Download Chapter Two: Practice Problem Key

Turner, J. Using statistics in small-scale language education research: Focus on non-parametric data
Chapter Two: Practice Problem Key
Practicing the Concepts—Variables and Scales
1. Here are some variables one might consider in doing research to explore the conditions for
optimal language learning. Be prepared to discuss what type of scale might be used in defining
the levels of each: nominal, ordinal, or interval.
I think you’ll notice when you discuss your responses with others that people define some of
these variables in different ways—the “right” answer for a variable should reflect an individual’s
understanding of what the variable is.
Variable name
Native language
of learner
Language
learning aptitude
Type of scale
nominal
Language ability
interval,
ordinal, or
nominal
Learner’s age
interval,
ordinal, or
nominal
interval,
ordinal, or
nominal
Comments
An individual can be placed into a category representing
his or her native language.
Language learning aptitude is a person’s potential ability
to learn a second or foreign language. I envision this
variable as being interval scale, because I am familiar with
several interval-scale tests of language learning aptitude;
however, others may envision the variable as being
ordinal, particularly if they are aware of how the outcomes
of the Defense Language Aptitude Battery may be used to
place foreign language students into an easier or more
difficult language of study. Some people may even
envision the variable as being nominal—one person told
me he thought of the variable as having two categorical
levels, “able to learn foreign languages” and “not able to
learn foreign languages.” A researcher who includes this
variable in a study should provide a definition that makes
good sense in the context of the study, taking into
consideration the research question as well as features of
the research design.
There are many types of tools for measuring or describing
language ability. Tests like the Internet Based TOEFL
(IBT) yield interval scale data; the American Council for
Teachers of Foreign Language has an Oral Proficiency
Interview that yields ordinal scale data; and some tests for
grades K – 12 provide results that allow teachers to
categorize their learners according to their language
ability—these tests yield nominal scale outcomes. A
researcher who includes language ability as a variable in a
study needs to provide a definition that makes sense in the
context of the study, taking into consideration the research
question as well as features of the research design.
Time, the concept that is typically used to define age, is an
interval scale variable. However, a researcher may decide
to define age as an ordinal variable, placing participants in
Turner, J. Using statistics in small-scale language education research: Focus on non-parametric data
Learner’s sex
nominal
Teacher’s
experience
interval,
ordinal, or
nominal
Teacher’s
education
interval,
ordinal, or
nominal
Teacher’s
personality
nominal;
perhaps ordinal
or interval?
Class meeting
time
nominal
Frequency of
class
nominal
Duration of class
nominal
Configuration of
seats
nominal
ordered levels. In some studies, age may even be defined
as a nominal variable; for example, in a study in which the
participants are placed into two groups, learners between
the ages of 21 and 35 and learners who are either younger
or older than students in the first group.
Sex is typically defined as a nominal variable with two
levels, male and female, in language education research.
Teacher’s experience could be defined in a number of
ways, including years of experience, which could be
defined on either an interval scale (simply number of
years) or an ordinal scale (with teachers placed into
ordered categories depending on the number of years’
experience they have). Teacher’s experience could be
defined using a nominal scale, too, perhaps as type of
experience.
Teacher’s education could be defined in a number of
ways, too, including years of education, expressed using
an interval scale (simply number of years), or an ordinal
scale (with teachers placed into ordered categories
depending on the number of years of education they’ve
had). Teacher’s education could be defined using a
nominal scale, too, with the levels being something like
high school diploma, associate degree, bachelor’s degree,
master’s degree, doctoral degree.
Teacher’s personality can be defined in many ways, too.
A researcher might choose to use a tool that categorizes or
characterizes individuals according to their personal
traits—a nominal scale. Alternatively, a single personality
trait might be identified, such as degree of extroversion or
introversion—and a data collection tool used that yields
an interval or ordinal degree of extroversion score.
Though time is an interval scale variable, I envision class
meeting time as a nominal variable—the time of day at
which a class meets is simply a category. The categories
might be given numerical labels (8 a.m., 6 p.m., etc.) or
semantic ones (morning, early afternoon, evening, etc.).
Just as classes might be characterized by when they meet,
they might also be categorized by how frequently they
meet (once a week; twice a week; five times a week, etc.).
Classes can be characterized or categorized by how many
hours each session is, or whether the course is a weekend
seminar or a full-length course.
