11, SD = .55) did not
differ significantly on levels of political agreement, t(1) = 1.03, p =
n.s.

Reporting a significant single sample t-test (μ ≠ μ0):
• Students taking statistics courses in Chemistry at the University of
Lagos reported studying more hours for tests (M = 121, SD = 14.2)
than did Unilag students in general, t(33) = 2.10, p = .034.
Reporting a significant t-test for dependent groups (μ1 ≠ μ2):
• Results indicate a significant preference for Meat pie (M = 3.45, SD
= 1.11) over chicken pie (M = 3.00, SD = .80), t(15) = 4.00, p = .001.

Reporting a significant t-test for independent groups (μ1 ≠
μ2):
• Unilag students taking statistics courses in Psychology had higher IQ
scores (M = 121, SD = 14.2) than did those taking statistics courses
in Chemistry (M = 117, SD = 10.3), t(44) = 1.23, p = .09.
• Over a two-day period, participants drank significantly fewer drinks
in the experimental group (M= 0.667, SD = 1.15) than did those in
the control group (M= 8.00, SD= 2.00), t(4) = -5.51, p=.005.

Non parametric variants of t-tests
• Where data is not normally distributed, statistical analyses that
assume a normal distribution may be inappropriate.
• This is especially a concern where the sample size is small (<50 in
total).
• Variables that are discrete (take only integer values) or have an upper
or lower limit are by definition non-normal.

• For one sample T tests that do not fulfil the normality assumption,
use the Wilcoxin One Sample Signed Rank T Test
• For 2 Related samples that do not fulfil the normality assumption,
use the McNemar test.
• For 2-Independent Groups, use Mann-Whitney U-test
• For Related Ordinal Data, i.e. ordered categorical or quantitative
variables that are not plausibly normal, the suggested procedure is to
use the Wilcoxon procedure

Summary
• A t-test is a parametric test used to compare the means of two
samples
• There are different t-tests we can use based on the assumptions we
can make – paired, pooled, and separate
• To test your null hypothesis you need to know the type of t-test to
use, the number of observations, the means of the two groups (or
with pairs, the difference between the pairs), and the variances of
the two groups

• Single sample t – we have only 1 group; want to test against a
hypothetical mean.
• Dependent (Paired) t – we have two means. Either same people in
both groups, or people are related, e.g., husband-wife, left hand-
right hand.
• Independent samples t – we have 2 means, 2 groups; no relation
between groups, e.g., people randomly assigned to a single group.

• You need to compare the variances of your groups using the F-
statistic to chose between the pooled or separate tests

• ANOVA is short for ANalysis Of VAriance
• The purpose of ANOVA is to determine whether the mean differences
that are obtained for sample data are sufficiently large to justify a
conclusion that there are mean differences between the populations
from which the samples were obtained.

• The difference between ANOVA and t tests is that ANOVA can be
used in situations where there are two or more means being
compared, whereas the t tests are limited to situations where only
two means are involved.

Assumptions for ANOVA
• ASSUMPTION 1: The dependent variable is normally distributed for
each of the populations as defined by the different levels of the factor
• ASSUMPTION 2: The variances of the dependent variable are the
same for all populations
• ASSUMPTION 3: The cases represent random samples from the
populations and the scores on the test variable are independent of
each other

Sample research questions for ANOVA
• Is there a difference in average pain levels during tooth extraction
among patients given 3 types of Local Anesthesia?
• Are there differences in children’s weight by type of school (Public vs.
missionary vs. private)?

• Experiment to check the effects of caffeine on reading time among
students.
• 45 volunteers from a university to participate. Randomly assigns an
equal number of students to three groups: placebo (group 1), mild
doses of caffeine (group 2), and heavy doses of caffeine (group 3).
• Before study, the average number of study hours for all the students
was taken over 2 months. The intervention over the next 2 months
with a cup of coffee or placebo every night and average hours of
study was documented again.

The basic ANOVA situation
**Two variables: 1 Categorical, 1 Quantitative
Main Question: Do the (means of) the quantitative variables
depend on which group (given by categorical variable) the
individual is in?
If categorical variable has only 2 values:
• 2-sample t-test
ANOVA allows for 3 or more groups

• Analysis of variance is necessary to protect researchers from
excessive risk of a Type I error in situations where a study is
comparing more than two population means.
• E.g., caffeine study with 3 populations taking coffee:
• Population 1 - No caffeine
• Population 2 - Mild dose
• Population 3 - Heavy dose

