This presentation includes the following subtopics • Norm- Referenced and Criterion Referenced Assessment • Measures of Central Tendency • Measures of Location/Point Measures • Measures of Variability • Standard Scores • Skewness and Kurtosis • Correlation

1. Ed. 310 – Assessment of Student Learning 1
CHAPTER 6

3. NORM-REFERENCED TESTS CRITERION-REFERENCED TESTS
1.Norm-referenced tests are used to
determine the
achievements of individuals in comparison
with the
achievements of other individuals who
take the same test.
1. Criterion-referenced tests are used to
determine the achievements of
individuals in comparison with criterion,
usually an absolute standard.
2. In a norm-referenced test, the quality
of achievements of a student is
determined by the distance of his score
from the mean or median
2. In a criterion-referenced test, the
quality of achievement of a student
is determined by the distance of his
score from the criterion established.
3. Norm-referenced tests are designed
to produce variability among
individuals.
3. In criterion-referenced tests,
variability is irrelevant.

4. NORM-REFERENCED TESTS CRITERION-REFERENCED TESTS
4. Norm-referenced tests are used for
selection and grouping purposes.
CRT are used to determine the level
of skill or knowledge of individuals if
they are capable of qualified to apply
such skill or knowledge.
5. Norm Referenced Tests, non
discriminating items such as items that
are easy, too difficult and improved.
5. In a criterion-referenced test,
too easy or too difficult items
are not removed, rather they
should be included if they truly
reflect the skill being measured.

6. Relates to a point in a distribution
around which the scores tend to center.
This point can be used as the most
representative value for a distribution of
scores. A measure of central tendency is
helpful in showing where the average or
typical score fails.

7. Is the point or score at the midpoint
of the distribution of scores arranged
from highest to lowest or vice versa.

8. Calculations of the Median for Ungrouped
Scores
When the number cases are odd,
arrange the scores from highest to lowest or
vice versa. Write down all the scores, the
median is the middlemost score.

9. Computation of the Median for Grouped
Data
Given this frequency distribution/ grouped data
X F
90-94 1
85-89 2
80-84 7
75-79 9
70-74 11
65-69 8
60-64 5
55-59 5
50-54 1
45-49 1
N= 50

10.
1. Use the formula
(N/2- Fl)
Mdn = LL + F x i
Where:
LL= the real lower limit of the median class
N/2=half sum
F1= partial sum
f= frequency of the class interval where the
median lies
N= the number of cases
i= the interval

11. 2. Find the values of the symbols:
•N/2 = 50/2 = 25
•FI = Add the frequencies of the score from the lower
score end upward until reaching half sum but not
exceeding it. (1 + 1 + 5 + 5 +8 =20). Twenty (20) is the
partial sum from the lower limit. The median (25th
score lies in the step-interval 70-74 and its frequency is
11)
•The value off is 11 ( the frequency of the interval
where the median lies)
•LL is 69.9 ( the real lower limit of 70 – 74 = the
interval where the median lies)
•i, the interval of the class limits, is 5

12. 3. Substitute the values for the symbols in the
formula and solve.
Mdn = 69.4 + (25-20) x 5
11
= 69.5 + 5 x 5
11
= 69.5 + (4545) x 5
= 69.5 + 2.2725
Mdn = 71.77

13.
4. Check the answer by using the formula:
Mdn =UI – (N/2-Fu) x i
f

14.
4.1 Find the values of the symbols and solve
4.1.1 N/2 =50/2 =25
4.1.2 Fu = Add the frequencies of the score from the upper score end
downward until reaching half sum but not exceeding it. (1+ 2+7+9=19)
Nineteen is the partial sum from the upper limit. The median (25th
score lies
in the step-interval and its frequency is 11)
4.1.3 The value of f is 11 (the frequency of the interval where the median
lies)
4.1.4 UL is 74.5 (the real lupper limit of 70-74= the interval where the
median lies)
4.1.5 I. the interval of the class limits, is 5
Mdn = 74.5 – (25- 19) x 5
11
Mdn = 74.5 – ( 6 ) x 5
11
Mdn = 74.5 – ( 5454 ) x 5
Mdn = 74.5 – 2.727

