UNIT-VI-EVALUATION

EVALUATION

Evaluation is a checking of what goes on. it is a continuous process. We test the individual’s development. By this we assess the achievement, interest, abilities, skills and attitudes.

Types of evaluation:

  1. Formative evaluation ( ongoing process of evaluation)
  2. Summative (at the end of each semester
  3. Diagnostic ( to know about the standard of
  4. Placement ( Group the students)

EXAMINATION

It is nothing but testing of knowledge for this, we use some test items.

Functions of evaluation:-

  1. We can diagnose the weakness of students.
  2. Diagnose the strength of students
  3. It helps to take remedial measures
  4. It provides information about the standard of the students
  5. It helps to classify the students
  6. It helps to assess teachers methodology and competence
  7. It provides data about student’s knowledge.
  8. It helps students to get admission and employment.

TEST

 Test is an evaluation device to test the  knowledge of students.

Evaluation devices:- (Evaluation tools)

  1. Lab
  2. Assignments   
  3. Home work
  4. cumulative and Anecdotal record.

Tests

Types of test

  1. Oral
  2. Written (Objective, short Assures, essay type)
  3. Practical

Objective type test:-

  1. Choose the best option
  2. Fill in the blanks
  3. True /False
  4. Matching
  5. Choose the right statement.

Merits:

  1. Easy to evaluate
  2. Easy to do judge
  3. Easy to frame the question
  4. Students cannot cheat
  5. Students need no language ability
  6. There is no place for hand writing
  7. There is no place for personal views
  8. Creativity is not needed
  9. Imagination not needed

Demerits:

  1. Very difficult to construct such type
  2. There is great scope for copying or malpractice
  3. By guessing answering is possible

Essay type test

Recalling his facts ideas expressing in his own words

Question settling is easy

Students are made easy to write

Originality

Linguistic fluency

Economical

Short Answer questions:

Short answer with precise way

Linguistic fluency

Merits:

Time Serving Method

Easy answering

Creating, expression and

Diagnostic test

Meaning:-

      We identify the weakness of our students in learning.

Why we have to identify: – Aim of the diagnostic

  • To find out difficult
  • To find out the weakness
  • To take some remedial measures
  • Change of teaching method
  • To test the knowledge
  • To classify the students

Types of diagnostic test:

  1. Standardaised test
  2. Teacher made test

How to prepare:

  1. Select the topic or unit
  2. Divide it into various learning points
  3. Don’t omit any learning point
  4. Use omit any learning point
  5. Use only short answers on objective type
  6. Easy
  7. Don’t give any time limit reasonable
  8. Instruction should be given
  9. Prepare test items
  10. Prepare scoring key (scheme evaluation)
  11. Printing
  12. Edit and fain it
  13. Administer the test
  14. Evaluation – item analysis ( qualitative, quantitative)
  15. Diagnostic chart

Remediation:-

  1. Review the lesson
  2. Reteach the lesson
  3. Motivational programme
  4. Ask them to change studying method
  5. Arrange for guest lecturer
  6. Team teaching
  7. Conduct supervise study
  8. Special class

ANALYSIS AND INTERPRETATION OF SCORES

  1. Measures of central tendency:

Measurements that describe the chief characteristics of the entire distribution, by being in the central parts of a classified distribution are called measures of central tendency.

There are several measures of central tendency. They are called Averages. The most important averages are:

  • Arithmetic mean
  • Median and
  • Mode

Arithmetic mean:

The mean is simple arithmetic average. The mean is defined as the sum of separate scores or other measures divided by their numbers. It is calculated adding together all the item and dividing this total by the number of items.

Calculation of mean for raw data:

Formula

Mean

Mean or 

Problem:   1

Consider the following scores

8, 14, 23, 10, 12, 5.

Solution:

The formula for calculating the mean for raw data is :

Where,

X =8+14+23+10+12+5

X =72

N =6

Apply the values in the formula,

CALCULATION OF MEAN FOR GROUPED DATA:

Here the mean is calculated by “Assumed mean method” or simply the “short method”.

Formula:

Mean =  

d (deviation) =

Where

AM = Assumed mean

Fd = product of the frequency and corresponding deviation

N = Total Number of scores

i = class interval size

d = deviation of midpoints form assumed mean.

