Index Numbers and Time Based Data

i) Correct. An index number is a ratio (or percentage) of two values, so all units cancel out, making it a pure (dimensionless) number.

ii) Correct. Because index numbers are ratios, they do not depend on the units in which prices or quantities are measured.

iii) Incorrect. An index number is always a ratio of two positive quantities multiplied by 100, so it is always a non-negative number.

iv) Correct. As the wholesale price index rises, the same amount of money buys fewer goods, meaning the purchasing power of money falls.

Correct Answer: iii) Weighted aggregative price index

A weighted aggregative price index uses the prices of items weighted by their quantities (or relative importance), giving a composite measure of price change. This distinguishes it from a simple aggregative index (no weights) or price relatives (individual ratios).

Correct Answer: iv) Paasche's index

Paasche's price index uses current-period quantities as weights:
$$P_{\text{Paasche}} = \frac{\sum p_1 q_1}{\sum p_0 q_1} \times 100$$
where $q_1$ denotes current-year quantities. Laspeyres, by contrast, uses base-period quantities.

Correct Answer: i) Quantity index

A quantity index measures the change in the volume/quantity of goods produced, consumed, or traded over time, keeping prices constant. It is defined as:
$$Q_{01} = \frac{\sum q_1 p_0}{\sum q_0 p_0} \times 100$$

Correct Answer: iii) Laspeyres method

Laspeyres price index uses base-period quantities ($q_0$) as weights:
$$P_{\text{Laspeyres}} = \frac{\sum p_1 q_0}{\sum p_0 q_0} \times 100$$

*(Note: The answer key in the textbook states i), but by definition Laspeyres uses base-period quantities. Fisher's method is the geometric mean of Laspeyres and Paasche's. The standard correct answer is iii) Laspeyres method.)*

Correct Answer: iv) 100

By convention, the index number of the base period is always taken as 100, since the price (or quantity) is compared with itself:
$$I_0 = \frac{p_0}{p_0} \times 100 = 100$$

Correct Answer: i) Average

An index number is a specialised average (typically a weighted average of price relatives or aggregates) that summarises the overall change in a group of related variables over time or space.

Correct Answer: i) Percentage

Index numbers are conventionally expressed as percentages. The formula multiplies the ratio by 100:
$$I = \frac{p_1}{p_0} \times 100$$
so the result is always a percentage figure (e.g., 120 means a 20% increase over the base).

Correct Answer: iii) Fisher

Fisher's index is called the Ideal Index because:
1. It is the geometric mean of Laspeyres and Paasche's indices, thus balancing the upward bias of Laspeyres and the downward bias of Paasche's.
2. It satisfies both the Time Reversal Test and the Factor Reversal Test.
$$P_{\text{Fisher}} = \sqrt{P_{\text{Laspeyres}} \times P_{\text{Paasche}}}$$

Correct Answer: i) Quantity in base year

Laspeyres price index:
$$P_L = \frac{\sum p_1 q_0}{\sum p_0 q_0} \times 100$$
The weight used is $q_0$, i.e., the quantity in the base year.

Correct Answer: ii) Quantity in current year

Paasche's price index:
$$P_P = \frac{\sum p_1 q_1}{\sum p_0 q_1} \times 100$$
The weight used is $q_1$, i.e., the quantity in the current year.

Correct Answer: ii) G.M. of Laspeyres and Paasche's

$$P_{\text{Fisher}} = \sqrt{P_L \times P_P} = \sqrt{\frac{\sum p_1 q_0}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_0 q_1}} \times 100$$
It is the geometric mean (G.M.) of Laspeyres ($P_L$) and Paasche's ($P_P$) indices.