15.
The mean or the arithmetic
mean is referred to as the
average of scores or measures.

16. It considered the best measure of central tendency due to the
following qualities:
1.Each score contributes its proportionate share in
computing the mean. The mean is more stable than the
median or the mode.
2.Since the mean average, it is best understood and
more widely used measure of central tendency
3.It is used as basis in computing other statistical
measures like the average deviation, standard
deviation, coefficient of variability, coefficient of
correlation, etc.

17. Computation of the Mean from Ungrouped
Data (when the number of cases is less than
30)
1.Use the formula: M=
( The sum of X divided b y N )
2.Write the scores in a column. They can be in
any order.
3.Count the number of scores to get N.
4.Add the scores to get the sum
5.Divide the sum by the number of cases.

18. The mean is:
X
68 M= 859/17
70 M= 50.529 or 50.53
56
45
60
54
63
48
35
29
45
63
36
49
36
55
47

19. Computation of the Mean for Grouped Data
1. The formula in finding Mean for Grouped Data is:
X=AM +
Where:
AM = assumed mean
= is the algebraic sum of the products of the
frequencies and their corresponding deviations from
the assumed mean
N = the number of cases
I = the class interval

20. 2. Steps in the Computation of the Mean:
2.1 Prepare a table of frequency or frequency distribution.
2.2 Assume a mean. The assumed mean can be in any part of the frequency
distribution, but it is advisable to get the midpoint of the class-interval at the
middle of the distribution, that one highest frequency.
2.3 Fill column D starting from the step where the assumed mean lies, assign
this a 0
deviation. Form 0, number the steps upward 1,2,3,4, and downward 1,2,3,4
etc. All deviations above the assumed mean have positive signs and all
deviations
below the assumed mean have negative signs.
2.4 Multiply the frequency by the deviation for each step to get the fd
column, and get
sum of fd. This is the algebraic sum of the fd column.
2.5 Divide summation fd by N and multiply by the class interval x i
2.6 Add the product to the assumed mean
2.7 Check the answer by assuming another mean

21. Example:
X f d fd
90-94 1 4 4
85-89 2 3 6
80-84 7 2 14
75-79 9 1 9 + 33
70-74 11 0 0
65-69 8 -1 -8
60-64 5 -2 -10
55-59 5 -3 -15
50-54 1 -4 -4
45-49 1 -5 -5 -42
N = 50 Efd= -9

22. 1. Assume a mean. Get the midpoint of the interval
where the assumed mean lies.
AM = 72
2. Fill in Column d (deviation). The deviation is the
spread of the score from a point of origin.
3. Fill in Column fd. The sum of the positive values is
+33 and that of the negative values is -42. The sum of fd
is -9.
4. Substituting the formula:
M = 72 + ( -9/50 ) 5
M = 72 + ( -0.18 ) 5
M = 72 + ( -0.9 )
M = 72 + 0.9
M = 71.10

23. 5. Check your answer by assuming another mean
X f d fd
90-94 1 5 5
85-89 2 4 8
80-84 7 3 21
75-79 9 2 18
70-74 11 1 11 + 63
65-69 8 0 -0
60-64 5 -1 -5
55-59 5 -2 -10
50-54 1 -3 -3
45-49 1 -4 -4 22
N = 50 Efd + 41

24. Given:
AM = 67
Efd = + 41
I = 5
N = 50
M= 67 + (+41/50) 5
M= 67 + (0.82) 5
M= 67 + 4.1
M= 71.10

25. Another method of computing the mean is through the midpoint method. The
formula is:
M= EFM
N
X f M fM
90-94 1 92 92
85-89 2 87 174
80-84 7 82 574
75-79 9 77 693
70-74 11 72 792
65-69 8 67 536
60-64 5 62 310
55-59 5 57 285
50-54 1 52 52
45-49 1 47 47
N= 50 EfM= 3555