Steps for calculating the mean:

  1. Find the midpoints (X) of the class intervals.
  2. Take one or the midpoints preferably the midpoint of an interval somewhere near the centre of the frequency distribution as assumed mean (A.M).
  3. Find the deviations (d) of the assumed mean from the midpoints using the formula

d =     However‘d’ values can be assigned almost mechanically.

  • Multiply the product of frequency and the corresponding deviation.
  • Find         by adding up the product Fd. 
  • Substitute the value of  AM, , N and i in the formula:

Problem:2

Class interval             : 80-90       70-80      60-70      50-60

      F                          :     1               2             6            10

Class frequency         : 40-50    30-40     20-30     10-20

    F                            :     8             6             4            3

Solution:

Class intervalFMidpoint (x)d = Fd
80-9018577
70-80275612
60-70665530
50-601055440
40-50845324
30-40635212
20-3042514
10-2031500
 N = 40  

Where,

Am =15

 =129

N=40

i =10

 = 15 + (3.225 x 10)

 = 15 + 32.25

 = 47.25

Mean = 47.25

Problem: 3

                 Class intervalmidpoint(f)
45-49478
40-444210
35-393720
30-343230
25-292720
20-242210
15-19172

Find Mean ?

Solution : 32.9

Merits of mean:

  • It is easy to understand and simple to calculate.
  • It is most commonly used in further statistical computations.
  • It is a reliable measure.

Demerits of mean:

  • It is unduly affected by the extreme items, ie: very small and very large values pull up or pull down the values of the average.
  • Its value cannot be determined graphically.

(ii) Median:

Median is the middle most point in an arranged series. It is called as averages of location.

Problem: 1

Find the median for the following data = (odd series)

10, 8, 12, 22, 15, 3, 7

Solution :

Arrange the data in ascending order :

3, 7, 10, 12, 15, 22

Median =10

Problem  : 2

Find the median for the following data: (even series)

10, 8, 12, 22, 15, 3, 7, 17

Solution:

Arrange the data in ascending order:

3, 7, 8, 10, 12, 15, 17, 22

There are two mid values available. So take the average of the values.

Median =

Median =  =  = 11

Median = 11

Calculation of median from a frequency distribution :

Formula :

Median = l + 

Where,

l           = exact lover limit of the class interval in which median lies.

cf         = Cumulative frequency up to the lower limit of the class Interval 

                containing  median

f           = frequency of the class interval containing median

i           = size of class interval

Problem : 3

Class interval80-8970-7960-6950-5940-4930-3920-2910-19
Frequency126108643

Solution :

Class IntervalFrequency (F)Cumulative frequency (Cf)Exact Limits
80-8914079.5 – 89.5
70-7923969.5 – 79.5
60-6963759.5 – 69.5
50-59103149.5 – 59.5
40-4982139.5 – 49.5
30-3961329.5 – 39.5
20-294719.5 – 29.5
10-19339.5 – 19.5
 N = 40  

N = 40

 =  = 20

Here,

L          = 39.5

M         = 13

F          = 8

i           = 10

Median = 39.5 + 

            = 39.5 +

            = 39.5 +  

            = 39.5 + 8.75

Median = 48.25

Steps for calculating the median:

  1. Write down the exact limits of the class intervals.
  2. Compute the cumulative frequency
  3. Determine
  4. Identify the class interval in which the th item lies.
  5. Find the value of t the exact limit of the median class.
  6. Find the cumulative frequency m up to the lower limit of the class interval, which contains the median.
  7. Read the frequency fm of the interval which obtains the median.
  8. Substitute the values of l, m, fm, N/2 and I in the formula

Median =

Merits of median:

  • Median is simple to understand.
  • It is easy to compute.
  • It is not affected by extreme items.
  • It is positional average.

Demerits of median:

  • It is not always rigidly defined.
  • It is generally not used in further statistical work.

Problem: 4

Marks85-10070-8555-7040-5525-4010-25
Frequency132644206

Solution : median = 48.18

(iii) Mode:

Mode is the most common item of a series. Mode is the value which occurs the greatest number of frequency in a series.

