List of Practical Experiments (Official Syllabus)

1. Fitting of modified exponential curve (three selected points).
2. Fitting of modified exponential curve (partial sums method).
3. Fitting of Gompertz curve (three selected points).
4. Fitting of Gompertz curve (partial sums method).
5. Fitting of logistic curve (partial sums method).
6. Index numbers by shifting the base.
7. Fixed-base and chain-base index numbers.
8. Splicing two index number series.
9. Deflating index numbers.
10. Income and price elasticities of demand.
11. Fitting Pareto's curve to income distributions.
12. Test reliability computations.

Phase 1 — Empirical Fitting of Growth Curves

Experiment 1 — Modified Exponential Curve (Three Selected Points)

1. Problem

The values of a series at three equidistant times (coded \(t = 0, 1, 2\)) are \(Y_0 = 120,\; Y_1 = 130,\; Y_2 = 145\). Fit the modified exponential trend \(Y_t = a + bc^{t}\).

2. Aim

To estimate \(a, b, c\) of the modified exponential curve from three equidistant selected points.

3. Formula

Applying it:

1. Choose three points separated by a fixed interval \(k\) (here \(k = 1\)).
2. Compute \(c\) from the ratio of successive differences, then \(a\), then \(b\).

4. Calculation

Blank working table:

Quantity	Value
\(c\)
\(a\)
\(b\)

Quantity	Value
\(c = (145-130)/(130-120)\)	1.5
\(a = (120\cdot145 - 130^2)/(120+145-260)\)	100
\(b = 120 - 100\)	20

\(a = (17400 - 16900)/5 = 500/5 = 100\).

5. Result

The fitted curve is \(Y_t = 100 + 20(1.5)^{t}\), with asymptote (lower bound) \(a = 100\).

Experiment 2 — Modified Exponential Curve (Partial Sums)

1. Problem

Fit \(Y_t = a + bc^{t}\) to the six successive values (\(t = 0,\dots,5\)): 105, 110, 120, 140, 180, 260 (\(n = 3k,\; k = 2\)).

2. Aim

To fit the modified exponential curve by the method of partial (group) sums, which smooths irregular variation.

3. Formula

Applying it:

1. Split the series into three equal groups of \(k = 2\) and form the group sums \(S_1, S_2, S_3\).
2. Compute \(c^{k}\), then \(c\), \(b\) and \(a\).

4. Calculation

Blank working table:

Group	Sum
\(S_1 = Y_0 + Y_1\)
\(S_2 = Y_2 + Y_3\)
\(S_3 = Y_4 + Y_5\)
\(c,\; b,\; a\)

Group	Sum
\(S_1 = 105 + 110\)	215
\(S_2 = 120 + 140\)	260
\(S_3 = 180 + 260\)	440

\(c^{2} = (440-260)/(260-215) = 180/45 = 4 \Rightarrow c = 2\); \(b = (45)(1)/(3)^2 = 45/9 = 5\); \(a = \tfrac12[215 - 5\cdot 3/1] = \tfrac12(200) = 100\).

5. Result

The fitted curve is \(Y_t = 100 + 5(2)^{t}\).

Experiment 3 — Gompertz Curve (Three Selected Points)

1. Problem

Fit the Gompertz curve \(Y_t = k\,a^{\,b^{t}}\) to three equidistant points (\(t = 0, 1, 2\)) with \(Y_0 = 10,\; Y_1 = 158.5,\; Y_2 = 478.6\).

2. Aim

To estimate \(k, a, b\) of the Gompertz S-curve by linearising it with logarithms.

3. Formula

Applying it:

1. Take logs: \(U_t = \log Y_t = \log k + (\log a)\,b^{t}\) — a modified exponential in \(U\).
2. Apply the three-point formulas to \(U_0, U_1, U_2\) to get \(b\), \(\log k\) and \(\log a\).

