Tools in Presenting, Analysing and Interpreting Classroom Data

7/30/2019 Tools in Presenting, Analysing and Interpreting Classroom Data

1/15

Evaluation in Home Economics Page 1

TOOLS IN PRESENTING, ANALYSING AND

INTERPRETING CLASSROOM DATAMeasures of Central Tendency

Introduction

Measures of central tendency are measures of the location of the middle or the center of a

distribution. The definition of "middle" or "center" is purposely left somewhat vague so that the

term "central tendency" can refer to a wide variety of measures.

Measures of central tendency attempt to quantify what we mean when we think of as the

"typical" or "average" score in a data set. The concept is extremely important and we encounter it

frequently in daily life.

The concept of Measures of Central Tendency:

Describe the center point of a data set with a single value

Valuable tool to help us summarize many pieces of data with a single number

The ways to measure the central tendency of our data are mean, median and mode.


2/15


Mean

For ungrouped data,

n

xxxxxMean n

...)( 321

Example 1

Find the mean for the set data: 3, 7, 2, 1, 7

Solution

5

71273 x

=

For grouped data,

n

nn

ffff

fxfxfxfxxMean

...

...)(

321

332211

Example 2

a) Find the mean of the set of data: 25, 36, 42, 38, 36b) Find the mean from the set of grouped data

Class mark 10.5 30.5 50.5 70.5 90.5 110.5

Frequency 19 6 3 2 1 2

Solution

a) mean =


3/15


b)

x x

10.5 19 199.530.5 6

50.5 3

70.5 290.5 1110.5 2

sum 33

mean =

Median

For ungrouped data,

Median = the middle datum, when n is odd.

Median = the mean of the two middle data, when n is even.

e.g.1 For the set of data

2, 4, 7, 9, 21

middle datum

median = 7

e.g.2 For the set of data

3, 5, 7, 7

middle of two data

median = (5 + 7) 2= 6

For grouped data,

Step 1: Draw the cumulative frequency polygon.

Step 2: The median is the datum corresponding to the middle value of the cumulative

frequency.

Example

a) Find the median of 2, 3, 10, 12, 999.b) Find the median of 2, 3, 10, 12, 22, 123.c) The cumulative frequency polygon for Home Economics marks of a class is given below,

find the median mark.


4/15


Solution

a) Median =

b) Median =

c) Total frequency= 40

The rank of median

= 40/2 =

From the cumulative

polygon,median =

Example

Marks of Home Economics class (57 students) were tabulated as follows. Draw a cumulative

frequency polygon and determine the median age score.

Marks 0 9 10 19 20 29 49 50 59 60 69 > 70

Frequency 2 3 7 30 5 1

Solution

Marks

x

Frequency cumulative

frequency

x < 10 2

10x < 20 3

20x < 30 5

30x < 40 4

40x < 50 7

50x < 60 30

60x < 70 5

70


5/15


Mode

For ungrouped data, mode is the datum that has the highest frequency.For grouped data, modal class is the class that has the highest frequency.

Example

a) Find the mode of the data:1, 2, 2, 2, 3, 3, 9

b) Find the modal class

Class 10 - 14 15 - 19 20 - 24 25 - 29

Frequency 2 8 7 3

Solution

a) The mode is

b) The modal class is

Comparison of the Mean, Median and Mode

The arithmetic mean is the most widely used measure of central tendency. It is usually preferred

over the median and the mode. It is the most reliable measure provided there are no extremevalues in the data because all the values in the data are used in calculating the mean unlike the

mode and the median. Whenever the set of data contains extreme values, the median or mode

will probably be a better indicator of the whole set of data because they are not influenced byextreme values.

The median may be the preferred measure of central tendency for describing economic,

sociological and educational data. It is popular in the study of the social sciences because muchof the data in the social sciences contain extreme values.

The mode is more useful in business planning as a measure of popularity that reflects centraltendency or opinion.


6/15


Measure of Variability

Apart from using a measure of central tendency to summarise a set of data, we need a quantity tomeasure the degree of dispersion of the set of data (so that we can determine the reliability of the

set of data).

Measures of variability:

a. rangeb. variancec. standard deviationd. quartiles

Range

For ungrouped data, the range is the difference between the largest datum and the smallest

datum.

For grouped data, the range is the difference between the highest class boundary and thelowest boundary.

