EDF 6472
Introduction to Data Analysis in Education
Assignments Due September 24, 2012 (Chapter 3) – Solutions
Green, et al.
Ann wants to describe the demographic characteristics of a sample of 25 individuals
who completed a large-scale survey. She has demographic data on the participants’
gender (two categories), educational level (four categories), marital status (three
categories), and community population size (eight categories).
Before answering any of the questions, we want to open Lesson 20 Exercise File 1, the
file that contains the data described above. When we do this, we see the Data View
window shown below.
1. Conduct a frequency analysis on the gender and marital status variables. From the
output, identify the following. 1) Percent of men, b) Mode for marital status, c)
Frequency of divorced people in the sample.
To carry out the frequency analysis we will use the Frequencies function of SPSS.
To find it, first click on the Analyze menu on top of the data view screen. Now choose
1
Descriptive Statistics from the menu and finally choose Frequencies from the submenu
that appears. Note the demonstration shown below.
You should now see the Frequencies dialog box. Select the variables gender and
marital and move these over to the Variable(s): window on the right. The window
should look like the one below.
Click on the OK button
to obtain the output
shown on the next page.
2
Frequencies
Statistics
N
Valid
Missing
Gender
25
0
Marital Status
25
0
Frequency Table
Gender
Valid
Men
Women
Total
Frequency
13
12
25
Percent
52.0
48.0
100.0
Valid Percent
52.0
48.0
100.0
Cumulative
Percent
52.0
100.0
Ma rita l Sta tus
Valid
Married
Divorced
Never Married
Total
Frequency
9
11
5
25
Percent
36.0
44.0
20.0
100.0
Valid Percent
36.0
44.0
20.0
100.0
Cumulative
Percent
36.0
80.0
100.0
a. The percent of men in the sample is found by observing the Valid Percent value for
Men in the first frequency table. We see that 52% of the participants are men.
b. The mode is the value that has the highest frequency. In this sample, as we can see
from the second frequency table, the mode for marital status is Divorced with a
frequency of 11 participants. We could get the same information by looking at the
value that has the highest percent.
c. The second frequency table tells us that 11 people in this sample are diverced.
2. Create a frequency table to summarize the data on the educational level variable.
We will use the Frequencies facility as we did in Question 1 to create a frequency
table to summarize the data on educational level. First click on the Analyze menu on top
of the data view screen. Now choose Descriptive Statistics from the menu and finally
choose Frequencies from the submenu that appears.
You should now see the Frequencies dialog box. Select the variable educ and
move it over to the Variable(s): window on the right. The window should look like the
one on the next page.
3
Now, click on the OK button and
obtain the output shown below.
This is a frequency table of the
data on the educational level
variable.
Frequencies
Ma rita l Sta tus
Valid
Married
Divorced
Never Married
Total
Frequency
9
11
5
25
Percent
36.0
44.0
20.0
100.0
Valid Percent
36.0
44.0
20.0
100.0
Cumulative
Percent
36.0
80.0
100.0
Julie asks 50 men and 50 women to indicate what type of books they typically read for
pleasure. She codes the responses into 10 categories: drama, mysteries, romance,
historical nonfiction, travel books, children’s books, poetry, autobiographies, political
science, and local interest books. She also asks participants how many books they read
in a month. She categorizes their responses into four categories: no readers (no books a
month), light readers (1-2 books a month), moderate readers (3-5 books a month), and
heavy readers (more than 5 books a month). Julies SPSS data file contains two
variables, book, a 10-category variable for type of books read, and readers, a fourcategory variable indicating the number of books read per month.
We must first open file Lesson 20 Exercise File 2 to view this data. It is shown in
the data screen pictured below.
4
5. Create a table to summarize the types of books that people report reading.
We can create a frequency table using the variable book as we did in Question 1.
The Frequency dialog box should look like the one show below.
Next we click on
the OK button and obtain
the frequency distribution
on the next page.
5
Frequencies
Statistics
Book Type
N
Valid
Missing
100
0
Book Type
Valid
Drama
Mysteries
Romance
Historical Nonfiction
Travel Books
Children's Literature
Poetry
Autobiographies
Political Science
Local Interest
Total
Frequency
16
7
10
6
13
10
9
9
8
12
100
Percent
16.0
7.0
10.0
6.0
13.0
10.0
9.0
9.0
8.0
12.0
100.0
Valid Percent
16.0
7.0
10.0
6.0
13.0
10.0
9.0
9.0
8.0
12.0
100.0
Cumulative
Percent
16.0
23.0
33.0
39.0
52.0
62.0
71.0
80.0
88.0
100.0
6. Crate a pie chart to describe how many books per month Julie’s sample reads.
The easiest way to construct a pie chart is to use the Graphs menu on the top of
the Data View Screen in SPSS. Click on this menu and then on the Pie… choice as
shown on the next page.
6
This will give us the Bar Chart dialog box shown below. Note that Summaries
for groups of cases is the
default for the arrangement
of data in the pie chart. Take
this default and click on the
Define button to obtain the
Define Pie: Summaries for
Groups of Cases dialog box
shown on the next page.
Select the variable
reader and move it into the
Define Slices by: window
using the right arrow since
this is the variable that will determine the area of each of the slices of the pie chart. Now,
click on the OK button to obtain the output shown on the next page.
7
Graph
Your pie chart
may look a little different
than the one to the left.
It may have pieces of
solid colors or it may
have different filler
patterns. This is because
of user settings that can
be found under the Edit
menu in SPSS. Just click
on the Edit menu and
then click on Options… .
Then click on the Charts
tab to see how these
colors and other
characteristics of charts
can be changed in SPSS.
8
Hinkle, et al,
1. A college dean examines the cumulative grade-point averages of 120 sophomore
students. The following frequency distribution is derived using class intervals of width
0.3.
Class interval
3.8 – 4.0
3.5 – 3.7
3.2 – 3.4
2.9 – 3.1
2.6 – 2.8
2.3 – 2.5
2.0 – 2.2
1.7 – 1.9
1.4 – 1.6
1.1 – 1.3
0.8 – 1.0
f
4
8
15
18
20
17
12
12
10
4
0
a. Draw a histogram and a frequency polygon for the distribution.
This will be easier to do if we expand the frequency distribution table shown
above to include the exact limits and the midpoint of each of the intervals. See the table
below.
Class interval Exact limits Midpoint f
cf
%
c%
3.8 – 4.0
3.75 – 4.05
3.9
4 120 3.3 100.0
3.5 – 3.7
3.45 – 3.75
3.6
8 116 6.7 96.7
3.2 – 3.4
3.15 – 3.45
3.3
15 108 12.5 90.0
2.9 – 3.1
2.85 – 3.15
3.0
18 93 15.0 77.5
2.6 – 2.8
2.55 – 2.85
2.7
20 75 16.7 62.5
2.3 – 2.5
2.25 – 2.55
2.4
17 55 14.2 45.8
2.0 – 2.2
1.95 – 2.25
2.1
12 38 10.0 31.6
1.7 – 1.9
1.65 – 1.95
1.8
12 26 10.0 21.6
1.4 – 1.6
1.35 – 1.65
1.5
10 14 8.3 11.6
1.1 – 1.3
1.05 – 1.35
1.2
4
4 3.3
3.3
0.8 – 1.0
0.75 – 1.05
0.0
0
0 0.0
0.0
A histogram representing distribution in the table is shown below.
9
The corresponding frequency polygon is shown below.
b. Find P10 ,P45 , P60 , and P95 .
10
 Np  cf 
w where
The percentiles can be found by using the formula PX    
f i 