26. Procedure:
1. Prepare a frequency distribution
2. Place column M which represents the
midpoints of each class interval
3. Fill in Column Fm by multiplying each
frequency by each corresponding midpoint
4. Find the sum of the data in Column M
5. Divide this by N.
M = 3555/50 = 71.10

27. The mode is the most
frequency occurring score in the
distribution. It is the score with the
highest frequency.

28. Determining the Mode from Ungrouped Scores
( Crude or Rough Mode )
Procedure:
1. Arrange the scores from highest to lowest
2. The score that occurs most often is the
crude mode.

29. Data:
25 30 37 41 52 52 30 37 42 37
X
52
52
42
41
37 Mode = 37
37
37
37
30
30
30
25

30. Determining the Crude Mode from Grouped
Scores (Frequency Distribution)
The crude mode is the midpoint of
the interval with the highest frequency.

31. X F
90-94 1
85-89 2
80-84 7
75-79 9
70-74 11 Crude Mode = 72
65-69 8
60-64 5
55-59 5
50-54 1
45-49 1
N= 50

32. Computation of the True Mode
The formula for the True Mode is:
Mo= 3Mdn- 2M
In which:
Mo = the mode
Mdn = the median
M= the mean

34. The measures of location or point measures
are the quartiles, deciles and percentiles are
points dividing the distribution into les. The
quartiles (Q1, Q2, Q3, and Q4) are points dividing
the distribution into four equal parts. The
percentiles (P1, P2, P3, etc) are points which
divide the score distribution into one hundred
equal parts.
The procedure in finding the point measures is
almost the same as that of the median.

35. Quartiles
The first quartile (Q1) is located at one- fourth of the
number of case, such as 25% of all the cases lie at or
below it and 75% at or above it.
The value of third quartile corresponds to the value
of the seventy-percentile. Seventy-five percent of all
the cases lie at or above it and 25% lie at or below it.
The value of the second quartile is equal to the value
of the median, such that 50%of all the cases lie at or
below it and 50% lie at or above it.

37. Finding Q1
X F CM
90-94 1 50
85-89 2 49
80-84 7 47
75-79 9 40
70-74 11 40
65-69 8 20
60-64 5 12
55-59 5 7
50-54 1 2
45-49 1 1
N=50
Procedure:
1. Add Column CM in the Frequency
Distribution. It stands for the
cumulative frequencies, is done
by adding the scores from the
lower score end upward.
2. Find N/4. 50/4 = 12.5. The twenty-
fifth score lies in the interval 65-
69.
3. Determine the partial sum (F).
That is the sum of the frequencies
upward which totals 25 (Q/4) but
not exceeding it. In the given
distribution, the partial sum (F) is
12.
4. The value of f is 8 since it is the
frequency of the interval where
Q1 lies.
5. The value of LL or lower limit is
64.5.

38. Substituting the formula:
Q1=64.5 + (12.5-12) x 5
8
Q1=64.5 + (0.5) x 5
8
Q1=64.5 + (0.06) x 5
Q1=64.5 + .30
Q1= 64.80
Third Quartile
Formula:
Q3=LL + (3N-F)
4 l
F
LL= 74.5+ (37.5-31) x 5
9
Q3=74.5 + (6.5/9) x5
Q3=74.5 + (.72) x 5
Q3= 74.5 + 3.6
Q3= 78.1

40. Finding the Percentiles
X F CM
90-94 1 50
85-89 2 49
80-84 7 47
75-79 9 40
70-74 11 40
65-69 8 20
60-64 5 12
55-59 5 7
50-54 1 2
45-49 1 1
N=50
Procedure:
1.Determined the desired percentile.
E.g. P20.
2.Find the percentile sum by
multiplying the number of cases
(N) of 5 by the percentage desired
20% of 50=50 *.20=10
3.Find the partial sum by adding the
frequencies of the scores from the
lower score end upward until
reaching the percentile sum but
not exceeding it. (1+1+5=7.
Percentile 20 or the 10th
score lies
at the interval 60-64.
4.Determine f= the frequency of 60-
64 is 5.
5.Determine LL. The exact or real
lower limit of 60-64 is 59.5
6.The interval is 5.

