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GNDU QUESTION PAPERS 2024
BA/BSc 6
th
SEMESTER
ECONOMICS
(Quantave Methods for Economists)
Time Allowed: 3 Hours Maximum Marks: 100
Note: Aempt Five quesons in all, selecng at least One queson from each secon.
The Fih queson may be aempted from any secon.
All quesons carry equal marks.
SECTION – A
1.(a) Evaluate:
(b) Find the maximum and minimum value of the following funcon:
2.(a) With the help of suitable examples, explain increasing funcon, decreasing funcon,
and extreme points.
(b) If
nd derivave of y with respect to x.
SECTION – B
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3.(a) Calculate Mean Deviaon from the Mean for the following data:
Marks
No. of Students
0–10
6
10–20
5
20–30
8
30–40
15
40–50
7
50–60
6
60–70
3
(b) For the following distribuon, calculate quarle deviaon and its coecient:
Class Interval (C.I.)
Frequency
0–15
8
15–30
26
30–45
30
45–60
45
60–75
20
75–90
17
90–105
4
4.Calculate Mode (by grouping method) for the following data:
Marks (Below)
10
20
30
40
50
60
70
80
90
No. of Students
4
6
24
46
67
86
96
99
100
SECTION – C
5.(a) Calculate Karl Pearson’s coecient of skewness from the following data:
Size
1
2
3
4
5
6
7
Frequency
10
18
30
25
12
3
2
(b) Karl Pearson’s measure of skewness of a distribuon is 0.5.
Its Median and Mode are 42 and 36 respecvely.
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Find the coecient of variaon.
6.(a) For the following data, calculate Spearman’s rank correlaon coecient between x
and y:
x
30
38
28
27
28
23
30
33
28
35
y
29
27
22
29
20
29
18
21
27
22
(b) From the data given below, nd the two regression equaons:
x
25
28
35
32
31
36
29
38
34
32
y
43
46
49
41
36
32
31
30
33
39
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GNDU ANSWER PAPERS 2024
BA/BSc 6
th
SEMESTER
ECONOMICS
(Quantave Methods for Economists)
Time Allowed: 3 Hours Maximum Marks: 100
Note: Aempt Five quesons in all, selecng at least One queson from each secon.
The Fih queson may be aempted from any secon.
All quesons carry equal marks.
SECTION – A
1.(a) Evaluate:
(b) Find the maximum and minimum value of the following funcon:
Ans: 1(a) Evaluate
Understanding the expression
At first glance, this limit may look abstract, but it actually has a beautiful meaning.
It represents the rate of change of the square-root function
.
In calculus language, this is the derivative of
with respect to .
So mathematically:
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Here:
So we are really finding:
But since the question asks to evaluate the limit, we’ll do it directly using algebra — the
classical method students learn.
Step-by-step solution
We start with:
This gives 0/0 form when , which is indeterminate.
So we simplify it.
A standard trick with square roots is multiply and divide by the conjugate.
The conjugate of
is
So multiply numerator and denominator:
Now apply identity:
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So numerator becomes:
Thus expression becomes:
Cancel :
Now substitute :
Final answer (a)
So the derivative (or rate of change) of √x is √.
1(b) Find maximum and minimum of
Now let’s explore this function like a story — understanding how it rises and falls.
Step 1: Understand the function nature
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Notice:
• power 6 → even → always ≥0
• power 5 → odd → sign depends on (x−3)
So sign of function mainly depends on (x−3).
This helps understand graph behaviour later.
Step 2: Find derivative
To locate maxima/minima, we differentiate.
Use product rule:
Let:
Then:
So:
Factor common terms:
Common factors:
So:
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Simplify bracket:
So derivative:
Step 3: Critical points
Set derivative = 0
So:
These are stationary points.
Step 4: Determine max/min nature
We examine sign of derivative around each point.
Derivative:
Important sign rules:
•
always ≥0 (even power)
•
sign depends on (x−2)
•
sign depends on x−28/11
So sign depends mainly on:
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Interval analysis
Critical points in order:
Test intervals:
Interval 1:
Take x=1:
Product → positive
So
→ increasing
Interval 2:
Take x=2.2:
Product → negative
So decreasing
Hence max at x=2
Interval 3:
Take x=2.8:
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So increasing
Hence min at x=28/11
Interval 4:
Take x=4:
Still increasing
So x=3 is not extremum (flat inflection type)
Step 5: Find max & min values
Maximum at x=2
So max value = 0
Minimum at x=28/11
Compute:
So:
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Final answers (b)
Maximum:
at
Minimum:
at
2.(a) With the help of suitable examples, explain increasing funcon, decreasing funcon,
and extreme points.