4. Calculation

Blank working table (\(U = \log_{10} Y\)):

\(t\)	\(Y_t\)	\(U_t = \log_{10} Y_t\)
0	10
1	158.5
2	478.6

\(t\)	\(Y_t\)	\(U_t = \log_{10} Y_t\)
0	10	1.00
1	158.5	2.20
2	478.6	2.68

\(b = (2.68-2.20)/(2.20-1.00) = 0.48/1.20 = 0.4\); \(\log k = (1\cdot2.68 - 2.20^2)/(1 + 2.68 - 4.40) = (-2.16)/(-0.72) = 3 \Rightarrow k = 1000\); \(\log a = 1 - 3 = -2 \Rightarrow a = 0.01\).

5. Result

The fitted Gompertz curve is \(Y_t = 1000\,(0.01)^{\,0.4^{t}}\), with upper asymptote (ceiling) \(k = 1000\).

Experiment 4 — Gompertz Curve (Partial Sums)

1. Problem

Fit \(Y_t = k\,a^{\,b^{t}}\) by partial sums to the six values (\(t = 0,\dots,5\)): 10, 158.5, 478.6, 744.7, 888.9, 954.3 (\(k = 2\) per group).

2. Aim

To fit the Gompertz curve by applying the partial-sums modified-exponential method to the logarithms of the data.

3. Formula

Applying it:

1. Convert to logs \(U_t = \log Y_t\).
2. Form the three group log-sums \(SU_1, SU_2, SU_3\).
3. Apply the partial-sums formulas (as in Experiment 2) to get \(b\), \(\log a\) and \(\log k\).

4. Calculation

Blank working table:

Group of \(U = \log_{10} Y\)	Sum
\(SU_1 = U_0 + U_1\)
\(SU_2 = U_2 + U_3\)
\(SU_3 = U_4 + U_5\)

Logs: \(U = 1.00,\; 2.20,\; 2.68,\; 2.872,\; 2.949,\; 2.980\).

Group of \(U = \log_{10} Y\)	Sum
\(SU_1 = 1.00 + 2.20\)	3.200
\(SU_2 = 2.68 + 2.872\)	5.552
\(SU_3 = 2.949 + 2.980\)	5.928

\(b^{2} = (5.928-5.552)/(5.552-3.200) = 0.376/2.352 = 0.16 \Rightarrow b = 0.4\); \(\log a = (2.352)(-0.6)/(0.16-1)^2 = -2 \Rightarrow a = 0.01\); \(\log k = \tfrac12[3.200 - (-2)(-0.84)/(-0.6)] = \tfrac12(6) = 3 \Rightarrow k = 1000\).

5. Result

The fitted Gompertz curve is \(Y_t = 1000\,(0.01)^{\,0.4^{t}}\), matching Experiment 3 but using all six data points.

Experiment 5 — Logistic Curve (Partial Sums)

1. Problem

Fit the logistic curve \(Y_t = \dfrac{k}{1 + b\,e^{-ct}}\) to the six values (\(t = 0,\dots,5\)): 100, 181.8, 307.7, 470.6, 640.0, 780.5.

2. Aim

To fit the symmetric logistic (S-shaped) growth curve by reducing it, via reciprocals, to a modified exponential and applying partial sums.

3. Formula

Applying it:

1. Take reciprocals \(Z_t = 1/Y_t = \tfrac1k + \tfrac{b}{k}e^{-ct}\) — a modified exponential in \(Z\) with ratio \(r = e^{-c}\).
2. Form the group sums \(SZ_1, SZ_2, SZ_3\) and solve for \(r\), \(A = 1/k\) and \(B = b/k\).
3. Recover \(k = 1/A,\; b = B/A,\; c = -\ln r\).

4. Calculation

Blank working table (\(Z = 1/Y\)):

Group of \(Z = 1/Y\)	Sum
\(SZ_1 = Z_0 + Z_1\)
\(SZ_2 = Z_2 + Z_3\)
\(SZ_3 = Z_4 + Z_5\)

Reciprocals: \(Z = 0.01000,\; 0.00550,\; 0.00325,\; 0.002125,\; 0.0015625,\; 0.00128\).