I understand this variable, configuration of seats, as the
manner in which a classroom is set up. I consider it a
nominal variable because I can categorize or characterize
classrooms by the way the furniture is arranged (rows of
Turner, J. Using statistics in small-scale language education research: Focus on non-parametric data
desks, small tables with seating for 4, conference table,
etc.).
2. No key for this question.
Study A. A Spanish vocabulary test was given to a group of Spanish-speaking first-graders. The
school district requires you to provide the mean and standard deviation for your students.
Complete the chart below to guide your computation of these values.
First the mean score must be determined using this formula: X =
X .
n
The sum of the scores (ΣX) is 326. The number of test takers (n) is 13. Divide 326 by 13 to
determine the mean score; 326/13 = 25.07692. I rounded that number off to 25.08 in my work
below: X = 25.08.
X
(X - X )
(X - X )2
41
41 - 25.08 = +15.92
(+15.92)2 = +253.45
33
33 - 25.08 = +7.92
(+7.92)2 = +62.73
32
32 - 25.08 = +6.92
(+6.92) 2 = +47.89
29
29 - 25.08 = +3.92
(+3.92) 2 = +15.37
27
27 - 25.08 = +1.92
(+1.92) 2 = +3.69
27
27 - 25.08 = +1.92
(+1.92) 2 = +3.69
26
26 - 25.08= +0.92
(+0.92) 2 = +0.85
24
24 - 25.08= -1.08
(-1.08) 2 = +1.17
19
19 - 25.08= -6.08
(-6.08) 2 = +36.97
19
19 - 25.08= -6.08
(-6.08) 2 = +36.97
18
18 - 25.08= -7.08
(-7.08) 2 = +50.13
17
17 - 25.08= -8.08
(-8.08) 2 = +65.29
14
14 - 25.08= -11.08
(-11.08) 2 = +122.77
The sum of the values in the third column, Σ(X - X )2, is 700.97
(X
The standard deviation formula is:
 X )2
n 1
.
The sum of the (X - X )2 values in the third column above is 700.97 and the number of
participants (n) is 13, so n – 13 = 12.
Turner, J. Using statistics in small-scale language education research: Focus on non-parametric data
(X
So
 X )2
n 1
=
700.97
13  1
=
700.97
12
.
Carrying out the math, I find that the standard deviation is 7.64 points.
700.97
12
=
58.41 = 7.64
3. What is the mean? 25.07692 (which I rounded off to 25.08)
4. What is the mode? This distribution is bimodal; there are two scores that are most frequently
occurring, 19 and 27.
5. What is the median? There are 13 scores. The median is the score in the physical center of
the distribution when the scores are arranged from highest to lowest or vice versa so the median
(which is highlighted in blue in the table above) is 26.
6. What is the range? The range is the highest score minus the lowest score so this distribution
has a range of 27 points (41 – 14 = 27).
7. What is the standard deviation? 7.64 points
8. Here’s my representation of the distribution...
X
... 14
X
X X X
16 18 20 22
X
X X X X
24 26 28 30
X X
32 34
X
36
38
40
42 ...
Turner, J. Using statistics in small-scale language education research: Focus on non-parametric data
Study B
9. Here are the R commands I used to enter the data and provide the information needed to
answer questions 9 through 14.
Enter the data.
data = c (88, 87, 54, 97, 34, 78, 56, 99, 87, 73,
74, 69, 85, 87, 86, 87)
Use the summary
command to calculate the
mean (blue highlighting)
and identify the median
(green highlighting). (The
minimum and maximum
scores are highlighted in
yellow.)
Use the subset (table)
commands to determine
the mode.
summary (data)
Here’s the output of that command:
Min. 1st Qu. Median Mean 3rd Qu. Max.
34.00 72.00 85.50
77.56 87.00
99.00
subset (table (data),
table(data)==max(table(data)))
Here’s the output—it means that the most
frequent score is 87, and that there are 4 scores of
87.
87
4
To calculate the standard
deviation, I used the
shortcut command sd
(pink highlighting).
I used the hist command
to make a frequency
distribution of the scores.
(I added color and a
label).