44. Introduction
The measures of central tendency represented by the mean,
median and mode are valuable statistical measures, but they
describe only the typical score representing the whole
distribution. They describe only tendency of the scores to pile
up or near the middle of the distribution. The measures of
variability or dispersion are important. They show the tendency
of the scores to spread or scatter above or below the central
point of dispersion. They show how close or how far the scores
are from each others. These measures also show the
homogeneity or heterogeneity of different sets of scores. The
higher the measure of variability the more homogenous is the
group; the lower the measure of variability, the more
heterogenous is the group.
The most common measures or variability are the range, the
standard deviation, the mean deviation and quartile deviation.
The most important and most often used in measurement and
research and in advanced statistics is the standard deviation.

45. Range
The range is the difference between the
highest and lowest scores. It provides a quick
approximation of the spread of the scores, but it
is not a dependable measure of variability
because it is calculated from only two values.
Example: Highest Score =78; lowest score is 25.
The range is 53.

46. Standard Deviation
The standard is the square root of the mean of the
squared deviation of all scores from the mean. It is basically a
measure of how far each score is from the mean. It is basically
a measure of how far each score is from the mean. Since the
standard deviation is based on deviations from the mean, these
two statistics are used together to give meaning to test scores.
Computation of the Standard Deviation from Ungroup scores

47. Procedure:
List the scores under X column.
Find the mean of the scores.
Place column (deviations); get the values by subtracting the mean
from each of the scores. When the scores are less than the mean,
the negative sign precedes the difference between the raw score
and the mean.
Place column (); square each of the values.
Find the sum of the squared deviation and divide it by the number
of cases.

48. X
43 7 49
41 5 25
40 4 16
38 2 4
37 1 1
33 -3 9
30 -6 36
29 -7 49
24 -12 144
24 -12 144
21 -15 225
∑ X = 360 702
N= 10
X=36
SD=
SD= = 4.415

49. Mean Deviation or Average Deviation
The mean deviation is not very much used in
statistical work. Nevertheless, there are times when it
becomes necessary to compute the mean or average
deviation. The mean deviation is the square root of
absolute values of the difference between the mean and
the raw scores.
MD= ∑/X-X/
N
The symbol / / means that the signs are disregarded.

50. Example:
X
43 7
41 5
40 4
38 2
37 1
33 -3
30 -6
29 -7
24 -12
24 -12
21 -15
∑ X = 360 =74
N= 10
X=36
AM=74/10= 7.4

51. Standard Deviation from Group Scores
The formula for standard deviation using the short method is:
SD=
______- _____
N N
Where SD is standard deviation using the short method is:
l is class interval
∑ is the sum of the products of the frequencies by the
deviations of the score from the mean, squared.
∑ is the sum of the products of the frequencies by the deviations
of the score from mean.
N is the number of cases.

52. X F d fd
90-94 1 4 4 16
85-89 2 3 6 18
80-84 7 2 14 28
75-79 9 1 9+33 9
70-74 11 0 0 0
65-69 8 -1 -8 8
60-64 5 -2 -10 20
55-59 5 -3 -15 45
50-54 1 -4 -4 16
45-49 1 -5 -4 25
N=50 ∑ fd= -9 ∑=185
SD=5
SD= 5
SD= 5
SD= 5
SD= 5 x 1.9150
SD= 9.575

53. Quartile Deviation (Q)
When using the statistics of percentiles, deciles, quartiles, or the
median which are based on the scores, the standard deviation cannot be
used as a measure of variability, since the deviation are based on the
mean. The variability of distribution of scores can be used by using the
two points, Q3 and Q1. A measure of the variability of the middle 50
percent of the scores is considered to be a good estimate, because extreme
scores or erratic spacing between scores in the upper 25 percent and lower
25 percent are excluded in the computation. This is the quartile deviation.
This is the value that is equal to the half the distance from Q1 to Q3.

