The document discusses different statistical analysis techniques for interpreting data, including t-tests, ANOVA, and correlation. It provides examples of when to use paired t-tests, independent t-tests, one-way ANOVA, and correlation. For one-way ANOVA, it demonstrates how to check assumptions, run the test in SPSS, and interpret the results, finding a significant difference between group means. It also provides an example of how to conduct post-hoc comparisons after finding significant ANOVA results.

1. Chapter 5 - Data Analysis and Research
B. Interpreting, Analyzing, and Reporting the Results from Data
William Allan Kritsonis, PhD
INTRODUCTION
The purpose of this chapter is to interpret, analyze, and report the
results from data. This chapter will introduce the methods and examples of
the paired samples t-test, independent samples t-test, one-way ANOVA, and
Bivariate-Pearson-Correlation. Steps to the tables and External Links for
online tutorial are provided for each test. Software package SPSS 10.0 was
used to analyze the data.
THE T-TEST
The t-test provides the probability that the null hypothesis is true
when examining the difference between the means of two groups. Normally
we use this test when data sets are small.
There are two different t-tests
♦ The “paired samples t-test,” and
♦ The “independent samples t-test.”
THE PAIRED SAMPLES T-TEST
When do we use it? We assume that the confidence interval is at 95% all
the time.

2. 1. When there is a natural relationship between the subjects from whom the
two sets of scores are obtained.
Example 1: Looking at differences between pre- and post-tests of one
group, you would choose the paired samples t-test, the scores of both
data sets came from the same persons.
Example 2: A teacher diagnostically tests her students at the
beginning of the year. After intensive instruction, the test is repeated
at the end of the semester. She is interested in knowing if the students
have made significant gains.
THE INDEPENDENT SAMPLES T-TEST
(We use this t-test more often)
When do we use it?
1. When there is no natural relationship between subjects whose scores are
being contrasted. Comparing scores obtained from two different groups
of people, you would use this t-test.
2. The data descriptions are normally distributed and of the groups are
homogeneous.
Example 1: Two groups of students are identified: an experimental
and a control group. Both groups are pretested (Both groups are
posttested), an intervention is used with the experimental group and is
withheld from the control group. Both groups are posttested.

3. Example 2: TLI scores are collected on students who have attended
school using block scheduling and students who have attended
schools with traditional scheduling. We want to know if the TLI
scores are significantly different according to the schedule
experienced by the students. (We reject the null when p<0.05; we fail
to reject the null when p>0.05).
ONE-WAY ANALYSIS OF VARIANCE (ANOVA)
About the one-way ANOVA,
1. The probability that the null hypothesis is true when examining the mean
differences among three or more groups. This procedure is equal to the
test, except that it handles more than two groups.
2. The assumptions for one-way ANOVA are the same as for the t-test:
normal distributions and homogeneity of variances. We have to run
Levene’s test for homogeneity. There are two situations: (1) use
Bonferroni when the data fail to reject the null and (2) use Tamhane
when the data reject the null.
3. A probability value p < 0.05 indicates that a significant difference exists
among the various means, but it does not indicate which means are
significantly different and which are chance differences.

4. Example 1: If the GPA averages were significantly different according to
undergraduate majors. We would input all of the GPAs into a variable (this
would be the dependent variable), and in a second variable (an independent
variable often called a “factor”) assigning a “1” if the GPA belonged to an
English major, a “2” for History majors, a “3” for Psychology majors, etc..
(Don’t use “0” for anything because it does not work sometimes.)
Example 2: TAAS scores are collected to describe scores of students. Three
different methods of teaching were used after the students had been divided
into three equal groups. The socioeconomic level of each student was
identified. The hypothesis was used: “was there a significant difference in
TAAS scores according to method used.”
Example 3: Professors who are primarily university administrators, regular
tenured professors, and regular non-tenured professors are rated by students
according to enthusiasm displayed in the classes they teach. The null
hypothesis is, “there is no significant difference in the degree of enthusiasm
displayed among the three groups of professors.”