• We have at least 3 means to test, e.g.,
• H0: 1 = 2 = 3.
• Could take them 2 at a time, but really want to test all 3 (or more) at
once.
• Although each t test can be done with a specific α-level (risk of Type I
error), the α-levels accumulate over a series of tests so that the final
experiment-wise α-level can be quite large

• ANOVA allows researcher to evaluate all of the mean differences in a
single hypothesis test using a single α-level and, thereby, keeps the
risk of a Type I error under control no matter how many different
means are being compared

Example of an ANOVA situation
Subjects: 25 patients with blisters
Treatments: Treatment A, Treatment B, Placebo (categorical)
Measurement: Number of days until blisters heal (quantitative)
Data [and means]:
• A: 5,6,6,7,7,8,9,10 [7.25]
• B: 7,7,8,9,9,10,10,11 [8.875]
• P: 7,9,9,10,10,10,11,12,13 [10.11]
Are these differences significant?

Informal Investigation
Graphical investigation:
• side-by-side box plots
• multiple histograms
ANOVA determines P-value from the F statistic

What does ANOVA do?
At its simplest ANOVA tests the following hypotheses:
H0: The means of all the groups are equal. 1 = 2 = 3.
Ha: Not all the means are equal
• doesn’t say how or which ones differ.
• Can follow up with “multiple comparisons”
Note: we usually refer to the sub-populations as “groups” when doing
ANOVA.

Assumptions of ANOVA
• Each group is approximately normal
· check this by looking at histograms and/or normal quantile plots, or
use assumptions
· can handle some non-normality, but not severe outliers

Normality Check
We should check for normality using:
• assumptions about population
• histograms for each group
• normal quantile (or QQ) plot for each group
With such small data sets, there really isn’t a really good way to
check normality from data, but we make the common assumption
that physical measurements of people tend to be normally
distributed.

• The test statistic for ANOVA is an F-ratio, which is a ratio of two
sample variances.
• In the context of ANOVA, the sample variances are called mean
squares, or MS values.

Recall example of an ANOVA situation
Subjects: 25 patients with blisters
Treatments: Treatment A, Treatment B, Placebo (categorical)
Measurement: Number of days until blisters heal (quantitative)
Data [and means]:
• A: 5,6,6,7,7,8,9,10 [7.25]
• B: 7,7,8,9,9,10,10,11 [8.875]
• P: 7,9,9,10,10,10,11,12,13 [10.11]

Notation for ANOVA
• n = number of individuals all together (25)
• I = number of groups (3)
• = mean for the entire data set is (??)
Group A has
• nA = # of individuals in group A
• xAw = value for individual w in group A
• = mean for group A
• sA = standard deviation for group A
A
x
A
x

Number of individuals all together (25)
Group A Group B Group P
N = 8 N = 8 N = 9

How ANOVA works
ANOVA measures two sources of variation in the data and compares
their relative sizes
• variation BETWEEN groups
• for each data value look at the difference between its group
mean and the overall mean
• variation WITHIN groups
• for each data value we look at the difference between that
value and the mean of its group
( )2
A
Aw x
x −
( )2
x
xA −

The ANOVA F-statistic is a ratio of the Between Group
Variation divided by the Within Group Variation:
MSW
MSB
Within
Between
F =
=

• The top of the F-ratio MSbetween measures the size of mean differences
between samples.
• The bottom of the ratio MSwithin measures the magnitude of
differences that would be expected without any outside effects.
MSW
MSB
Within
Between
F =
=

• A large value for the F-ratio indicates that the obtained sample mean
differences are greater than would be expected if the samples had no
outside effect.
• A large F is evidence against H0, since it indicates that there is more
difference between groups than within groups.
• F is another distribution like z and t. There are tables of F used for
significance testing.

126
The Logic and the Process of Analysis of Variance
• The two components of the F-ratio can be described as follows:
• Between-Treatments Variability: MSbetween measures the size of the
differences between the sample means. For example, suppose that
three treatments, each with a sample of n = 5 subjects, have means of
M1 = 1, M2 = 2, and M3 = 3.
• Notice that the three means are different; that is, they are variable.

127
The Logic and the Process of ANOVA (cont.)
• By computing the variance for the three means we can measure
the size of the differences
• Although it is possible to compute a variance for the set of
sample means, it usually is easier to use the total, T, for each
sample instead of the mean, and compute variance for the set of
T values.

128
Logically, the differences (or variance) between means can be
caused by two sources:
1. Outside or Treatment Effects: If the treatments have different
effects, this could cause the mean for one treatment to be
higher (or lower) than the mean for another treatment.
2. Chance or Sampling Error: If there is no treatment effect at all,
you would still expect some differences between samples.
Mean differences from one sample to another are an example
of random, unsystematic sampling error.