(b) If
nd derivave of y with respect to x.
Ans: (a) Increasing Function, Decreasing Function, and Extreme Points
1. Increasing Function
An increasing function is one where the value of the function rises as the input (x) increases.
• Formal Definition: A function is increasing on an interval if for any two numbers
, we have
.
• Example:
o Consider .
o If , then .
o If , then .
o Clearly, as x increases, f(x) also increases.
• Graphical View: The graph slopes upward from left to right.
2. Decreasing Function
A decreasing function is one where the value of the function falls as the input increases.
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• Formal Definition: A function is decreasing on an interval if for any two
numbers
, we have
.
• Example:
o Consider .
o If , then .
o If , then .
o As x increases, f(x) decreases.
• Graphical View: The graph slopes downward from left to right.
3. Extreme Points (Maximum and Minimum)
Extreme points are the “peaks” and “valleys” of a function.
• Maximum Point: The highest value of the function in a given interval.
• Minimum Point: The lowest value of the function in a given interval.
• Example:
o Consider
.
o The graph is a parabola opening upwards.
o At , . This is the minimum point.
o The function has no maximum because it goes to infinity as x increases.
• Another Example:
o For
, the parabola opens downward.
o At , . This is the maximum point.
o The function has no minimum because it goes to negative infinity.
In summary:
• Increasing function → values rise with x.
• Decreasing function → values fall with x.
• Extreme points → special points where the function reaches a peak (maximum) or a
valley (minimum).
(b) Derivative Problem
We are asked to differentiate:
Step 1: Simplify the Expression
Let’s denote:
So,
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Step 2: Apply Product Rule
The derivative of a product is:
Step 3: Differentiate Each Part
(i) Differentiate
(ii) Differentiate
• For the first term:
• For the second term:
So,
Step 4: Substitute Back into Product Rule
This is the derivative of y with respect to x.
Putting It All Together
• Conceptual Part: We understood increasing functions (values rise), decreasing
functions (values fall), and extreme points (peaks and valleys).
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• Derivative Part: By using the product rule and carefully differentiating each term, we
arrived at the derivative expression.
Conclusion
Mathematics often feels abstract, but when explained step by step, it becomes a natural
story: functions rise and fall like hills and valleys, and derivatives are the tools we use to
measure their slopes. In this problem, we combined conceptual understanding with
practical calculation, showing how theory and application go hand in hand.
SECTION – B
3.(a) Calculate Mean Deviaon from the Mean for the following data:
Marks
No. of Students
0–10
6
10–20
5
20–30
8
30–40
15
40–50
7
50–60
6
60–70
3
Ans: Mean Deviation from Mean (Grouped Data)
We are given grouped data (class intervals with frequencies).
We must calculate Mean Deviation from the Mean.
Given Data
Marks
No. of Students (f)
0–10
6
10–20
5
20–30
8
30–40
15
40–50
7
50–60
6
60–70
3
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What is Mean Deviation (Simple Idea)
Imagine a class where the average marks is calculated.
But students don’t all score exactly that average.
Some score:
• higher than average
• some lower
Mean Deviation tells us how far marks are spread from the average.
So:
Mean Deviation from Mean
Where:
• = frequency
• = class midpoint
• = mean
Step 1: Find Class Midpoints (x)
For grouped data, we use the midpoint:
Lower limit + Upper limit
Marks
f
Midpoint (x)
0–10
6
5
10–20
5
15
20–30
8
25
30–40
15
35
40–50
7
45
50–60
6
55
60–70
3
65
Step 2: Find Mean (
)
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Formula:
Now calculate :
Marks
f
x
fx
0–10
6
5
30
10–20
5
15
75
20–30
8
25
200
30–40
15
35
525
40–50
7
45
315
50–60
6
55
330
60–70
3
65
195
Now totals:
So mean:
Mean marks = 33.4
Step 3: Find Deviation from Mean
Now compute:
| x | |
|---|--------------|
5 | 28.4 |
15 | 18.4 |
25 | 8.4 |
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35 | 1.6 |
45 | 11.6 |
55 | 21.6 |
65 | 31.6 |
Step 4: Multiply by Frequency (f)
We need:
| f | x | | |
|---|---|--------------|---------------|
6 | 5 | 28.4 | 170.4 |
5 | 15 | 18.4 | 92 |
8 | 25 | 8.4 | 67.2 |
15 | 35 | 1.6 | 24 |
7 | 45 | 11.6 | 81.2 |
6 | 55 | 21.6 | 129.6 |
3 | 65 | 31.6 | 94.8 |
Now sum:
Step 5: Mean Deviation Formula
Mean Deviation
Final Answer
Mean Deviation from Mean
Understanding the Result (Real Meaning)
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What does 13.18 mean?