54. Where:
Q = quartile deviation
Q3 =75th
percentile
Q1 = 25th
percentile
Finding the quartile deviation
X F CM
90-94 1 50
85-89 2 49
80-84 7 47
75-79 9 40
70-74 11 40
65-69 8 20
60-64 5 12
55-59 5 7
50-54 1 2
45-49 1 1
N=50

55. Q1= 64.5+ (12.5-12) x 5
8
Q1= 64.5 + (0.5) x5
8
Q1= 64.5 + (0.06) x 5
Q1=64.5 + .30
Q1=68.80

56. Third Quartile
Formula:
(3N-F)
Q 3=LL+ 4 l
f
3N/4 = 3 x 50 = 150/4 = 37.5
4
LL = 74.5
F= 31
f= 9
I =5

57. Q3= 74.5 + (37.5-31) x 5
9
Q3 = 74.5 + (6.5/9) x 5
Q3= 74.5 + (.72) x 5
Q3 = 74.5 + 3.6
Q3 = 78.1
Q= 78.1 – 64 = 13.3
2

59. A standard score is one of many derived scores
used in testing today. Derived scores are valuable to the
classroom teacher. Since scores are differ from different
tests, the teacher can make them comparable by
expressing them in the same scale. For norm-referenced
test, it is meaningful to interpret classroom test scores
by locating a student’s score with reference to the
average for the class and to describe the distance
between the score and the average in terms of the
spread of the scores in the distribution.

60. Tristan’s raw score on an English achievement
test was 50. In the same class of students Tristan scored
70 on the Mathematics achievement test. To compare
the raw score on one test with a raw score on another
test to obtain a total or average score is meaningless.
The units are not comparable because he tests may
have different possible total scores, the units become
comparable, and can be interpreted properly.

61. Using the deviation of a score from the
mean (X – X) and the standard deviation (SD), a
teacher can build what is called a z-score.
z=(X-M)/SD
Z= a standard score
X= any raw score
M= the mean
SD= the standard deviation

62. For example, the means and
standard deviation for Tristan’s
two test scores are as follow:
Tristan’s Raw Score Mean
Standard
Deviation
English test 50 45 5.6
Mathematics test 70 75 7

63. Comparison can be made between the two
scores because the scores were earned in the
same group of students.
Substituting the formula:
For English For Mathematics
Z=(50-45)/5.6 Z=(70-75)/7
= 5/5.6 = .89 Z= -5/7 = -0.71

64. The two scores of Tristan can now be compared.
Even if he got a higher score in mathematics than in
English, he still did well in English as shown by the
higher value of the standard score in that subject.

66. Skewness is the degree of symmetry of the
scores.
- Refers to the degree of symmetry
attached to the occurrence of the scores along
the score interval.

67. When the scores tend to center around one point
with those on both sides of that point balancing each
other, the distribution is said to have no skewness. If
the atypical scores are above the measure of central
tendency (in the positive direction), the distribution is
said to be positively skewed. Likewise, if the atypical
scores are below the measure of central tendency (in
the negative direction), the distribution is said to be
negatively skewed.

68. SK=(3 (M-Md))/SD

69. The characteristic of kurtosis is very closely
related to the characteristics of variability. It can give an
indication of the degree of homogeneity of the group
being tested in regard to the characteristic being
measured. if students tend to be much alike, the scores
will generate a leptokurtic frequency polygon; if
students are very different, a platykurtic distribution is
generated. A mesokurtic distribution is neither
platykurtic nor leptokurtic .

70. The kurtosis for the normal distribution is
approximately 0.263. hence if the Ku is greater than
0.263, the distribution is most likely platykurtic; while if
the Ku is less than 0.263, the distribution is most likely
leptokurtic (Garett, 1973).
K=Q/((P90-P10))

71. Kurtosis is the degree of peakedness of a
distribution. A normal disribion is a mesokric
distribution. A pure leptokuric distribution has a
higher peak than the normal distribution and has
heavier tails. A pure platykurtic distribution has a
lower peak than a normal distribution and lighter
tails.