Correct Answer: iii) Price index

A price index measures the relative change in the price of a commodity (here, rice) over time. It is computed as:
$$\text{Price Index} = \frac{p_1}{p_0} \times 100$$

Correct Answer: iii) Consumer price index

The Consumer Price Index (CPI) measures the average change in prices paid by consumers for a basket of goods and services. The purchasing power of money is the reciprocal of CPI:
$$\text{Purchasing Power} = \frac{1}{\text{CPI}} \times 100$$

Correct Answer: ii) Consumer price index

The Consumer Price Index (CPI), also called the cost-of-living index, is specifically designed to compare the cost of maintaining a given standard of living across different time periods or different cities/regions.

Given:
Base year = 1995, Current year = 2005.

Formula (Simple Aggregative Method):
$$P_{01} = \frac{\sum p_1}{\sum p_0} \times 100$$

Calculation:
$$\sum p_0 = 80 + 50 + 90 + 30 = 250$$
$$\sum p_1 = 95 + 60 + 100 + 45 = 300$$

$$P_{01} = \frac{300}{250} \times 100 = \boxed{120}$$

Interpretation: Prices in 2005 are 20% higher than in 1995.

Given: Base year = 2008 ($p_0, q_0$), Current year = 2010 ($p_1, q_1$).

Step 1: Prepare the calculation table.

| Commodity | $p_0$ | $p_1$ | $q_0$ | $q_1$ | $p_1q_0$ | $p_0q_0$ | $p_1q_1$ | $p_0q_1$ |
|---|---|---|---|---|---|---|---|---|
| P | 2 | 4 | 8 | 5 | 32 | 16 | 20 | 10 |
| Q | 5 | 6 | 12 | 10 | 72 | 60 | 60 | 50 |
| R | 4 | 5 | 15 | 12 | 75 | 60 | 60 | 48 |
| S | 2 | 4 | 18 | 20 | 72 | 36 | 80 | 40 |
| Total | | | | | 251 | 172 | 220 | 148 |

i) Laspeyres' Index:
$$P_L = \frac{\sum p_1 q_0}{\sum p_0 q_0} \times 100 = \frac{251}{172} \times 100 \approx \boxed{146}$$

ii) Paasche's Index:
$$P_P = \frac{\sum p_1 q_1}{\sum p_0 q_1} \times 100 = \frac{220}{148} \times 100 \approx \boxed{149}$$

iii) Fisher's Ideal Index:
$$P_F = \sqrt{P_L \times P_P} = \sqrt{146 \times 149} = \sqrt{21754} \approx \boxed{147}$$

Given: Base year = 1995, $p_0 = 12$.

Formula (Price Relative):
$$I = \frac{p_n}{p_0} \times 100$$

Calculations:

| Year | Price ($p_n$) | Index $= \frac{p_n}{12} \times 100$ |
|---|---|---|
| 1995 | 12 | $\frac{12}{12} \times 100 = \mathbf{100}$ |
| 1997 | 14 | $\frac{14}{12} \times 100 = \mathbf{117}$ |
| 1999 | 13 | $\frac{13}{12} \times 100 = \mathbf{108}$ |
| 2001 | 20 | $\frac{20}{12} \times 100 = \mathbf{167}$ |
| 2003 | 25 | $\frac{25}{12} \times 100 = \mathbf{208}$ |
| 2005 | 21 | $\frac{21}{12} \times 100 = \mathbf{175}$ |

Result: The relative index numbers are 100, 117, 108, 167, 208, 175 for the respective years.

Given: Current year prices ($p_1$), Base year prices ($p_0$), Weights ($W$).

Formula:
$$P_{01} = \frac{\sum p_1 W}{\sum p_0 W} \times 100$$

Calculation table:

| Variable | $p_1$ | $p_0$ | $W$ | $p_1 W$ | $p_0 W$ |
|---|---|---|---|---|---|
| X | 5 | 4 | 60 | 300 | 240 |
| Y | 3 | 2 | 50 | 150 | 100 |
| Z | 2 | 1 | 30 | 60 | 30 |
| Total | | | | 510 | 370 |

$$P_{01} = \frac{510}{370} \times 100 = 137.84 \approx \boxed{137}$$

Given: Base year = 1990, Current year = 2010.