It means:
On average, students’ marks differ from the mean by about 13 marks.
So although average = 33.4,
most students are roughly between:
≈ 20 to 46 marks
Which matches the data — many students are in the 20–40 range.
(b) For the following distribuon, calculate quarle deviaon and its coecient:
Class Interval (C.I.)
Frequency
0–15
8
15–30
26
30–45
30
45–60
45
60–75
20
75–90
17
90–105
4
Ans: Step 1: Understanding the Problem
We are given a frequency distribution:
Class Interval (C.I.)
Frequency
0–15
8
15–30
26
30–45
30
45–60
45
60–75
20
75–90
17
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90–105
4
We need to calculate:
1. Quartile Deviation (Q.D.) =
2. Coefficient of Quartile Deviation =
So the task is to find the first quartile (
) and the third quartile (
) using the formula for
quartiles in grouped data.
Step 2: Prepare the Cumulative Frequency Table
We first compute cumulative frequencies (CF):
Class Interval
Frequency (f)
Cumulative Frequency (CF)
0–15
8
8
15–30
26
34
30–45
30
64
45–60
45
109
60–75
20
129
75–90
17
146
90–105
4
150
Total frequency .
Step 3: Locate Quartile Positions
• Position of
• Position of
So,
lies at the 37.5th value, and
lies at the 112.5th value.
Step 4: Identify Quartile Classes
• For
: Look at CF. The 37.5th value lies in the 30–45 class (since CF up to 30 is 34,
and CF up to 45 is 64).
• For
: The 112.5th value lies in the 60–75 class (since CF up to 60 is 109, and CF up
to 75 is 129).
Step 5: Apply Quartile Formula
The formula for quartiles in grouped data is:
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Where:
• = lower boundary of quartile class
• = total frequency
• = cumulative frequency before quartile class
• = frequency of quartile class
• = class width
(i) Calculate
• Quartile class = 30–45
•
•
•
•
• Position = 37.5
(ii) Calculate
• Quartile class = 60–75
•
•
•
•
• Position = 112.5
Step 6: Quartile Deviation and Coefficient
• Quartile Deviation (Q.D.):
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• Coefficient of Quartile Deviation:
Coefficient
Final Answer
• Quartile Deviation = 15.44
• Coefficient of Quartile Deviation ≈ 0.33
Explanation in Simple Terms
Think of quartiles as “markers” that divide your data into four equal parts.
•
is the point where 25% of the data lies below.
•
is the point where 75% of the data lies below.
• The difference between them tells us how spread out the middle 50% of the data is.
• Dividing that difference by 2 gives the quartile deviation, a measure of variability.
• The coefficient of quartile deviation is a relative measure—it tells us how big the
spread is compared to the average of
and
.
In this case, the quartile deviation of 15.44 shows a moderate spread in the middle half of
the distribution, and the coefficient of 0.33 indicates that the spread is about one-third of
the average of the two quartiles.
4.Calculate Mode (by grouping method) for the following data:
Marks (Below)
10
20
30
40
50
60
70
80
90
No. of Students
4
6
24
46
67
86
96
99
100
Ans: Given Data
We are given a “Marks (Below)” cumulative distribution:
Marks (Below)
10
20
30
40
50
60
70
80
90
No. of Students
4
6
24
46
67
86
96
99
100
This means:
• 4 students scored below 10
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• 6 students scored below 20
• 24 students scored below 30
• … and so on
But to find the mode by grouping method, we first need the simple frequency distribution
(class frequencies).
Step 1: Convert Cumulative Frequency → Simple Frequency
We subtract successive values:
Marks Class
Cumulative Freq
Simple Freq
0–10
4
4
10–20
6
6−4 = 2
20–30
24
24−6 = 18
30–40
46
46−24 = 22
40–50
67
67−46 = 21
50–60
86
86−67 = 19
60–70
96
96−86 = 10
70–80
99
99−96 = 3
80–90
100
100−99 = 1
So the frequency table becomes:
Class
Frequency
0–10
4
10–20
2
20–30
18
30–40
22
40–50
21
50–60
19
60–70
10
70–80
3
80–90
1
Step 2: Grouping Method Table
The grouping method helps when frequencies are irregular or close.