Formula:

Mode = 3 median – 2mean

  1. Find the mode:      15, 70, 75, 70, 76, 20, 16, 18.

Mode means the repeated numbers

Mode = 70

  • Find mode for the data where the mean and the median of the distribution are 56 and 58 respectively.

Solution:

Mean               = 56

Median            = 58

Mode               = 3 x 58 – 2 x 56

Mode               = 174 – 112

Mode              = 62

Merits of mode:

  1. It is simple to understand.
  2. Its value is not affected by the presence of extreme items
  3. Mode can be determined proper ally.

Demerits of mode:

  1. It is a measure having very limited practical value.
  2. It is not used for the further statistical measures.

II. Measures of variation (or) measures of Dispersion:

The central tendency has its own limitations and its representation is best as single score curve. It is apteral a single numerical value and may fail to reveal the variability in a numerical value indicating the amount of variation about a central value.

The measures of central tendency was supported of variation. A measure of dispersion is designed to state the extent to which the individual items differ from their arithmetic mean. In measuring dispersion we shall be interested in the amount of the variation or its degree, but not in the direction.

There are three measures of variability. They are

  • Range
  • Quartile deviation (Q.R) semi-inter quartile  range
  • Standard deviation (S.D)
  • Range:

Range is the difference between the minimum and maximum measurement. It is usually calculated by the formula.

Range = H – L

H = highest score

L = lowest score

The range is the most simple and easily understood measure of dispersion.

Problem:

Find out the range of the distribution: 40, 51, 22, 45, 37, 65, 30 and 56.

Solution:

Range = highest score (H) – lowest score (L)

Range = 65 – 22 =43

Range = 43

  • Quartile deviation (Q . D) (or)  semi inter quartile range:

                             The quartile deviation is one half of the distance both third and first quartiles in a frequency distribution.

                                                  Q1                  Q2                    Q3

The range which includes the middle fifty percent of the distribution i.e one quarter of the observations at the lower end and another quarter of the observations at the upper and are excluded in computing the inter-quartile range.

In notation inter-quartile range = Q3 – Q1

Half of the inter quartile range or semi-inter-quartile range is called the quartile deviation (Q.D).

Formula:

Q.D =

Where,

Q .D = Quartile deviation

Q   = First quartile in the frequency distribution.

Q3 = Third quartile in the frequency distribution.

Q3 =  

Q1 =

Where,

Q, = l1+

L1 =exact lower limit of the first variable.

F1 = frequency in the first quartile.

i= class interval

M1 = cumulative frequency which is below the first quartile.

Q, = l3+

L3 = exact lower limit in the third quartile

F3 = frequency in the third quartile

i = class interval

M3 = cumulative frequency which is below the third quartile.

Problem:1

Class Interval45-4940-4435-3930-3425-2920-2415-1910-145-9
Frequency232688759

Solution:

Class IntervalFrequency (f)Cumulative frequency (fm)
45-49250
40-44348
35-39245
30-34643
25-29837
20-24829
15-19721
10-14514
5-999
 N = 50 

Calculation of Q1:

Here  

l1          = 9.5

F1           = 5

M1         =9

i           = 5

Substituting these values in the formula for Q1, we have

Q1 = 9.5 +

Q1 = 9.5 +

Q1 = 9.5 + 3.5

Q1 = 13

Calculation of Q3:

l3        = 29.5

F3       =   6

M3      = 37

i          =  5

Substituting these values in the formula for Q3

Q3 = 29.5 +

Q3 = 29.5 +

Q3 = 29.5 + 0.42 = 29.32

Q3 = 29.92

Calculation of Q.D:

Substituting the values Q1 and Q3 in the formula Q.D , we get

Q.D

Q.D =

Q.D = 8.46

Merits of quartile deviation:

  • Specially useful in case of open end classes.
  • Not affected by the presence of extreme values.
  • Superior  to  range as a rough measure of dispersion.

Limitations of quartile deviation:

  • Affected by sampling fluctuations.
  • It gives only a position on the scale rather than the scatter.
  • As the value does not depend upon every item of the series, it cannot be regarded as a good method of dispersion.