5. ♦ The standard way to report the one-way ANOVA:
The null hypothesis is that there will be no significant differences in
_____________________________________________________________
_. To test this hypothesis, the one-way ANOVA from SPSS (10.0) was
used. The null hypothesis is accepted/ not accepted F=(n-1, N-n), p = ____<
or > than 0.05.
We reject the null when p< 0.05; we fail to reject the null when p> 0.05.
Example 4: It was hypothesized that students who excel in fine arts are also
the best students in the academic subjects. A measure of fine arts
achievement and a measure of academic achievement were collected. The
relationship of the two measures was analyzed statistically. (Bivariate-
Pearson-Correlation)
Example 5: There will not be a significant relationship between the percent
of students passing all TAAS tests and the size of the school districts.
(Bivariate-Pearson-Correlation)
♦ The correlation coefficient is between –1 and 1. The closer to the
positive/negative 1 the stronger the relationship. The closer to the 0 the
weaker the relationship.

6. ♦ The standard way to report the Bivariate-Pearson-Correlation:
The null hypothesis is that there is no significant relationship between the
________________________________________________________.
To test this hypothesis, the Bivariate-Person-Correlation from SPSS
(10.0) is used. The null hypothesis is accepted/not accepted. (We need
to interpret some more details about the data.)
Following question was the review question for cohort VII in 2001.

7. JOINT UNIVERSITY DOCTORAL PROGRAM
REVIEW FOR THE COMPREHENSIVE EXAMINATION
Review question for Cohort VII
A Single Factor ANOVA
To compare the effectiveness of three different methods of
teaching reading, 26 children of equal reading aptitude were divided
into three groups. Each group was instructed for a give period of time
using one of the three methods. After completing the instruction
period, all students were tested. The test results are shown in the
following table. Is the evidence sufficient to reject the hypothesis that
all three instruction-methods are equally effective? Use α = 0.05.
Method I Method II Method III
Test scores: 45 45 44
51 44 50
48 46 45
50 44 55
46 41 51
48 43 51
45 46 45
48 49 47
47 44

8. To do the following:
1) Test the Normality Assumption.
2) Test the Equality of Variance Assumption.
3) Run the ANOVA test and produce the ANOVA Table.
4) Run Post-hoc comparisons.
SPSS Data Entry: (check on scale)
readscr teachmth
45 1
51 1
48 1
50 1
46 1
48 1
45 1
48 1
47 1
45 2
44 2
46 2
44 2
41 2
43 2
46 2
49 2
44 2
44 3
50 3
45 3
55 3
51 3
51 3
45 3
47 3
Run the Analysis
1) Check Normality.

9. Steps to the tables:
1. Analyze → Descriptive Statistics → Explore → Dependent : readscr
Factor List: teachmth
Go to and check: Statistics → Descriptives
Go to and check: Plots → Box plots
♦ Factor Levels to get that
♦ Normality plots with tests
Explore
TEACHMTH
Case Processing Summary
Cases
Valid Missing Total
TEACHMTH N Percent N Percent N Percent
READSCR 1.00 9 100.0% 0 .0% 9 100.0%
2.00 9 100.0% 0 .0% 9 100.0%
3.00 8 100.0% 0 .0% 8 100.0%
Tests of Normality
a
Kolmogorov-Smirnov Shapiro-Wilk
TEACHMTH Statistic df Sig. Statistic df Sig.
READSCR 1.00 .193 9 .200* .933 9 .490
2.00 .173 9 .200* .950 9 .667
3.00 .193 8 .200* .919 8 .437
*. This is a lower bound of the true significance.
a. Lilliefors Significance Correction
Test these two assumptions for each of the three groups:

10. (a) Normality
(b) Homogeneity (equality) of variance
Write a short paragraph in which you describe the results.
Analyzing the data:
(a) The assumption of Normality was analyzed using two tests of
significance: the Kolmogorov-Smirnov test and the Shapiro-Wilk
test. The Kolmogorov-Smirnov test showed a probability coefficient
of 0.2 for each group since this value is greater than 0.05, the
Kolmogorov-Smirnov test did not reject the null hypothesis that the
scores for each group is normally distributed.
The Shapiro-Wilk test showed a probability coefficient of 0.49 for
method 1, .667 for method 2, and 0.437 for method 3. In all three
cases the coefficient is greater than 0.05. Therefore, the Shapiro-
Wilk test did not reject the null hypothesis that the scores for each
group are normally. The results for both the Kolmogorov-Smirnov
test and Shapiro-Wilk test provide support for the assumption of
normality.
(b) The assumption of homogeneity of variance was tested using the
Levene test. The results for the Levene test showed a probability
coefficient of 0.042. Since this value is less than 0.05, the null

11. hypothesis is rejected. The assumption of homogeneity of variance is
not supported.
Post-hoc comparisons among the groups could be tested with either the
Bonferroni or Tamhane test, depending on whether or not the homogeneity
of variance assumption was rejected. The Bonferroni test is appropriate if
the homogeneity of variance assumption is supported and the Tamhane test
is appropriate when it is not supported. Since the Levene test showed that
the homogeneity of variance assumption was not supported, the Tamhane
test was used to test differences between the means of the three groups.

12. Descriptives
TEACHMTH Statistic Std. Error
READSCR 1.00 Mean 47.5556 .6894
95% Confidence Lower Bound 45.9657
Interval for Mean Upper Bound
49.1454
5% Trimmed Mean 47.5062
Median 48.0000
Variance 4.278
Std. Deviation 2.0683
Minimum 45.00
Maximum 51.00
Range 6.00
Interquartile Range 3.5000
Skewness .335 .717
Kurtosis -.651 1.400
2.00 Mean 44.6667 .7454
95% Confidence Lower Bound 42.9479
Interval for Mean Upper Bound
46.3855
5% Trimmed Mean 44.6296
Median 44.0000
Variance 5.000
Std. Deviation 2.2361
Minimum 41.00
Maximum 49.00
Range 8.00
Interquartile Range 2.5000
Skewness .450 .717
Kurtosis 1.300 1.400
3.00 Mean 48.5000 1.3628
95% Confidence Lower Bound 45.2776
Interval for Mean Upper Bound
51.7224
5% Trimmed Mean 48.3889
Median 48.5000
Variance 14.857
Std. Deviation 3.8545
Minimum 44.00
Maximum 55.00
Range 11.00
Interquartile Range 6.0000
Skewness .429 .752
Kurtosis -.887 1.481

13. To analyzing, interpreting and reporting the results from data:
Method I has 9 scores ranging from 45 as the lowest score to the
highest score of 51. The mean of the distribution is 47.56, the median is 48,
and the standard deviation is 2.07. The skew and Kurtosis coefficients are
0.34 and –0.65, respectively. Method I can be considered as a normal
distribution.
Method II has 9 scores ranging from 41 as the lowest score to the
highest score of 49. The mean of the distribution is 44.67, the median is 44,
and the standard deviation is 2.24. The skew and Kurtosis coefficients are
0.45 and 1.3, respectively. Method II can be considered as a normal
distribution.
Method III has 8 scores ranging from 44 as the lowest score to the
highest score of 55. The mean of the distribution is 48.5, the median is 48.5,
and the standard deviation is 3.85. The skew and Kurtosis coefficients are
0.43 and –0.887, respectively. Method III can be considered as a normal
distribution.

14. 2) Finish the Analysis: run the ANOVA test/table, and Post-hoc
comparisons.
Steps to the tables:
1) Analyze → Compare means → one way ANOVA
Dependent: readscr
Factor: teachmth
Go to: Post-hoc, check:
♦ Bonferroni
♦ Tamhane T2
Go to: Options, check:
♦ Statistics
♦ Descriptives
♦ Homogeneity of Variance
Oneway
Descriptives
READSCR
95% Confidence
Interval for Mean
Std. Lower Upper
N Mean Deviation Std. Error Bound Bound Minimum Maximum
1.00 9 47.5556 2.0683 .6894 45.9657 49.1454 45.00 51.00
2.00 9 44.6667 2.2361 .7454 42.9479 46.3855 41.00 49.00
3.00 8 48.5000 3.8545 1.3628 45.2776 51.7224 44.00 55.00
Total 26 46.8462 3.1457 .6169 45.5756 48.1167 41.00 55.00