We create 6 columns:
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Original frequencies
Sum of 2 successive frequencies (start from 1st)
Sum of 2 successive frequencies (start from 2nd)
Sum of 3 successive frequencies (start from 1st)
Sum of 3 successive frequencies (start from 2nd)
Sum of 3 successive frequencies (start from 3rd)
Column 1: Original Frequencies
Highest frequency = 22 → class 30–40
Column 2: Pairs (start 1st)
4+2 = 6
18+22 = 40
21+19 = 40
10+3 = 13
Highest = 40 → classes 20–30 & 30–40 and 40–50 & 50–60
Column 3: Pairs (start 2nd)
2+18 = 20
22+21 = 43
19+10 = 29
3+1 = 4
Highest = 43 → classes 30–40 & 40–50
Column 4: Triples (start 1st)
4+2+18 = 24
22+21+19 = 62
10+3+1 = 14
Highest = 62 → classes 30–40, 40–50, 50–60
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Column 5: Triples (start 2nd)
2+18+22 = 42
21+19+10 = 50
3+1 = 4
Highest = 50 → classes 40–50, 50–60, 60–70
Column 6: Triples (start 3rd)
18+22+21 = 61
19+10+3 = 32
Highest = 61 → classes 20–30, 30–40, 40–50
Step 3: Analysis Table
Now count how many times each class appears in the highest groups.
Class
Tally
20–30
2
30–40
5
40–50
4
50–60
2
others
0
So the modal class = 30–40 (appears most).
Step 4: Mode Formula (Grouped Data)
Where:
• = lower limit of modal class = 30
•
= frequency of modal class = 22
•
= frequency of preceding class = 18
•
= frequency of succeeding class = 21
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• = class width = 10
Step 5: Substitute Values
Final Answer
Mode
Easy Understanding
Imagine a classroom of 100 students.
Their marks are grouped into ranges like 0–10, 10–20, 20–30, etc.
When we converted the cumulative data into actual class frequencies, we saw:
• Most students scored between 30–40
• That range had 22 students — the highest
• So naturally, the most common marks lie near the middle of that class
The grouping method confirms that 30–40 repeatedly appears as dominant, meaning it is
truly the modal class.
Then the formula refines the estimate inside that class → giving Mode ≈ 38.
So we can interpret:
The most common marks scored by students are around 38.
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SECTION – C
5.(a) Calculate Karl Pearson’s coecient of skewness from the following data:
Size
1
2
3
4
5
6
7
Frequency
10
18
30
25
12
3
2
(b) Karl Pearson’s measure of skewness of a distribuon is 0.5.
Its Median and Mode are 42 and 36 respecvely.
Find the coecient of variaon.
Ans: Part (a) Karl Pearson’s Coefficient of Skewness
We are given the distribution:
Size (x)
Frequency (f)
1
10
2
18
3
30
4
25
5
12
6
3
7
2
Step 1: Compute Mean
The formula for mean is:
• Total frequency .
• Compute :
o
o
o
o
o
o
o
.
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So,
Step 2: Find Mode
Mode is the value with the highest frequency.
• Maximum frequency = 30 (for size = 3). So, Mode = 3.
Step 3: Find Standard Deviation
Formula:
• Compute
:
o
o
o
o
o
o
o
.
So,
Step 4: Karl Pearson’s Coefficient of Skewness
Formula:
Mode
Answer for Part (a): Karl Pearson’s coefficient of skewness ≈ 0.14
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Part (b) Coefficient of Variation
We are told:
• Karl Pearson’s skewness = 0.5
• Median = 42
• Mode = 36
We need to find the coefficient of variation (CV).
Step 1: Recall Karl Pearson’s Skewness Formula
Mode
But there’s also a relation between mean, median, and mode:
Mode Median
Step 2: Use Given Values
Median = 42, Mode = 36. So,
So, Mean = 45.
Step 3: Use Skewness Formula to Find Standard Deviation
Given skewness = 0.5,
So, Standard Deviation = 18.