To calculate the range, I
retrieve the maximum
and minimum scores
from the output for the
summary command and
find the difference. (The
maximum and minimum
scores are highlighted in
yellow in the summary
output.)
sd (data)
Here’s the output:
17.23163
hist (data, col = "lavender", main =
"Frequency Distribution of Reading Quiz
Scores")
The histogram is presented below in question #9.
99 – 34
R presents the difference like this:
[1] 65
Turner, J. Using statistics in small-scale language education research: Focus on non-parametric data
9. Make a frequency distribution of the data.
10. Calculate the mean (highlighted in blue in the chart of R commands above). 77.56
11. Calculate the standard deviation (highlighted in pink in the chart of R commands above)—I
round it off to 17.23 here.
13. Report the median (highlighted in green in the R commands). 85.5
14. What is the range? The range is 65 points (the difference between the maximum and the
minimum scores).
Standardized Scores
When scores are in a normal distribution, one can use that fact to ‘regularize’ them; that is, to
convert the scores from the raw score scale to a recognized ‘common’ scale (a standardized
scale), so the scores from different tests (or test forms) can be directly compared even when the
number of points on the different tests, the averages, and the standard deviations are different.
Keep in mind that scores can be standardized only when they are in a normal distribution
because the process relies on the defined characteristics of a normal distribution.
Here is some test performance information.
Midterm test
Mean = 47
s=3
n = 45
k = 100
Rob’s score = 53
Final test
Mean = 51
s=6
n = 43
k = 100
Rob’s score = 53
Turner, J. Using statistics in small-scale language education research: Focus on non-parametric data
Michael’s score = 39
Michael’s score = 49
15. Calculate the percentages for Rob’s and Michael’s midterm scores and final scores.
Midterm
Final
Rob
53/100 = 53%
53/100 = 53%
Michael
39/100 = 39%
49/100 = 49%
16. Calculate the z-scores for Rob’s and Michael’s midterm scores and final scores.
The z-score formula resets the mean at zero and resets the standard deviation to 1.
The formula looks like this: z =
Midterm
Final
Rob
X  X 53  47
=
=
3
s
X X
53  51
=
=
6
s
X X
s
6
= +2
3
2
= + .33
6
Michael
X X
39  47 6
=
=
= -2
3
3
s
X  X 49  51 2
=
=
= - .33
6
6
s
17. Convert the z-scores to the T-scale or the CEEB scale.
Here’s the formula for converting z-scores to T-scores:
T = 10z + 50
Here’s the formula for converting z-scores to CEEB-scores:
CEEB = 100z + 500
T-scale conversion:
Midterm
Final
Rob
T = 10z + 50 =
(10)(+2) + 50 =
20 + 50 = 70
T = 10z + 50 =
(10)(+.33) + 50 =
3.3 + 50 = 53.30
Michael
T = 10z + 50 =
(10)(-2) + 50 =
- 20 + 50 = 30
T = 10z + 50 =
(10)(-.33) + 50 =
-3.3 + 50 = 46.70
Turner, J. Using statistics in small-scale language education research: Focus on non-parametric data
CEEB scale conversion
Midterm
Final
Rob
CEEB = 100z + 500 =
(100)(+2) + 500 =
200 + 500 = 700
CEEB = 100 z + 500 =
(100)(+.33) + 500 =
33 + 500 = 533
Michael
CEEB = 100z + 500 =
(100)(-2) + 500 =
- 200 + 500 = 300
CEEB = 100z + 500 =
(100)(-.33) + 500 =
-33 + 500 = 467
I believe I’d use the T-scale scores to discuss Rob’s and Michael’s performance with them and
their parents.
For Rob and his parents, I think I’d stress that in comparison to his peers he was doing very well
at midterm—as well as or better than 84% of them. However, on the final he didn’t do so well—
his score on the final test is only slightly above the mean. Unless he was ill when he took the
final test or there was some problem during the test administration, his performance on the final
(in comparison to his peers’ performance) suggests that he’s having difficulty with the material.
He may need some extra tutoring, a remedial summer class, or a curfew!
For Michael and his parents, I believe I’d note that he is working below the level of most of his
peers, but that he had made astounding progress. On the midterm, he performed the same as or
better than 16% of his peers. However, his score on the final test approaches the mean score—he
performed as well as or better than nearly half of his peers. I’d congratulate him on his progress
and encourage him to continue his hard work. He might benefit from extra tutoring and a
remedial summer class.