 is the real lower (exact) limit of the interval that has the percentile we are looking for,
N is the number of subjects in the distribution, p is the proportion of subjects with scores
below the percentile we are looking for, cf is the cumulative frequency of the interval
below the interval that contains the desire percentile, fi is the frequency of the interval
that contains the percentile we are looking for, and w is the width of the interval.
The interval 1.4 – 1.6 contains the 10th percentile. We can see this by noting that
this interval has 11.6% of the scores in or below it. The next lowest interval only has a
cumulative percentage of 3.3%. Clearly, the score that has 10% of the scores at or below
it is in the interval 1.4 – 1.6. The real or exact lower limit  of this interval is 1.35.
There are 120 scores in this distribution, so N equals 120. Ten percent of the scores are at
or below the 10th percentile, so p=.10. The cumulative frequency of the interval below
the interval that contains the 10th percentile (i.e., the interval 1.1 – 1.3) is 4. The
frequency of the interval containing P10 (1.4 – 1.6) is 10, so f1 = 10. Finally, w is the
width of the interval which is, in this case, 3. So, we find that P10 , the 10th percentile, is
 Np  cf
P10    
fi


 120.10  4 
w  1.35  
.3  1.35  0.24  1.59
10



The rest of the percentiles in this question are found in a similar manner.
 Np  cf
P45    
fi


 120.45  38 
w  2.25  
.3  2.25  0.28  2.53
17



 Np  cf
P60    
fi


 120.60  55 
w  2.55  
.3  2.55  0.26  2.81
20



 Np  cf
P95    
fi


 120.95  108 
w  3.45  
.3  3.45  0.23  3.68
8



c. Find the percentile ranks of cumulative grade point overages of 2.40, 2.75, 3.25, and
3.60.
The percentile ranks of raw scores can be found using the formula
 Np  cf 
 120.60  55 
w  2.55  
P60    
.3  2.55  0.26  2.81
f
20


i


11
where all the symbols mean the same thing as in Part b and X corresponds to the raw
score for which we are trying to find the percentile rank. In the first example, we see that
a cumulative grade point average of 2.40 is in the interval 2.3 – 2.5 which has a real
lower limit of 2.25 and a frequency 17. So,   2.25 and fi = 17. The cumulative
frequency of the interval below 2.3 – 2.5 (that is, 2.0 – 2.2) is 38, so cf = 38. We are
looking for the percentile rank of a score of 2.40, so X = 2.40. The width of the intervals
(w) is 3 and there are 120 scores in this distribution, so N = 120. Therefore, the percentile
rank of a cumulative grade point average of 2.40 in this distribution is
PR2.40
X  
2.40  2.25