Group A Group B Group P
N = 8 N = 8 N = 9

130
• Within-Treatments Variability: MSwithin measures the size of the
differences that exist inside each of the samples.
• Because all the individuals in a sample receive exactly the same
treatment, any differences (or variance) within a sample cannot
be caused by different treatments.

132
The Logic and the Process of ANOVA
Thus, these differences are caused by only one source:
Chance or Error: The unpredictable differences that exist between
individual scores are not caused by any systematic factors and are
considered to be random, chance, or error.

133
• Considering these sources of variability, the structure of the F-ratio
becomes,
treatment effect + chance/error
F = ──────────────────────
chance/error

134
• When the null hypothesis is true and there are no differences
between treatments, the F-ratio is balanced.
• That is, when the "treatment effect" is zero, the top and bottom
of the F-ratio are measuring the same variance.
• In this case, you should expect an F-ratio near 1.00. When the
sample data produce an F-ratio near 1.00, we will conclude that
there is no significant treatment effect.

135
• On the other hand, a large treatment effect will produce a large
value for the F-ratio. Thus, when the sample data produce a
large F-ratio we will reject the null hypothesis and conclude that
there are significant differences between treatments.
• To determine whether an F-ratio is large enough to be
significant, you must select an α-level, find the df values for the
numerator and denominator of the F-ratio, and consult the F-
distribution table to find the critical value.

A sample class question:
• Do students' performance in exams vary depending on where they
live? – Research Question
• Independent variable: Where students live eg, Ikoyi, Ojota and Badagry
• Dependent variable : Performance in exams (standard quiz)
Ho: There is no difference in the mean score of students based on different
residence i.e. mean of Ikoyi = mean of Ojota = mean of Badagry
Ha: There is a difference in the mean score of students based on different
residence i.e. mean of Ikoyi ≠ mean of Ojota ≠ mean of Badagry

(Each score is over 10)
• Ikoyi – scores of 3 students in tests:
3, 2, 1
• Ojota – scores of 3 students in tests:
5, 3, 4
• Badagry – scores of 3 students in tests:
5, 6, 7

Manual procedure for computing 1-way ANOVA for
independent samples
• Step 1: Complete the table
• Calculate the sum in each group
• Calculate the mean in each group
Ikoyi
A
Ojota
B
Badagry
C
3 5 5
2 3 6
1 4 7

independent samples
Ikoyi
A
Ojota
B
Badagry
C
3 5 5
2 3 6
1 4 7
Total 6 12 18

independent samples
• Mean for A = 6/3 = 2
• Mean for B = 12/3 = 4
• Mean for C = 18/3 = 6
Ikoyi
A
Ojota
B
Badagry
C
3 5 5
2 3 6
1 4 7
Total 6 12 18
Mean 2 4 6

Procedure for computing 1-way ANOVA for
independent samples
• Step 2: Calculate the Grand mean
X =
=
= 6 + 12 + 18 = 4
9
Or the mean of group A + mean of group B + mean of group C
k (number of groups)
= 2 + 4 + 6 = 12/3 = 4
3
(X)
N
(XA+XB+XC)
N
Ikoyi
A
Ojota
B
Badagry
C
3 5 5
2 3 6
1 4 7
Total 6 12 18
Mean 2 4 6
x

Procedure for computing 1-way ANOVA for independent
samples
• Step 3: Compute total Sum of Squares
SStotal= (every value – grand mean)2
This is done for each value in all the
groups and added together to give the SStotal
= (3-4)2 + (2-4)2 + (1-4)2 + (5-4)2 + (3-4)2 + (4-4)2
+ (5-4)2 + (6-4)2 + (7-4)2
= 1 + 4 + 9 + 1 + 1 + 0 + 1 + 4 +9 = 30
SStotal= 30, dftotal = 8
Degree of freedom = (N – 1) = 9 – 1 = 8 (Recall that “N” is the
number of sample size from ALL the population i.e. 3 students
each from the different location)
Ikoyi
A
Ojota
B
Badagry
C
3 5 5
2 3 6
1 4 7
Total 6 12 18
Mean 2 4 6
Grand mean = 4

independent samples
• Step 4: Compute between groups
Sum of Squares
• Compare the means of EACH group with
the grand mean
i.e for group A = (mean of grp A – grand mean)2
for the number of variables in the group
This is the same as the value multiplied by
the number of variables in each group.
The same for grp B and Group C
SSbet = Total sum of all the values i.e value for Grp A + value for
Grp B + value for Grp C etc
Ikoyi
A
Ojota
B
Badagry
C
3 5 5
2 3 6
1 4 7
Total 6 12 18
Mean 2 4 6
Grand mean = 4