Example

a) Find the range of the data:

1, 2, 2, 2, 3, 3, 9b) Find the range of the grouped data

Class 10 - 14 15 - 19 20 - 24 25 - 29

Frequency 2 8 7 3

Solution

a) The range =

b) The range =

Inter quartile range

Inter quartile range = Q3Q1

where Q1, Q2, Q3 are called quartiles which divide the data (which have been ranked, i.e.

arranged in order) into four equal parts.

Moreover,

Q2 is the median of the whole set of data (50%),Q1 is the median of the lower half (25%),Q3 is the median of the upper half (75%).


7/15


Example

Find the inter quartile range ofa) 1, 2, 3, 5, 11, 12, 13.b) 1, 2, 3, 4, 11, 12, 13, 14.

Solution

a) inter-quartile range =

b) inter-quartile range =

Example

The following frequency distribution gives the marks of a sample of 50 students:

marks Frequency Cumulativefrequency

20 to under 30 2

30 to under 40 4

40 to under 50 6

50 to under 60 14

60 to under 70 13

70 to under 80 7

80 to under 90 4

Find the median and the inter-quartile range of the data.

The rank of median

= 50 =The median score is

The rank of upper quartile

= 50 == 38 , to the nearest integer

The upper quartile Q3 is

The rank of lower quartile= 50 == 13 , to the nearest integer

The lower quartile Q1 is

The inter-quartile range = Q3 Q1=

0

8

16

24

32

40

48

56Fre

uenc

Less than cumulative frequency polygon

for marks of 50 students

20 30 40 50 60 70 80 90

Marks (upper bound)


8/15


Standard deviation and Variance

For ungrouped datax1,x2,,xn, with a mean x , the standard deviation (s) is

For grouped data with class marksx1,x2,,xn; corresponding frequenciesf1,f2,,fn, and a mean

x , the standard deviation (s) is

Example

Find the standard deviation for

a) the ungrouped data 8, 9, 10, 10, 11b) the grouped data

x 17 22 27 32 37 42 47

f 2 4 7 8 7 4 2

Solution

a) mean i

i

n

xx = standard deviation

n

xx

s ii

2)(

=

c) mean =

n

nn

ffff

fxfxfxfxxMean

...

...)(

321

332211

d) standard deviation =

n

nn

ffff

fxxfxxfxxfxx

...

)(...)()()(

321

2

3

2

32

2

21

2

1

nxxxxxxxx n

22

3

2

2

2

1 )(.. .)()()(

n

nn

ffff

fxxfxxfxxfxx

...

)(...)()()(

321

2

3

2

32

2

21

2

1


9/15


Variance

Variance is the squared standard deviation (s2)

Data organization and Presentation

Generally speaking, DATA refers to facts that are gathered in some way, and whichconstitute a body of information. In order to be useful, data must not only be organized, but

it should also be able to be represented in some way that comparisons, trends and orrelationships may easily be visualized. Graphs and charts provide such representations.

There are a wide number of types of graphs and charts, but the most common ones are line

graphs, bar graphs, pie charts and histograms.

Line graphs

Line graphs connect data points that are somehow related. For example, a mother may

measure the height of her child every year on that child's birthday. The age of the child and

her height are data that are clearly related, whereas measurements of any child at any timewould not be particularly meaningful. The same thing with students performance over timeor student progress. The graph immediately suggests that the child undergoes a regular

growth between the ages of 1 and 8. It also allows us to have a reasonable estimate of what

the height of the child was at an intermediate age, say 4.5 years, when the height, read from

the graph, would have been about 102 cm.


10/15


Bar graphs

Bar graphs are commonly used to represent qualitative data, consisting of separated, vertical

bars of equal widths whose heights are drawn proportional to the frequencies of the classes

they represent. The example below present 40 students preferred method of teaching.

Method no. students

Lecture 2

Group work 6

Student centred 8

Teacher centered 20

Other 40

5

10

15

20

25

Lecture Group

work

Student

centred

Teacher

centered

Other

Method

Prefered Teaching

Method


11/15


Pie charts

Pie charts are useful to show the proportion of certain types of data as parts of a whole. The

"pie" represents the whole, and the segments of the pie represent the proportion of various

data points to that whole. For an example, let us take a school tuckshop that is making an

analysis of its sales. In a given week, it solds goods to a total of P13 414. The amountscollected for each category of items are tabulated, and the percentage contribution of each

class of items worked out. The resultant pie chart is shown above on the right.