73. Most departures from normality display
combinations of both skewness and kurtosis
different from a normal distribution.

75. Correlation is a measure of relationship or
association between two or more paired variable or sets
of data. The degree of correlation is indicated
numerically by the correlation coefficient ® and
graphically by a scotterplot. The r telis us the strength
(weak or strong) and direction (negative or positive) of
the relationship between distributions. The closer a
coefficient gets to -1.0 or 1.0, the stronger the
relationship. A perfect correlation is either -1.0 or + 1.0
and a complete lack of correlation is zero (0).

76. Computations
a. ScatterplotThe strength and direction of relationships between
variable A and B can be determined by inspection with
the use of a scatterplot. Consider the following:
Perfect Negative Correlation

77. Perfect Positive
Correlation
Weak Negative Correlation

78. Weak Positive
Correlation
No Correlation

79. The Scotterplots in Different Directions
The scatterplots range from straight lines (perfect
correlations) to ellipses (weaker correlations ) to
circles (no correlations).
1. Spearman Rank Order Correlaion
2.Pearson’s product-Moment Coefficient of correlation

80. Spearman Rank Order Correlation
Let us consider a relationship between learners’ rank in A and B
subjects. The r is the easiest method of estimating relationship orᵨ
association. Let us compute the following pairs of data using the
Spearman rank order correlation (r ) starting from the raw score asᵨ
follows:
Learner A B
A 85 60
B 80 65
C 78 70
D 75 66
E 70 72

81. Now, let us use the rank as follows:
Difference in
Learner A B Rank (D) D²
A 5 1 4 16
B 4 2 2 4
C 3 4 1 1
D 2 3 1 1
E 1 5 4 16
----------------
SD² = 38

82. To compute for rᵨ’ use this formula:
rᵨ = 1 - 6ED²/N (N²-1)
where: rᵨ = rank difference correlation
SD² = the sum of the squared difference between ranks
N = number of learners
rᵨ = 1- 6(38)/ 5(25-1)
= 1 – 228/ 5(24)
= 1 – 228/ 120
= 1- 1.9
= -0.9

83. The Spearman rank order correlation can be used
with small number of cases, hence can be easily
determined. It is acceptable for ordinal data only.
Difference in rank (D) may carry a negative sign as in
the case of 3- 4 = -1. However, since D is squared (D x
D), it follows that negative times negative equals
positive.

84. Pearson’s Product-Moment Coefficient of
Correlation (r)
The Pearson’s- Product Moment Coefficient ® is
the most commonly used and the most precise
coefficient of correlation. It may be calculated by
converting the raw scores (Z) and finding the mean
value of their products or by use of the raw score
method.

85. Let us consider the raw score method between
paired variables X and Y.
Learner X Y X²Y² Y² XY
A 85 60 7225 3600 5100
B 80 65 64004225 5200
C 78 70 60844900 5460
D 75 66 5625 4356 4950
E 70 72 49005184 5040

86. The raw score method has five columns as illustrated
above. To compute, follow the formula below:

87. where: r = Pearson’s Product- Moment
Coefficient of Correlation
= sum of X scores
= sum of Y scores
= the sum of squared X score
= the sum of squared Y score
= the sum of the products of paired X and Y
scores
N = the number of scores

88. =
= -0.87

89. The correlation value with the rank order method is – 0.9,
while that with the raw score method is – 0.87.
Correlations are interpreted in different ways. A crude
method of interpreting the degree of correlation is shown
below:
Coefficient ( r ) Relationship
± 0.00 to 0.20 Negligible
± 0.20 to 0.40 Low
± 0.40 to 0.60 Moderate
± 0.70 to 0.80 Substantial
± 0.80 to 1.00 High to very high