 
17
 cf  w  f1   38 
.3


100  38.75
N
120

 


 

The other three percentile ranks are found in a similar manner.
PR2.75
X  
2.75  2.55

 
20
 cf  w  f1   55 
.3


100  56.94
N
120

 


 

PR3.25
X  
3.25  3.15

 
15
 cf  w  f1   93 
.3


100  81.67
N
120

 


 

PR3.60
X  
3.60  3.45 

 
8
 cf  w  f1  108 
.3


 
100  93.33
N
120

 


 

12
10. The following are the final examination score of 40 students in a basic statistics
class. These scores were randomly selected from the records of all students who
have taken the same course over the past 10 years and have taken the standardized
final examination.
a. Determine the mean and median.
The data are
58 86 70 80 82 The mean of a set of scores is found using the formula
88 60 80 72 75
X
89 61 72 76 80 X   . In this case We add up the values to find ΣX,
N
63 73 82 81 89
75 65 82 86 90
 X 3069
75 63 65 84 82 which is equal to 3096. So, X  N  40  77.4 .
76 68 82 91 94
68 74 79 84 96 The median will be the center score in the distribution.
Since there are 40 scores this means that the median will be the score between the 20th
and the 21st score. To find these we must first rank order the data. In this case we rank in
ascending order and obtain the following:
Rank Score Rank Score
1
58
21
80
2
60
22
80
3
61
23
80
4
63
24
81
5
63
25
82
6
65
26
82
7
65
27
82
8
68
28
82
9
68
29
82
10
70
30
84
11
72
31
84
12
72
32
86
13
73
33
86
14
74
34
88
15
75
35
89
16
75
36
89
17
75
37
90
18
76
38
91
19
76
39
94
20
79
40
96
In the table to the left we can see that the 20th score is 79
and the 21st score is 80. The median score is 79.5 since it
is half way between these two scores.
13
b. Determine the variance and standard deviation.
X
XX
X  X 
58
88
89
63
75
75
76
68
86
60
61
73
65
63
68
74
70
80
72
82
82
65
82
79
80
72
76
81
86
84
91
84
82
75
80
89
90
82
94
96
-19.40
10.60
11.60
-14.40
-2.40
-2.40
-1.40
-9.40
8.60
-17.40
-16.40
-4.40
-12.40
-14.40
-9.40
-3.40
-7.40
2.60
-5.40
4.60
4.60
-12.40
4.60
1.60
2.60
-5.40
-1.40
3.60
8.60
6.60
13.60
6.60
4.60
-2.40
2.60
11.60
12.60
4.60
16.60
18.60
376.36
112.36
134.56
207.36
5.76
5.76
1.96
88.36
73.96
302.76
268.96
19.36
153.76
207.36
88.36
11.56
54.76
6.76
29.16
21.16
21.16
153.76
21.16
2.56
6.76
29.16
1.96
12.96
73.96
43.56
184.96
43.56
21.16
5.76
6.76
134.56
158.76
21.16
275.56
345.96
The variance of a distribution of scores is found
2
 X  X 

2
using the formula S
2
. Note that
n 1
we use the denominator n-1 since we are told
that this is a sample of a larger population (all
the students who have taken the final exam over
a period of years. We want to estimate the
population standard deviation from the standard
deviation this sample, so we subtract one from
the sample size to adjust for the fact that
samples tend to have smaller variation than the
populations they are drawn from. In part a. of
this problem we found that the mean of this
distribution of scores was 77.4, so we will begin
by subtracting 77.4 from each score as shown in
the table to the left.

Now we compute X  X

2
by squaring the
differences and placing them in the appropriate
column of the table. We find that
X  X 
2
 3735.60 . So,
 X  X 

2
S
2
n 1

3735.60 3735.60

 95.78
40  1
39
The standard deviation is simply the square
root of the variance, so
S  S 2  95.78  9.79 .
X  X 
2
 3735.60
14
22. John took three standardized tests.
a. Compute a z score for each of John’s test scores.
The three standardized test scores obtained by John were
Test
Score X s
Chemistry
85
82 10
Mathematics
80
75 8
English
80
90 12
The z score for a particular score is the number of
standard deviations that that score is from the mean of
the distribution. It is, therefore, found using the
XX
formula z 
.
s
85  82 3

 .3 .
a. For Chemistry the z score is z 
10
10
80  75 5
  .625 .
For Mathematics the z score is z 
8
8
80  90  10

 -.833.
For English, the z score is z 
12
12
b. What appears to be John’s strongest subject area?
Johns strongest subject area on the basis of these tests in mathematics. He had the
highest z score on this examination.
15