15. Test of Homogeneity of Variances
READSCR
Levene
Statistic df1 df2 Sig.
3.641 2 23 .042
ANOVA
READSCR
Sum of Mean
Squares df Square F Sig.
Between Groups 69.162 2 34.581 4.463 .023
Within Groups 178.222 23 7.749
Total 247.385 25
The null hypothesis (H0) is that there is no significant difference in the
effectiveness of three different methods of teaching reading. To test this
hypothesis, the one-way ANOVA from SPSS (10.0) was used. The null
hypothesis is rejected F(2/23) = 4.463, p = 0.023<0.05.

16. Post Hoc Tests
Multiple Comparisons
Dependent Variable: READSCR
95% Confidence
Mean Interval
Difference Lower Upper
(I) TEACHMTH (J) TEACHMTH (I-J) Std. Error Sig. Bound Bound
Bonferroni 1.00 2.00 2.8889 1.3122 .114 -.4993 6.2771
3.00 -.9444 1.3526 1.000 -4.4369 2.5480
2.00 1.00 -2.8889 1.3122 .114 -6.2771 .4993
3.00 -3.8333* 1.3526 .028 -7.3258 -.3408
3.00 1.00 .9444 1.3526 1.000 -2.5480 4.4369
2.00 3.8333* 1.3526 .028 .3408 7.3258
Tamhane 1.00 2.00 2.8889* 1.3122 .035 .1815 5.5963
3.00 -.9444 1.3526 .909 -5.2769 3.3880
2.00 1.00 -2.8889* 1.3122 .035 -5.5963 -.1815
3.00 -3.8333 1.3526 .091 -8.2019 .5352
3.00 1.00 .9444 1.3526 .909 -3.3880 5.2769
2.00 3.8333 1.3526 .091 -.5352 8.2019
*. The mean difference is significant at the .05 level.
Post-hoc comparisons among the groups could be tested with either
the Bonferroni or Tamhane test, depending on whether or not the
homogeneity of variance assumption was rejected. The Bonferroni test is
appropriate if the homogeneity of variance assumption is supported and the
Tamhane test is appropriate when it is not supported. Since the Levene test
showed that the homogeneity of variance assumption was not supported, the
Tamhane test was used to test differences between the means of the three
groups.
The Tamhane test indicated the following:

17. ♦ There was a statistically significant difference between the mean of
Method I and Method II (sig.=0.035). The mean for Method I was 2.889
higher than the mean for Method II.
♦ There were no statistically significant differences between Methods I and
III or Methods II and III.
Links for SPSS 10.0 Tutorial
Statistical Package for the Social Sciences
It covers a broad range of statistical procedures that allow you to summarize
data (e.g., compute means and standard deviations), determine whether there
are significant differences between groups (e.g., t-tests, analysis of
variance), examine relationships among variables (e.g., correlation, multiple
regression), and graph results (e.g., bar charts, line graphs). Written by Gil
Einstein and Ken Abernethy
Analysis of Variance (ANOVA) Procedures
J. Cooper Cutting SPSS for Windows: Brief How-To's Introduction SPSS basics Graphs
Descriptive Statistics Crosstabulation and Chi-square Reliability Regression Procedures
T-tests ANOVAs Nonparametric tests Factor Analysis Cluster Analysis Discrimina
lilt.ilstu.edu
Using SPSS 10.0 for Windows (for statistical analysis)
This link introduces you how to collect, interpret, and report the data by using Windows
Excel program. ANOVA, Independent T-test, Levene’s test, and other statistical methods
for analysis are included.
SPSS Tutorial for Research and Analysis
This link will take you to the Confidence Interval, One sample T-test, Independent T-test,
Mann Whitney U test, Paired T-Test, and Wilcoxon Signed Rank Test and Sign Test and
interpreting the output data.
Computing a T-Test for Between-Subjects Designs
In this link, we will describe how to analyze the results of between-subjects designs.
It is important to distinguish between these two types of designs because they require
different versions of the t-test.