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Question
Using the following data, construct Fisher’s Ideal index and show how it satisfies Factor Reversal Test and Time Reversal Test?
| Commodity | Price in Rupees per unit | Number of units | ||
| Basic year | Current year | Base year | Current year | |
| A | 6 | 10 | 50 | 56 |
| B | 2 | 2 | 100 | 120 |
| C | 4 | 6 | 60 | 60 |
| D | 10 | 12 | 50 | 24 |
| E | 8 | 12 | 40 | 36 |
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Solution
| Commodity | Base year | Current year | p0q0 | p0q1 | p1q0 | p1q1 | ||
| p0 | q0 | p1 | q1 | |||||
| A | 6 | 10 | 50 | 56 | 300 | 336 | 500 | 560 |
| B | 2 | 2 | 100 | 120 | 200 | 240 | 200 | 240 |
| C | 4 | 6 | 60 | 60 | 240 | 240 | 360 | 360 |
| D | 10 | 12 | 50 | 24 | 500 | 240 | 600 | 288 |
| E | 8 | 12 | 40 | 36 | 320 | 288 | 480 | 432 |
| Total | `sum"p"_0"q"_0` = 1560 | `sum"p"_0"q"_1` = 1344 | `sum"p"_1"q"_0` = 2140 | `sum"p"_1"q"_1` = 1880 | ||||
Fisher’s Price Index Number
`"P"_01^"F" = sqrt((sum"p"_1"q"_0)/(sum"p"_0"q"_0) xx (sum"p"_1"q"_1)/(sum"p"_0"q"_1)) xx 100`
= `sqrt(2140/1560 xx 1880/1344) xx 100`
= `sqrt((40,23,200)/(20,96,640)) xx 100`
= `sqrt(1.92) xx 100`
= `1.385 xx 100`
= 138.5
Time Reversal Test: To prove P01 × P10 = 1
P01 × P10 = `sqrt((sum"p"_1"q"_0 xx sum"p"_1"q"_1)/(sum"p"_0"q"_0 xx sum"p"_0"q"_1)) xx sqrt((sum"p"_0"q"_1 xx sum"p"_0"q"_0)/(sum"p"_1"q"_1 xx sum"p"_1"q"_0))`
= `sqrt(2140/1560 xx 1880/1344 xx 1344/1880 xx 1560/2140)`
P01 × P10 = 1
Time reversal test is satisfied.
Factor Reversal Test: To prove P01 × Q01 = `(sum"p"_1"q"_1)/(sum"p"_0"q"_0)`
= `sqrt((sum"p"_1"q"_0 xx sum"p"_1"q"_1)/(sum"p"_0"q"_0 xx sum"p"_0"q"_1)) xx sqrt((sum"q"_1"P"_0 xx sum"q"_1"P"_1)/(sum"q"_0"p"_0 xx sum"q"_0"p"_1))`
= `sqrt(2140 /1560 xx 1880/1344 xx 1344/1560 xx 1880/2140)`
= `sqrt((1880 xx 1880)/(1560 xx 1560)`
= `1880/1560`
⇒ `"P"_01 xx "Q"_01 = (sum"p"_1"q"_1)/(sum"p"_0"q"_0)`
Factor reversal test is satisfied.
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| Group A | Group B |
| 1) Price Index | a) `(sump_1q_1)/(sump_0q_0)xx100` |
| 2) Value Index | b) `(sumq_1)/(sumq_0)xx100` |
| 3) Quantity Index | c) `(sump_1q_1)/(sump_0q_1)xx100` |
| 4) Paasche's Index | d) `(sump_1)/(sump_0)xx100` |