Formula:
$$P_{01} = \frac{\sum p_1}{\sum p_0} \times 100$$

Calculation:
$$\sum p_0 = 60 + 40 + 100 + 60 + 90 = 350$$
$$\sum p_1 = 140 + 60 + 205 + 70 + 100 = 575$$

$$P_{01} = \frac{575}{350} \times 100 = 164.28 \approx \boxed{164}$$

Interpretation: Prices in 2010 are approximately 64% higher than in 1990.

Given: Base year = 1990 ($p_0$), Current year = 2003 ($p_1$).

Formula (Simple Aggregative Method for Cost of Living Index):
$$\text{CLI} = \frac{\sum p_1}{\sum p_0} \times 100$$

Calculation:
$$\sum p_0 = 1400 + 200 + 400 + 200 + 250 = 2450$$
$$\sum p_1 = 1500 + 250 + 750 + 300 + 425 = 3225$$

$$\text{CLI} = \frac{3225}{2450} \times 100 = 131.63 \approx \boxed{132}$$

Interpretation: The cost of living in 2003 is approximately 32% higher than in 1990.

Given: Index of sales volume and index of sales value for each year.

Concept:
$$\text{Value Index} = \text{Price Index} \times \text{Volume Index}$$
$$\Rightarrow \text{Price Index} = \frac{\text{Value Index}}{\text{Volume Index}} \times 100$$

Calculations:

Year 1995:
$$\text{Price Index} = \frac{105}{101} \times 100 = 103.96 \approx \mathbf{104}$$

Year 1996:
$$\text{Price Index} = \frac{108}{113} \times 100 = 95.58 \approx \mathbf{96}$$

Year 1997:
$$\text{Price Index} = \frac{124}{106} \times 100 = 116.98 \approx \mathbf{117}$$

Result:

| Year | Index of Retail Prices |
|---|---|
| 1995 | 104 |
| 1996 | 96 |
| 1997 | 117 |

Given: Weights ($W$), Base year prices ($p_0$), Current year prices ($p_1$).

Formula:
$$P_{01} = \frac{\sum p_1 W}{\sum p_0 W} \times 100$$

Calculation table:

| Commodity | $W$ | $p_0$ | $p_1$ | $p_0 W$ | $p_1 W$ |
|---|---|---|---|---|---|
| P | 14 | 90 | 120 | 1260 | 1680 |
| Q | 20 | 10 | 17 | 200 | 340 |
| R | 35 | 40 | 60 | 1400 | 2100 |
| S | 15 | 50 | 93 | 750 | 1395 |
| Total | | | | 3610 | 5515 |

$$P_{01} = \frac{5515}{3610} \times 100 = 152.77 \approx \boxed{153}$$

Time Reversal Test: An index satisfies the time reversal test if $P_{01} \times P_{10} = 1$.

Let base year = 0 ($p_0, q_0$) and current year = 1 ($p_1, q_1$).

Step 1: Compute required sums.

| Commodity | $p_0$ | $q_0$ | $p_1$ | $q_1$ | $p_1q_0$ | $p_0q_0$ | $p_1q_1$ | $p_0q_1$ |
|---|---|---|---|---|---|---|---|---|
| P | 4 | 10 | 6 | 15 | 60 | 40 | 90 | 60 |
| Q | 6 | 15 | 4 | 20 | 60 | 90 | 80 | 120 |
| R | 8 | 5 | 10 | 4 | 50 | 40 | 40 | 32 |
| Total | | | | | 170 | 170 | 210 | 212 |

i) Paasche's Formula — Time Reversal Test:

$$P_{01}^{\text{Paasche}} = \frac{\sum p_1 q_1}{\sum p_0 q_1} = \frac{210}{212}$$

For $P_{10}^{\text{Paasche}}$, swap 0 and 1 (i.e., current becomes base and base becomes current):
$$P_{10}^{\text{Paasche}} = \frac{\sum p_0 q_0}{\sum p_1 q_0} = \frac{170}{170} = 1$$

$$P_{01} \times P_{10} = \frac{210}{212} \times 1 = \frac{210}{212} \neq 1$$

Conclusion: Paasche's formula does NOT satisfy the time reversal test.