Problem:2

Class Interval91-10081-9071-8061-7051-6041-5031-40
Frequency81218171393

Solution: 11.93

(iii) Standard deviation: (SD or σ)

The concept of the standard deviation was introduced by Karl Pearson in 1893. It is the root mean square deviation measured from the average. It is the square of the arithmetic average of the square of deviations.

Standard deviation commonly denoted by the Greek letter (sigma)

Calculation of the SD from ungrouped scores:

The formula for calculating S.D  for the ungrouped scores are:

S.D =

Where,

d = deviation

N = number of items

Problem: 1

Calculate the S.D of the series 35, 49, 32, 45 and 39

Solution:

First find the mean:

Mean = 

Mean= = 40

Scoresmeandeviation from mean (d)square of deviation (d)
3540-525
4940981
3240-864
4540525
3940-11
N =5   = 196

N=5,        =196

S.D    =  

             =    =

S.D      =         6.26

Problem: 2

Calculate the S.D for the following data  6, 8, 10, 12, 14

Solution:-

S.D (or) σ  = 2.83

Calculation of SD from Grouped Data:

Formula:

S.D (or) σ = i  x

           d  =  deviation

           f   =  frequency

           I   =  class interval

N =   Total number of values

Problem: 3

Class Interval45-4940-4435-3930-3425-2920-2415-1910-145-9
Frequency232688759

Solution:

Class intervalmidpointF  dFdFd2
45-4947286416128
40-4442374921147
35-393726361272
30-3432652530150
25-2927841632128
20-24228392472
15-19177241428
10-141251155
5-9790000
  N = 50  ∑Fd = 154∑Fd2 = 730

  = 196

     = 154 

S.D (or) σ = i  x

S.D (or) σ = 5  x

                       σ  = 5 x  2

                       σ  = 5 x  

                       σ  = 5 x

                       σ  = 5 x 2.26

                       σ  = 11.3

Computational steps

  1. Find out the midpoints (x) of all the class intervals
  2. Take a midpoint as an assumed mean.
  3. Find deviation(d) using the formula
  • Multiply columns f and d to obtain fd and sum up to obtain ∑fd
  • Multiply columns f and d2 to obtain fd and sum up to obtain ∑fd2
  • Substitute the values in the formula σ = i  x

Advantages of standard deviation:

  • It is useful for comparing the groups.
  • It is used in the normal curve.
  • It is a reliable meaure

Problem: 4

Class Interval90-9980-8970-7960-6950-5940-4930-3920-2910-190-9
Frequency1314101775166

Solution:

S.D=23.40

(iii).Rank correlation:

Correlation:

The statistical relation between two variables or two subjects is called correlation.

There are three types of correlation

  1. Positive correlation
  2.  Negative correlation
  3. Zero correlation
  4. Positive correlation:

If the value of the variable increases, then the value of another variable increase. And the value of one variable decreases, then the value of another variable decreases. It is positive correlation.

As the intelligence of the students increases their achievement will increase.

  • Negative correlation:

                              If the value of one variable increases, then the value of another variable will decrease. It is negative correlation.

Eg;

    As the production increases, the price reduces.

  • Zero correlation

If there is no relationship between the variables, it is called zero correlation.

Rank correlation:

        In rank correlation the scores are converted in to ranks to make calculations simple.

Formula:

R (or) P = 1 – 

Where,

D = difference in ranks

N = number of items

D = Rx – Ry  

Interpretation:

0.20 à Negligible

  •  0.40 à Low

0.41  0.60 à Moderate

0.61  à High

Problem: 1

               The scores of 6 pupils in social science and science are given below. Calculate correlation by the rank difference method.

Scores in Social Science455367403550
Scores in Science687670645466

Procedure:(solution)

             Convert the scores into ranks. In computer science C gets the first rank, and E gets the 6th rank. In physics B gets the first rank, C gets the second rank and E gets the last rank.

Calculation:

Scores in Social ScienceScores in ScienceRank in computer science (Rx)Rank in Physics (R1)Rank difference (D)Square of rank different D2
45684311
53762111
67701211
40645500
35546600
50663411
     ∑D2 = 730

∑D2 = 4

N = 6

    = 1 –     = 1 – 

  =     1 –     =   1 – 

  = 1- 0.11

  = 0.89

There is high positive correlation between the scores in computer science and those in physics. The attainment in computer science have positive influence in learning of physics.