independent samples
SSbet = Total sum of all the values i.e value for
Grp A + value for Grp B + value for Grp C
= (2-4)2 + (2-4)2 + (2-4)2 + (4-4)2 + (4-4)2 + (4-4)2
+ (6-4)2 + (6-4)2 + (6-4)2
= 4 + 4 + 4 + 0 + 0 + 0 + 4 + 4 + 4 = 24
SSbet = 24, dfbet = 2
Degree of freedom = (k – 1) = 3 – 1 = 2 (Recall that “K” is the
number of categories of the population)
Ikoyi
A
Ojota
B
Badagry
C
3 5 5
2 3 6
1 4 7
Total 6 12 18
Mean 2 4 6
Grand mean = 4

independent samples
• Step 5: Compute within groups Sum of Squares
• Note that: SStotal= SSwit + SSbet therefore,
SSwit= SStotal – SSbet
= 30 – 24 = 6
or
Compare the values in EACH group with the
group mean and square them
= (3-2)2 + (2-2)2 + (1-2)2 + (5-4)2 + (3-4)2 + (4-4)2
+ (5-6)2 + (6-6)2 + (7-6)2
= 1 + 0 + 1 + 1 + 1 + 0 + 1 + 0 + 1 = 6
SSwit= 6, dfwit = 6
Degree of freedom = (N – k) = 9 – 3 = 6
Ikoyi
A
Ojota
B
Badagry
C
3 5 5
2 3 6
1 4 7
Total 6 12 18
Mean 2 4 6
Grand mean = 4

independent samples
• Step 6: Determine the d.f. for each sum of squares
dftotal= (N - 1) = 8
dfbet= (k - 1) = 2
dfwit= (N - k) = 6
N = total number of records
K = number of groups

Systematic
Variance
(between means)
Error
Variance
(within means)
independent samples
• Step 7/8: Estimate the Variances & Compute F
=
=
SSbet
dfbet
SSwit
dfwit
MSbetween
MSwithin
MSW
MSB
Within
Between
F =
=

MSW
MSB
Within
Between
F =
=
SSbet
dfbet
SSwit
dfwit
= = 24/2
6/6
F = 12

In Summary (The formula)
MSW
MSB
F
DF
SS
MS
SST
SSB
SSW
x
x
n
x
x
SSB
df
s
x
x
SSW
DFT
s
x
x
SST
groups
i
i
obs
i
i
groups
i
obs
i
ij
obs
ij
=
=
=
+
−
=
−
=
=
−
=
=
−
=





;
;
)
(
)
(
)
(
)
(
)
(
)
(
2
2
2
2
2
2

independent samples
• Step 9: Consult F distribution table (it has 2 sides: d1 and d2)
d1 is your df for the numerator (i.e. systematic variance) dfbetween
d2 is your df for the denominator (i.e. error variance) dfwithin

• A large F is evidence against H0, since it indicates that there is more
difference between groups than within groups
• Calculated F- ratio was 12
• Reading from F-Table at 0.05 level was 5.14
• Conclusion: reject H0
Ho: There is no difference in the mean score of students based on different
residence i.e. mean of Ikoyi = mean of Ojota = mean of Badagry
Ha: There is a difference in the mean score of students based on different
residence i.e. mean of Ikoyi ≠ mean of Ojota ≠ mean of Badagry

Independent 1-way ANOVA: SPSS Output
ANOVA
Maths
Sum of Squares df Mean Square F Sig.
Between Groups 24.000 2 12.000 12.000 .008
Within Groups 6.000 6 1.000
Total 30.000 8

The computation of F statistic is illustrated in the following table, called ANOVA table.
Source of variance Degrees of freedom Sum of squares Mean square F ratio
Between k-1 SSB MSB=BSS/ (k-1) MSB/MSW
Within n-k SSW MSW=WSS/(n-k)
Total n-1 SST

158
Analysis of Variance and Post Tests
• The null hypothesis for ANOVA states that for the general population
there are no mean differences among the treatments being
compared; H0: μ1 = μ2 = μ3 = . . .
• When the null hypothesis is rejected, the conclusion is that there are
significant mean differences.
• However, the ANOVA simply establishes that differences exist, it does
not indicate exactly which groups, intervention, categories or
treatments are different.

• If the t-test is significant, you have a difference in population means.
• If the F-test is significant, you have a difference in population means.
But you don’t know where.
• With 3 means, could be A=B>C or A>B>C or A>B=C.