ii) Fisher's Formula — Time Reversal Test:

$$P_{01}^{\text{Fisher}} = \sqrt{\frac{\sum p_1 q_0}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_0 q_1}} = \sqrt{\frac{170}{170} \times \frac{210}{212}}$$

$$P_{10}^{\text{Fisher}} = \sqrt{\frac{\sum p_0 q_1}{\sum p_1 q_1} \times \frac{\sum p_0 q_0}{\sum p_1 q_0}} = \sqrt{\frac{212}{210} \times \frac{170}{170}}$$

$$P_{01} \times P_{10} = \sqrt{\frac{170}{170} \times \frac{210}{212}} \times \sqrt{\frac{212}{210} \times \frac{170}{170}}$$
$$= \sqrt{\frac{170 \times 210 \times 212 \times 170}{170 \times 212 \times 210 \times 170}} = \sqrt{1} = 1$$

Conclusion: Fisher's formula DOES satisfy the time reversal test.

Given: Annual rainfall data from 2001 to 2009.

Method: 3-year moving averages — each average is the mean of 3 consecutive years.

Formula: $\text{3-year MA} = \frac{y_t + y_{t+1} + y_{t+2}}{3}$, centred at year $t+1$.

Calculation table:

| Year | Rainfall (cm) | 3-Year Moving Total | 3-Year Moving Average |
|---|---|---|---|
| 2001 | 1.2 | — | — |
| 2002 | 1.9 | $1.2+1.9+2.0 = 5.1$ | $5.1/3 = 1.70$ |
| 2003 | 2.0 | $1.9+2.0+1.4 = 5.3$ | $5.3/3 = 1.77$ |
| 2004 | 1.4 | $2.0+1.4+2.1 = 5.5$ | $5.5/3 = 1.83$ |
| 2005 | 2.1 | $1.4+2.1+1.3 = 4.8$ | $4.8/3 = 1.60$ |
| 2006 | 1.3 | $2.1+1.3+1.8 = 5.2$ | $5.2/3 = 1.73$ |
| 2007 | 1.8 | $1.3+1.8+1.1 = 4.2$ | $4.2/3 = 1.40$ |
| 2008 | 1.1 | $1.8+1.1+1.3 = 4.2$ | $4.2/3 = 1.40$ |
| 2009 | 1.3 | — | — |

Result: The 3-year moving averages (trend values) are: 1.70, 1.77, 1.83, 1.60, 1.73, 1.40, 1.40 for years 2002 through 2008 respectively. The trend shows slight fluctuation with no strong upward or downward movement.

Given: Annual production data from 1980 to 1989.

Method: 4-year moving averages.

Step 1: Compute 4-year moving totals (sum of 4 consecutive years).

Step 2: Compute 4-year moving averages (divide by 4).

Step 3: Centre the averages (average of two consecutive 4-year averages to align with actual years).

| Year | Production | 4-yr Moving Total | 4-yr Moving Avg | Centred Moving Avg |
|---|---|---|---|---|
| 1980 | 2450 | — | — | — |
| 1981 | 1470 | — | — | — |
| 1982 | 2150 | $2450+1470+2150+1800=7870$ | $7870/4=1967.5$ | — |
| 1983 | 1800 | $1470+2150+1800+1210=6630$ | $6630/4=1657.5$ | $(1967.5+1657.5)/2=1812.5$ |
| 1984 | 1210 | $2150+1800+1210+1950=7110$ | $7110/4=1777.5$ | $(1657.5+1777.5)/2=1717.5$ |
| 1985 | 1950 | $1800+1210+1950+2300=7260$ | $7260/4=1815.0$ | $(1777.5+1815.0)/2=1796.25$ |
| 1986 | 2300 | $1210+1950+2300+2500=7960$ | $7960/4=1990.0$ | $(1815.0+1990.0)/2=1902.5$ |
| 1987 | 2500 | $1950+2300+2500+2480=9230$ | $9230/4=2307.5$ | $(1990.0+2307.5)/2=2148.75$ |
| 1988 | 2480 | $2300+2500+2480+2680=9960$ | $9960/4=2490.0$ | $(2307.5+2490.0)/2=2398.75$ |
| 1989 | 2680 | — | — | — |

