dr. william allan kritsonis - statistics

7/28/2019 Dr. William Allan Kritsonis - Statistics

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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.



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.



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.



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.”



♦ 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.



Dr. William Allan Kritsonis

Review for the Comprehensive PhD Examination

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



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

48 150 1

46 1

48 1

45 1

48 1

47 1

45 2

44 2

46 2

44 2

41 2

43 2

46 249 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.



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

9 100.0% 0 .0% 9 100.0%

9 100.0% 0 .0% 9 100.0%

8 100.0% 0 .0% 8 100.0%

TEACHMTH

1.00

2.00

3.00

READSCR

N Percent N Percent N PercentValid Missing Total

Cases

Tests of Normality

.193 9 .200* .933 9 .490

.173 9 .200* .950 9 .667

.193 8 .200* .919 8 .437

TEACHMTH

1.00

2.003.00

READSCR

Statistic df Sig. Statistic df Sig.

Kolmogorov-Smirnova

Shapiro-Wilk

This is a lower bound of the true significance.*.

Lilliefors Significance Correctiona.

Test these two assumptions for each of the three groups:



(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



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.



Descriptives

47.5556 .6894

45.9657

49.1454

47.5062

48.0000

4.278

2.0683

45.00

51.00

6.00

3.5000

.335 .717

-.651 1.400

44.6667 .7454

42.9479

46.3855

44.6296

44.0000

5.000

2.2361

41.00

49.00

8.00

2.5000.450 .717

1.300 1.400

48.5000 1 .3628

45.2776

51.7224

48.3889

48.5000

14.857

3.8545

44.0055.00

11.00

6.0000

.429 .752

-.887 1.481

Mean

Lower Bound

Upper Bound

95% Confidence

Interval for Mean

5% Trimmed Mean

Median

Variance

Std. Deviation

Minimum

Maximum

Range

Interquartile Range

Skewness

Kurtosis

Mean

Lower Bound

Upper Bound

95% Confidence

Interval for Mean

5% Trimmed Mean

Median

Variance

Std. Deviation

Minimum

Maximum

Range

Interquartile RangeSkewness

Kurtosis

Mean

Lower Bound

Upper Bound

95% Confidence

Interval for Mean

5% Trimmed Mean

Median

Variance

Std. Deviation

Minimum

Maximum

Range

Interquartile Range

Skewness

Kurtosis

TEACHMTH1.00

2.00

3.00

READSCRStatistic Std. Error



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,


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,


0.43 and –0.887, respectively. Method III can be considered as a normal

distribution.



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

9 47.5556 2.0683 .6894 45.9657 49.1454 45.00 51.00

9 44.6667 2.2361 .7454 42.9479 46.3855 41.00 49.00

8 48.5000 3.8545 1.3628 45.2776 51.7224 44.00 55.00

26 46.8462 3.1457 .6169 45.5756 48.1167 41.00 55.00

1.00

2.00

3.00

Total

N Mean

Std.

Deviation Std. Error

Lower

Bound

Upper

Bound

95% Confidence

Interval for Mean

Minimum Maximum



Test of Homogeneity of Variances

READSCR

3.641 2 23 .042

Levene

Statistic df1 df2 Sig.

ANOVA

READSCR

69.162 2 34.581 4.463 .023

178.222 23 7.749

247.385 25

Between Groups

Within Groups

Total

Sum of

Squares df

Mean

Square F Sig.

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.



Post Hoc Tests

Multiple Comparisons

Dependent Variable: READSCR

2.8889 1.3122 .114 -.4993 6.27

-.9444 1.3526 1.000 -4.4369 2.54

-2.8889 1.3122 .114 -6.2771 .49

-3.8333* 1.3526 .028 -7.3258 -.34

.9444 1.3526 1.000 -2.5480 4.43

3.8333* 1.3526 .028 .3408 7.32

2.8889* 1.3122 .035 .1815 5.59

-.9444 1.3526 .909 -5.2769 3.38

-2.8889* 1.3122 .035 -5.5963 -.18

-3.8333 1.3526 .091 -8.2019 .53

.9444 1.3526 .909 -3.3880 5.27

3.8333 1.3526 .091 -.5352 8.20

(J) TEACHMTH

2.00

3.00

1.00

3.00

1.00

2.00

2.00

3.00

1.00

3.00

1.00

2.00

(I) TEACHMTH

1.00

2.00

3.00

1.00

2.00

3.00

Bonferroni

Tamhane

Mean

Difference

(I-J) Std. Error Sig.

Lower

Bound

Uppe

Boun

95% Confidence

Interval

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:



♦ 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 GraphsDescriptive 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 WindowsExcel 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.

http://s9000.furman.edu/mellonj/spss1.htm

http://lilt.ilstu.edu/jccutti/138web/spss.html

http://www.urban.uiuc.edu/courses/Varkki/spss/

http://www.furman.edu/~dhaney/ch5.htm

http://s9000.furman.edu/mellonj/spss_3.htm

http://s9000.furman.edu/mellonj/spss1.htm

http://lilt.ilstu.edu/jccutti/138web/spss.html

http://www.urban.uiuc.edu/courses/Varkki/spss/

http://www.furman.edu/~dhaney/ch5.htm

http://s9000.furman.edu/mellonj/spss_3.htm

dr. william allan kritsonis - statistics

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