Histograms

Histograms display distributions of quantitative data. Imagine a class of 26 students who

have written a Home Economics test. The scores are arranged in data ranges 0-9, 10-19, andso on. Against these, the number of learners whose scores fall within these ranges are

tabulated as shown in the table above on the left. A histogram is then drawn, with the count

of learners whose scores fall in each range, as shown above on the right. The bars are alsovertical but joined as compared to bar graphs


12/15



13/15


Item Analysis of Classroom Tests

After you create your objective assessment items and give your test, how can you be sure that

the items are appropriate -- not too difficult and not too easy? How will you know if the testeffectively differentiates between students who do well on the overall test and those who do not?

An item analysis is a valuable, yet relatively easy, procedure that teachers can use to answer bothof these questions.

1. Level of Difficulty

To determine the difficulty level of test items, a measure called the Difficulty Index is used. Thismeasure asks teachers to calculate the proportion of students who answered the test itemaccurately. By looking at each alternative (for multiple choice), we can also find out if there are

answer choices that should be replaced. For example, let's say you gave a multiple choice quiz

and there were four answer choices (A, B, C, and D). The following table illustrates how manystudents selected each answer choice for Question #1 and #2.

Question A B C D

#1 0 3 24* 3

#2 12* 13 3 2

* Denotes correct answer.

For Question #1, we can see that A was not a very good distractor -- no one selected that answer.

We can also compute the difficulty of the item by dividing the number of students who choose

the correct answer (24) by the number of total students (30). Using this formula, the difficulty of

Question #1 (referred to as p) is equal to 24/30 or .80. A rough "rule-of-thumb" is that if the itemdifficulty is more than .75, it is an easy item; if the difficulty is below .25, it is a difficult item.

Given these parameters, this item could be regarded moderately easy -- lots (80%) of students

got it correct. In contrast, Question #2 is much more difficult (12/30 = .40). In fact, on Question#2, more students selected an incorrect answer (B) than selected the correct answer (A). This

item should be carefully analyzed to ensure that B is an appropriate distractor.

Difficulty Index = #of correctTotal #of question

2. Discrimination Test

Another measure, the Discrimination Index, refers to how well an assessment differentiatesbetween high and low scorers. In other words, you should be able to expect that the high-

performing students would select the correct answer for each question more often than the low-


14/15


performing students. If this is true, then the assessment is said to have apositive discrimination

index (between 0 and 1) -- indicating that students who received a high total score chose the

correct answer for a specific item more often than the students who had a lower overall score. If,however, you find that more of the low-performing students got a specific item correct, then the

item has a negative discrimination index (between -1 and 0). Let's look at an example.

Table 2 displays the results of ten questions on a quiz. Note that the students are arranged with

the top overall scorers at the top of the table.

StudentTotal

Score (%)

Questions

1 2 3

Asif 90 1 0 1

Sam 90 1 0 1

Jill 80 0 0 1

Charlie 80 1 0 1

Sonya 70 1 0 1

Ruben 60 1 0 0

Clay 60 1 0 1

Kelley 50 1 1 0

Justin 50 1 1 0

Tonya 40 0 1 0

"1" indicates the answer was correct; "0" indicates it was incorrect.

Follow these steps to determine the Difficulty Index and the Discrimination Index.

1. After the students are arranged with the highest overall scores at the top, count the

number of students in the upper and lower group who got each item correct. For Question

#1, there were 4 students in the top half who got it correct, and 4 students in the bottom

half.2. Determine the Difficulty Index by dividing the number who got it correct by the total

number of students. For Question #1, this would be 8/10 or p=.80.

3. Determine the Discrimination Index by subtracting the number of students in the lower

group who got the item correct from the number of students in the upper group who got

the item correct. Then, divide by the number of students in each group (in this case, there


15/15


are five in each group). For Question #1, that means you would subtract 4 from 4, and

divide by 5, which results in a Discrimination Index of 0.

4. The answers for Questions 1-3 are provided in Table 2.

discrimination index =#correct(upper) - #correct(lower)

#students per group

Item# Correct (Upper

group)

# Correct (Lower

group)

Difficulty

(p)

Discrimination

(D)

Question

14 4 .80 0

Question

20 3 .30 -0.6

Question

35 1 .60 0.8

Now that we have the table filled in, what does it mean? We can see that Question #2 had a

difficulty index of .30 (meaning it was quite difficult), and it also had a negative discrimination

index of -0.6 (meaning that the low-performing students were more likely to get this item

correct). This question should be carefully analyzed, and probably deleted or changed. Our

"best" overall question is Question 3, which had a moderate difficulty level (.60), and

discriminated extremely well (0.8).

Documents

Tools in Presenting, Analysing and Interpreting Classroom Data