Step 4: Coefficient of Variation
Formula:
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Answer for Part (b): Coefficient of Variation = 40%
Final Results
• (a) Karl Pearson’s coefficient of skewness ≈ 0.14
• (b) Coefficient of Variation = 40%
6.(a) For the following data, calculate Spearman’s rank correlaon coecient between x
and y:
x
30
38
28
27
28
23
30
33
28
35
y
29
27
22
29
20
29
18
21
27
22
(b) From the data given below, nd the two regression equaons:
x
25
28
35
32
31
36
29
38
34
32
y
43
46
49
41
36
32
31
30
33
39
Ans: 6(a) Spearman’s Rank Correlation Coefficient between x and y
We are given paired data:
x
30
38
28
27
28
23
30
33
28
35
y
29
27
22
29
20
29
18
21
27
22
We need Spearman’s rank correlation (ρ), which measures how strongly two variables are
related in terms of their rank order, not their actual values.
Think of it like this:
If higher x values generally go with higher y values → positive correlation
If higher x values go with lower y values → negative correlation
Step 1: Rank x values
Arrange x from smallest to largest:
23, 27, 28, 28, 28, 30, 30, 33, 35, 38
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Now assign ranks (average rank for ties):
x
Rank
23
1
27
2
28
3,4,5 → avg = 4
30
6,7 → avg = 6.5
33
8
35
9
38
10
So ranked x:
x
Rank
30
6.5
38
10
28
4
27
2
28
4
23
1
30
6.5
33
8
28
4
35
9
Step 2: Rank y values
y sorted:
18, 20, 21, 22, 22, 27, 27, 29, 29, 29
Ranks:
y
Rank
18
1
20
2
21
3
22
4.5
27
6.5
29
9
So ranked y:
y
Rank
29
9
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27
6.5
22
4.5
29
9
20
2
29
9
18
1
21
3
27
6.5
22
4.5
Step 3: Find d = (Rx − Ry) and d²
Rx
Ry
d
d²
6.5
9
−2.5
6.25
10
6.5
3.5
12.25
4
4.5
−0.5
0.25
2
9
−7
49
4
2
2
4
1
9
−8
64
6.5
1
5.5
30.25
8
3
5
25
4
6.5
−2.5
6.25
9
4.5
4.5
20.25
Step 4: Apply Spearman formula
Here:
n = 10
Σd² = 217.5
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Answer (a):
Interpretation:
There is a moderate negative rank correlation between x and y.
This means: as x increases, y tends to decrease slightly.
6(b) Find the Two Regression Equations
Given data:
x
25
28
35
32
31
36
29
38
34
32
y
43
46
49
41
36
32
31
30
33
39
We need:
Regression of y on x → predict y from x
Regression of x on y → predict x from y
Step 1: Means
Step 2: Regression coefficient
Formula:
After calculation:
Regression equation of y on x
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Step 3: Regression of x on y
Regression equation of x on y
Final Answers (b)
Regression of y on x:
Regression of x on y:
Understanding the Meaning (Intuition)
Let’s interpret the results like a story:
Imagine x and y represent two related measurements (like study hours and stress level).
Both regression slopes are negative.
That means:
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When x increases → y tends to decrease
When y increases → x tends to decrease
So these variables move in opposite directions.
This matches part (a), where Spearman correlation was also negative (−0.36).
So everything is consistent — which is a good statistical sign!
Final Summary
(a) Spearman rank correlation:
Moderate negative relationship.
(b) Regression equations:
7. Construct the Fisher’s Ideal Index for the following and show that it sases the factor
reversal test and me reversal test :
Commodity
Base Year
Current Year
Price
Quanty
Price
Quanty
A
2
10
4
12
B
4
5
4
8
C
5
10
7
15
D
10
12
12
10
E
15
15
15
10
Ans: Step 1: Recall Fisher’s Ideal Index Formula
Fisher’s Index is defined as the geometric mean of Laspeyres and Paasche indices:
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Where:
•
(Laspeyres Price Index)
•
(Paasche Price Index)
Here,
= price and quantity in base year,
= price and quantity in current year.
Step 2: Organize the Data
Commodity
A
2
10
4
12
20
40
24
48
B
4
5
4
8
20
20
32
32
C
5
10
7
15
50
70
75
105
D
10
12
12
10
120
144
100
120
E
15
15
15
10
225
225
150
150
Total
435
499
381
455
Step 3: Calculate Laspeyres and Paasche Indices
• Laspeyres Price Index:
So,
.
• Paasche Price Index:
So,
.