Problem: 2

X10151114162010879
y16162418222414101214

N = 10

 = 0.86

Interpretation:

The correlation between X and Y is very high and positive.

Computational steps:

  1. Compute ranks for each set of scores. (Rx and Ry)
  2. Find the difference between Rx and Ry and represent it by D (ie ,D = Rx – Ry)
  3. Compute the squares of D (ie . D2)
  4. Sum up d2 to get ∑D2
  5. Substitute the values in the formula.

    = 1 – 

  • Interpret the result.

Advantages:

  • It is used in the co relational studies.
  • It is used for establishing validity of a tool.
  • It is useful to establish the reliability of a tool.

Graphical Representation of Data:

  • Frequency Tabulation:

A frequency distribution refers to data classified on the basis of some variable that can be measured. A variable refers to characteristic that various in magnitude. It may be either continuous or discontinuous.

Divide the variable into various classes of uniform size.

About the observation one by one to the classes  which the values of the observation falls as in the table. Repeated value can be included in the upper limit of the class.

Eg:

Scores of 40 pupils:

40, 22, 51, 40, 48, 57, 53, 65, 10, 34,

37, 66, 70, 44, 56, 30, 35, 29, 50, 47,

53, 18, 25, 76, 45, 31, 42, 66, 40, 27,

14, 32, 44, 46, 51, 34, 25, 52, 77, 64,

Frequency Table:

Class IntervalTablesFrequency
10-191113
20-2911115
30-391111   117
40-491111   111110
50-591111  1118
60-6911114
70-791113
  • Histogram:

                                 Histogram is the most accurate graph that represents a frequency distribution. In histogram the scores are spread uniformly over the entire class-interval. The class-interval is marked on the x-axis and the frequencies on the y-axis.

Each interval in a histogram is represented by a separate rectangle. The area of each rectangle is directly proportional to the number of measures with in the class interval. The whole of the histogram is in proportion. With the whole of the statistical data.

Class IntervalFrequencyExact class limit
10-1939.5 – 19.5
20-29519.5 – 29.5
30-39729.5 – 39.5
40-491039.5 – 49.5
50-59849.5 – 59.5
60-69459.5 – 69.5
70-79369.5 – 79.5
  • Frequency polygon:

In frequency polygon, we mark the midpoint of each class on the x-axis and the frequencies on the y-axis. After line segments. We complete the polygon that is we close it by joining it to the x-axis.

Frequency polygon gives a less accurse representation of the distribution, than a histogram as it represents the frequency of each class by a single point not by the whole class-interval.

Class IntervalExact class limitFrequencyMid point
10-199.5 – 19.5314.5
20-2919.5 – 29.5524.5
30-3929.5 – 39.5734.5
40-4939.5 – 49.51044.5
50-5949.5 – 59.5854.5
60-6959.5 – 69.5464.5
70-7969.5 – 79.5374.5
  • Frequency curve:

The frequency polygon consists of sharp turns and ups and downs which are really not in conformity with actual conditions. To remove these sharp features of a polygon it becomes necessary to smooth it.

No definite rule for smoothing the polygon can be laid down. It should be borne in mind that the curve is as smooth as possible and does not in any way sharply deviate from the polygon. In order to draw a satisfactory frequency curve, we need first of all to draw a frequency histogram then the frequency polygon and ultimately the frequency curve.

  • Cumulative frequency graph (or) ogive:

It is a graphical representation of cumulative percentage distribution. The curves obtained by plotting cumulative frequencies percentage is called ogive curve. The curve obtained less than ‘F’ is called less than ogive curve and the curve obtained more than ‘F’ is called more than o give curves.

Class IntervalExact class limitFrequencyLess than ogiveGreater than ogive
10-199.5 – 19.53340
20-2919.5 – 29.55837
30-3929.5 – 39.571532
40-4939.5 – 49.5102525
50-5949.5 – 59.583315
60-6959.5 – 69.54377
70-7969.5 – 79.53403