160
Analysis of Variance and Post Tests
• With more than two treatments, this creates a problem.
• Specifically, you must follow the ANOVA with additional tests, called
post tests, to determine exactly which treatments are different and
which are not.
• The Scheffe test and Tukey’s HSD are examples of post tests.
• HSD means honestly significant difference

• These tests are done after an ANOVA where H0 is rejected with more
than two treatment conditions. The tests compare the treatments,
two at a time, to test the significance of the mean differences.

Multiple Comparisons
Dependent Variable: Maths
Tukey HSD
(I) Location (J) Location Mean
Difference
(I-J)
Std.
Error
Sig. 95% Confidence Interval
Lower
Bound
Upper
Bound
Ikoyi Ojota -2.00000 .81650 .109 -4.5052 .5052
Badagry -4.00000* .81650 .006 -6.5052 -1.4948
Ojota Ikoyi 2.00000 .81650 .109 -.5052 4.5052
Badagry -2.00000 .81650 .109 -4.5052 .5052
Badagry Ikoyi 4.00000* .81650 .006 1.4948 6.5052
Ojota 2.00000 .81650 .109 -.5052 4.5052
*. The mean difference is significant at the 0.05 level.
Mean of Ikoyi = 2. Ojota = 4 and Badagry = 6
POST-HOC TEST

One-Way Analysis of Variance with SPSS
• Select an independent variable with three or more levels and a
dependent variable.
• Develop the null hypothesis and the alternative hypothesis for main
effects.
• Using SPSS, calculate an ANOVA. Include a post hoc test.
• Report on the p value and the confidence interval.
• Interpret the confidence interval.
• Decide whether to reject or retain the null hypothesis based on main
effects and/or post-hoc statistical tests.

How to report ANOVA
• A one-way between subjects ANOVA was conducted to
compare the effect of (IV)______________ on
(DV)_______________ in _________________,
__________________, and __________________
conditions.”
• There was a significant (not a significant) effect of IV
____________ on DV ______________ at the p<.05 level for
the three conditions [F(df1, df2) = ___, p = ____].

For our class example:
There was a significant effect of the location of residence on the
performance of students at p < .05 [F(2, 6) = 12, p = .008]
ANOVA
Maths
Sum of Squares df Mean Square F Sig.
Between Groups 24.000 2 12.000 12.000 .008
Within Groups 6.000 6 1.000
Total 30.000 8

• The Kruskal-Wallis test is a nonparametric (distribution free) test, and
is used when the assumptions of one-way ANOVA are not met.
• Both the Kruskal-Wallis test and one-way ANOVA assess for significant
differences on a continuous dependent variable by a categorical
independent variable (with two or more groups)

• The most common use of the Kruskal–Wallis test is when you have
one nominal variable and one measurement variable, an experiment
that you would usually analyze using one-way ANOVA, but the
measurement variable does not meet the normality assumption of a
one-way ANOVA

Next Class
• Quantitative Data Analysis Lecture 4
• Regressions.
• Contingency Table Analysis

Sample SPSS Analysis
• I have a sample SPSS data with the following variables
• Age:Continuous
• Sex: Nominal (Binary)
• Basic Metabolic Index (BMI): Continuous
• Lipid Level: Continuous
• Stroke: Nominal (Binary)
• Hypertension: Nominal (Binary)
• Educational Level: Ordinal
• Income per day: Continuous

Comparison Tests
Differences among group means
Statistical tests Predictor variable Outcome variable Sample research question
Paired t-test • Categorical
• 1 predictor
• Quantitative
• Groups come from
the same
population
What is the difference in the average
knowledge of the choice of data analysis
before and after this lecture in this group
Independent sample t-
test
• Categorical
• 1 predictor
• Quantitative
• Groups come from
different
populations
What is the difference in the average
knowledge of the choice of data analysis
in group A and group B
ANOVA • Categorical
• 1 or more predictor
• Quantitative
• 1 outcome
What is the difference in average pain
levels during tooth removal among
patients given 3 types of Local
Anesthesia

One Sample T-
Test
• Continuous • Quantitative
• One population
Is there an improvement in lipids
after a vegetarian diet over 3
days? Cut off of 200
Paired t-test • Categorical
• 1 predictor
• Quantitative
• Groups come
from the same
population
What is the difference in the
average basic metabolic index
(BMI) before and after this
lecture in this group
Independent
sample t-test
• Categorical
• 1 predictor
• Quantitative
• Groups come
from different
populations
What is the difference in the
average basic metabolic index
(BMI) after this lecture in males
and females

Analysis of Variance (ANOVA)
ANOVA • Categorical
• 1 or more
predictor
• Quantitative
• 1 outcome
What is the difference in baseline
BMI among the different
Educational level

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