Result: The centred 4-year moving averages (trend values) for years 1983–1988 are: 1812.5, 1717.5, 1796.25, 1902.5, 2148.75, 2398.75 respectively. These represent the trend component of the time series.

Given: $n = 5$ years. Let $A = 2003$ (middle year) so that $\sum x = 0$.

Step 1: Set up the table.

| Year | $y$ | $x = \text{Year} - 2003$ | $x^2$ | $xy$ |
|---|---|---|---|---|
| 2001 | 160 | $-2$ | 4 | $-320$ |
| 2002 | 185 | $-1$ | 1 | $-185$ |
| 2003 | 220 | $0$ | 0 | $0$ |
| 2004 | 300 | $1$ | 1 | $300$ |
| 2005 | 510 | $2$ | 4 | $1020$ |
| Total | 1375 | 0 | 10 | 815 |

Step 2: Calculate $a$ and $b$.

Since $\sum x = 0$:
$$a = \frac{\sum y}{n} = \frac{1375}{5} = 275$$
$$b = \frac{\sum xy}{\sum x^2} = \frac{815}{10} = 81.5$$

Step 3: Equation of straight-line trend:
$$\boxed{y = 275 + 81.5x}$$

Step 4: i) Trend values:

| Year | $x$ | Trend value $\hat{y} = 275 + 81.5x$ |
|---|---|---|
| 2001 | $-2$ | $275 + 81.5(-2) = 275 - 163 = 112$ |
| 2002 | $-1$ | $275 + 81.5(-1) = 275 - 81.5 = 193.5$ |
| 2003 | $0$ | $275 + 0 = 275$ |
| 2004 | $1$ | $275 + 81.5 = 356.5$ |
| 2005 | $2$ | $275 + 163 = 438$ |

Step 5: iii) Expected trend for 2006:
$$x_{2006} = 2006 - 2003 = 3$$
$$\hat{y}_{2006} = 275 + 81.5(3) = 275 + 244.5 = \boxed{₹519.5 \text{ lakh}}$$

Given: $n = 6$ (even number of years). Use coding: $x = \frac{\text{Year} - 1994.5}{0.5}$, so that $\sum x = 0$.

Step 1: Set up the table.

| Year | $y$ | $x = \frac{\text{Year}-1994.5}{0.5}$ | $x^2$ | $xy$ |
|---|---|---|---|---|
| 1992 | 210 | $-5$ | 25 | $-1050$ |
| 1993 | 225 | $-3$ | 9 | $-675$ |
| 1994 | 275 | $-1$ | 1 | $-275$ |
| 1995 | 220 | $1$ | 1 | $220$ |
| 1996 | 240 | $3$ | 9 | $720$ |
| 1997 | 235 | $5$ | 25 | $1175$ |
| Total | 1405 | 0 | 70 | 115 |

Step 2: Calculate $a$ and $b$.

Since $\sum x = 0$:
$$a = \frac{\sum y}{n} = \frac{1405}{6} = 234.17$$
$$b = \frac{\sum xy}{\sum x^2} = \frac{115}{70} = 1.643$$

Step 3: Equation of straight-line trend:
$$\hat{y} = 234.17 + 1.643x \quad \text{where } x = \frac{\text{Year} - 1994.5}{0.5}$$

Step 4: Trend value for 1998:
$$x_{1998} = \frac{1998 - 1994.5}{0.5} = \frac{3.5}{0.5} = 7$$
$$\hat{y}_{1998} = 234.17 + 1.643 \times 7 = 234.17 + 11.50 = \boxed{245.67 \text{ tons}}$$