Term: 1^st Term

Week: 2

Class: Senior Secondary School 2

Age: 16 years

Duration: 40 minutes of 2 periods each

Date:

Subject: Economics

Topic:- Basic tools of economic analysis II

SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to

Give some examples of simple equations
Explain the meaning each of the measures of dispersion
State the advantages and disadvantages of each of the measures of dispersion

INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source

INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures

INSTRUCTIONAL PROCEDURES

PERIOD 1-2

PRESENTATION	TEACHER’S ACTIVITY	STUDENT’S ACTIVITY
STEP 1 INTRODUCTION	The teacher reviews the previous lesson on Basic tools for economic analysis	Students pay attention
STEP 2 EXPLANATION	She gives some examples of simple equations. She then explains the meaning of each of the measures of dispersion	Students pay attention and participates
STEP 3 DEMONSTRATION	She also states the advantages and disadvantages of each of the measures of dispersion and performs calculations on them	Students pay attention and participate
STEP 4 NOTE TAKING	The teacher writes a summarized note on the board	The students copy the note in their books

NOTE

BASIC TOOLS FOR ECONOMIC ANALYSIS

SIMPLE LINEAR EQUATION

Examples: y = f (x) or

q= 200 5p. where y is dependent variable which depends on the value of x and q is also dependant variable which depends on the value of x.

MEASURES OF DISPERSION

These tell us whether the values in the distribution are clustered or spread out. The spread can be determined in many ways by using the following: range, mean deviation, standard deviation and variance.

THE RANGE

This is the difference between the highest and the lowest values in the observation. It indicates the limits within which the value falls.

For example, the range of the observation of data

5, 15, 18, 22, 20, 24, 9 is:

solution Highest value = 24 Lowest value = 5

Range = Highest Lowest = 24 5 Range = 19

ADVANTAGES OF RANGE

It is easy to understand
Its computation is simple
It is useful in weather forecast
It is used in industrial quality control

DISADVANTAGES OF RANGE

It is not exact measure of dispersion
It depends only on extreme value not on all items.
It cannot be computed when a distribution has open-ended classes.
It is not amendable to algebraic manipulation
It does not state anything about the distribution values in the series relative to the measure of central tendency.

MEAN DEVIATION

This is the mean of the absolute values of the deviation from some measures of central tendency.

The formular to calculate mean deviation is

[Σ |X – µ|]/N

Here,

Σ represents the addition of values

X represents each value in the data set

µ represents the mean of the data set

N represents the number of data values

| | represents the absolute value, which ignores the “-” symbol

Example 1:

Determine the mean deviation for the data values 5, 3,7, 8, 4, 9.

Solution:

Given data values are 5, 3, 7, 8, 4, 9.

We know that the procedure to calculate the mean deviation.

First, find the mean for the given data:

Mean, µ = ( 5+3+7+8+4+9)/6

µ = 36/6

µ = 6

Therefore, the mean value is 6.

Now, subtract each mean from the data value, and ignore the minus symbol if any

(Ignore”-”)

5 – 6 = 1

3 – 6 = 3

7 – 6 = 1

8 – 6 = 2

4 – 6 = 2

9 – 6 = 3

Now, the obtained data set is 1, 3, 1, 2, 2, 3.

Finally, find the mean value for the obtained data set

Therefore, the mean deviation is

= (1+3 + 1+ 2+ 2+3) /6

= 12/6

= 2

Hence, the mean deviation for 5, 3,7, 8, 4, 9 is 2.

Example 2:

In a foreign language class, there are 4 languages, and the frequencies of students learning the language and the frequency of lectures per week are given as:

Language	Yoruba	Igbo	Hausa	English
No. of students(x_i)	6	5	9	12
Frequency of lectures(f_i)	5	7	4	9

Calculate the mean deviation about the mean for the given data.

Solution

The following table gives us a tabular representation of data and the calculations

ADVANTAGES OF MEAN DEVIATION

It is easy to understand
It is less affected by extreme values.
It is better than the range since it is base on values in the distribution
It’s computation is simpler when compared with standard of deviation.

DISADVANTAGES OF MEAN DEVIATION

It is not suitable for further mathematical analysis.
It lacks those algebraic properties.

VARIANCE

This is the mean of squared deviations. It can be derived by finding the square of the standard deviation.