Step 4: Fisher’s Ideal Index
Fisher’s Ideal Price Index = 117.1
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This means prices in the current year are about 17.1% higher than in the base year,
according to Fisher’s balanced measure.
Step 5: Factor Reversal Test
The factor reversal test states:
Where
is Fisher’s Quantity Index:
• Laspeyres Quantity Index:
• Paasche Quantity Index:
• Fisher’s Quantity Index:
Now check factor reversal:
On the other hand:
Both values match (small rounding difference). Factor Reversal Test satisfied.
Step 6: Time Reversal Test
The time reversal test states:
We already have
. Now compute
(reverse base and current year).
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• Laspeyres (reverse):
• Paasche (reverse):
• Fisher (reverse):
Now check:
Time Reversal Test satisfied.
Conclusion
• Fisher’s Ideal Index = 117.1
• It satisfies both the Factor Reversal Test and the Time Reversal Test.
8. Interpolate the missing values from the following data :
Year
Producon (in millions)
1990
76.6
1991
78.7
1992
?
1993
77.7
1994
78.7
1995
?
1996
80.6
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Ans: Interpolation of Missing Values in Time Series Data
Imagine you are looking at agricultural production data for several years. Most values are
available, but for two years — 1992 and 1995 — the production figures are missing.
Your task is to estimate these missing values using the surrounding data. This process is
called interpolation.
Given Data
Year
Production (millions)
1990
76.6
1991
78.7
1992
?
1993
77.7
1994
78.7
1995
?
1996
80.6
What is Interpolation?
Interpolation means estimating a missing value within the range of known data.
Since 1992 lies between 1991 and 1993, and 1995 lies between 1994 and 1996, we can
estimate them using nearby values.
In simple terms:
We assume the change from year to year is gradual.
So the missing value should lie between its neighboring years.
Method Used: Simple Average Interpolation
Because only one value is missing between two known values, the easiest and most
common method is:
Missing Value
Previous Year Value Next Year Value
This assumes a steady change between the two known years.
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Step 1: Find Production for 1992
Known values:
• 1991 = 78.7
• 1993 = 77.7
So 1992 lies between them.
Apply formula:
Estimated production for 1992 = 78.2 million
Step 2: Find Production for 1995
Known values:
• 1994 = 78.7
• 1996 = 80.6
Apply formula:
Estimated production for 1995 = 79.65 million
Final Completed Table
Year
Production (millions)
1990
76.6
1991
78.7
1992
78.2
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1993
77.7
1994
78.7
1995
79.65
1996
80.6
Understanding the Logic (Conceptual Explanation)
Let’s understand this in a more intuitive way.
Imagine production changes like temperature over days — it usually does not jump
suddenly unless something unusual happens. So if production was:
• 78.7 in 1991
• 77.7 in 1993
then in 1992 it should logically lie between them.
Similarly:
• 78.7 in 1994
• 80.6 in 1996
so 1995 should fall between these.
This assumption of gradual change is the basis of interpolation.
Why This Method Works Here
This simple average method is suitable because:
✔ Only one year missing between known years
✔ No sharp irregular fluctuations
✔ Time intervals equal (1 year each)
✔ Data trend smooth
If many years were missing or the trend was curved, we would use:
• Newton’s interpolation
• Parabolic interpolation
• Polynomial methods
But here, simple interpolation is sufficient and correct.
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Interpretation of Results
Now let’s see the pattern after filling missing values:
Production trend:
• 1990 → 76.6
• 1991 → 78.7 (rise)
• 1992 → 78.2 (slight drop)
• 1993 → 77.7 (drop)
• 1994 → 78.7 (rise)
• 1995 → 79.65 (rise)
• 1996 → 80.6 (rise)
So overall trend shows:
Slight fluctuation in early 1990s
Steady growth after 1993
Highest production in 1996
This looks realistic and smooth — confirming interpolation is reasonable.
Exam-Style Conclusion
The missing production values were estimated using simple interpolation by averaging the
production of adjacent years. The estimated production for 1992 is 78.2 million and for
1995 is 79.65 million. The completed data shows a generally increasing production trend
from 1990 to 1996 with minor fluctuations. This method assumes gradual change between
consecutive years and is appropriate when only one value is missing between two known
observations.
Final Answers
Production in 1992 = 78.2 million
Production in 1995 = 79.65 million
“This paper has been carefully prepared for educaonal purposes. If you noce any
mistakes or have suggesons, feel free to share your feedback.”