Formular to calculate variance is

The population variance formula is given by:

Here,

σ² = Population variance

N = Number of observations in population

X_i = ith observation in the population

μ = Population mean

The sample variance formula is given as:

Here,

s² = Sample variance

n = Number of observations in sample

x_i = ith observation in the sample

= Sample mean

STANDARD DEVIATION

This is the square root of the variance and also referred to as the root mean square deviation.

The population standard deviation formula is given as:

Here,

σ = Population standard deviation

Similarly, the sample standard deviation formula is:

Here,

s = Sample standard deviation

ADVANTAGES OF STANDARD DEVIATION

It is not affected by variation in sample
It subjects itself to rigorous algebraic treatment
It is rigidly defined and it is based on all observations in the distribution.
It is insensitive to sample size provided the size in large enough.

DISADVANTAGES OF STANDARD DEVIATION

Huge amount of work is involved in its computation.
It attaches large weight to extreme values because of the process of
squaring involved in its calculation

SOME CALCULATIONS ON VARIANCE AND STANDARD DEVIATION

Example 1:

Find the variance of heights (in mm) are 610, 450, 160, 420, 310.

Solution

Let’s say the heights (in mm) are 610, 450, 160, 420, 310.

Mean and Variance is interrelated.

The first step is finding the mean which is done as follows,

Mean = ( 610+450+160+420+310)/ 5 = 390

So the mean average is 390 mm.

To calculate the Variance,

compute the difference of each from the mean, square it and find then find

the average once again.

So for this particular case the variance is :

= (220² + 60² + (-230)² +30² + (-80)²)/5

= (48400 + 3600 + 52900 + 900 + 6400)/5

Variance = 22440

EXAMPLE 2:

Find the variance of the numbers 3, 8, 6, 10, 12, 9, 11, 10, 12, 7.

Solution:

Given,

3, 8, 6, 10, 12, 9, 11, 10, 12, 7

Step 1: Compute the mean of the 10 values given.

Mean = (3+8+6+10+12+9+11+10+12+7) / 10 = 88 / 10 = 8.8

Step 2: Make a table with three columns, one for the X values, the second

for the deviations and the third for squared deviations. As the data is not

given as sample data so we use the formula for population variance. Thus,

the mean is denoted by μ.

Value X	X – μ	(X – μ)²
3	-5.8	33.64
8	-0.8	0.64
6	-2.8	7.84
10	1.2	1.44
12	3.2	10.24
9	0.2	0.04
11	2.2	4.84
10	1.2	1.44
12	3.2	10.24
7	-1.8	3.24
Total	0	73.6

Step 3:

= 73.6 / 10

= 7.36

Example 3

If a die is rolled, then find the variance and standard deviation of the

possibilities.

Solution

When a die is rolled, the possible outcome will be 6.

So the sample space, n = 6 and the data set = { 1;2;3;4;5;6}.

To find the variance, first, we need to calculate the mean of the data set.

Mean, x̅ = (1+2+3+4+5+6)/6 = 3.5

We can put the value of data and mean in the formula to get;

σ² = Σ (x_i – x̅)²/n

σ²= ⅙ (6.25+2.25+0.25+0.25+2.25+6.25)

σ²= 2.917

Now, the standard deviation,

σ = √2.917 = 1.708

EVALUATION: 1. Give an example of simple equations

Define measures of dispersion
Explain the meaning of

a. Range

b. mean deviation

c. variance

d. Standard deviation

4. State two advantages and disadvantages of

a. Range

b. mean deviation

c. variance

d. Standard deviation

5. Find the variance and standard deviation of the following scores on an exam: 92, 95, 85, 80, 75, 50

6. Find the standard deviation of the average temperatures recorded over a five-day period last winter: 18, 22, 19, 25, 12

MORE PRACTISE PROBLEMS

Find the variance and standard deviation for the five states with the most

covered bridges:

Oregon: 106 Vermont: 121 Indiana: 152 Ohio: 234 Pennsylvania: 347

Find the variance and standard deviation of the heights of five tallest

skyscrapers in the United States:

Sears Tower (Willis Building): 1450 feet

Empire State Building: 1250 feet

One World Trade Center: 1776 feet

Trump Tower: 1388 feet

World Trade Center: 1340 feet

Find the variance and standard deviation of the scores on the most

recent reading test: 7.7, 7.4, 7.3, and 7.9 4.

Find the variance and standard deviation of the highest temperatures

recorded in eight specific states: 112,100, 127, 120, 134, 118, 105, and

110.

CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively

Lesson Notes By Weeks and Term - Senior Secondary 2