Group of \(Z = 1/Y\)	Sum
\(SZ_1\)	0.015500
\(SZ_2\)	0.005375
\(SZ_3\)	0.002844

\(r^{2} = (0.002844-0.005375)/(0.005375-0.015500) = 0.25 \Rightarrow r = 0.5\), so \(c = -\ln 0.5 = 0.693\); \(A = 0.001 \Rightarrow k = 1000\); \(B = 0.009 \Rightarrow b = B/A = 9\).

5. Result

The fitted logistic curve is \(Y_t = \dfrac{1000}{1 + 9\,e^{-0.693\,t}}\), with carrying capacity \(k = 1000\).

Phase 2 — Index Numbers

Experiment 6 — Index Numbers by Shifting the Base

1. Problem

An index series with base 2010 = 100 is 100, 110, 121, 133, 146 for 2010–2014. Shift the base to 2012.

2. Aim

To recompute an index series onto a new base year.

3. Formula

Applying it:

1. Identify the old index value in the new base year (2012 = 121).
2. Divide every old index by that value and multiply by 100.

4. Calculation

Blank working table:

Year	Old index (2010 = 100)	New index (2012 = 100)
2010	100
2011	110
2012	121
2013	133
2014	146

Year	Old index (2010 = 100)	New index (2012 = 100)
2010	100	82.6
2011	110	90.9
2012	121	100.0
2013	133	109.9
2014	146	120.7

Each value is (old index / 121) × 100, e.g. 2010: \(100/121\times100 = 82.6\).

5. Result

On the 2012 base the series becomes 82.6, 90.9, 100, 109.9, 120.7 — the same trend expressed relative to 2012.

Experiment 7 — Fixed-Base and Chain-Base Index Numbers

1. Problem

Prices for 2018–2022 are 100, 120, 150, 180, 216. Construct the fixed-base index (2018 = 100) and the chain-base index, and compare.

2. Aim

To build fixed-base and chain-base index series and confirm they agree for a single-commodity series.

3. Formula

Applying it:

1. Fixed base: \((P_t/P_0)\times 100\).
2. Link relatives: \(L_t = (P_t/P_{t-1})\times 100\).
3. Chain index: \(\text{CBI}_t = \text{CBI}_{t-1}\times L_t/100\), starting at 100.

4. Calculation

Blank working table:

Year	Price	Link relative	Fixed base	Chain base
2018	100	—
2019	120
2020	150
2021	180
2022	216

Year	Price	Link relative	Fixed base	Chain base
2018	100	—	100	100
2019	120	120.0	120	120
2020	150	125.0	150	150
2021	180	120.0	180	180
2022	216	120.0	216	216

5. Result

The fixed-base and chain-base indices coincide (100, 120, 150, 180, 216), as they must for a single price series computed consistently.

Experiment 8 — Splicing Two Index Number Series

1. Problem

An old series (2010 = 100) is 100, 120, 150 for 2010–2012; a new series (2012 = 100) is 100, 110, 125 for 2012–2014. Splice the new series onto the old base to form one continuous series.

2. Aim

To join two index series with different base years into a single continuous timeline by forward splicing.

3. Formula

Applying it:

1. Find the overlap year (2012), where the old index = 150.
2. Forward splice: multiply each new-series index by (old index in overlap year)/100.

4. Calculation

Blank working table:

Year	New index (2012 = 100)	Spliced to 2010 base
2012	100
2013	110
2014	125

Year	New index (2012 = 100)	Spliced to 2010 base
2012	100	150.0
2013	110	165.0
2014	125	187.5

Each new index × 150/100, e.g. 2013: \(110\times 1.5 = 165\).

5. Result

The continuous series on the 2010 base is 100, 120, 150, 165, 187.5 — the old series extended seamlessly by the spliced new values.

Experiment 9 — Deflating Index Numbers

1. Problem

Nominal annual wages for 2018–2021 are ₹20,000, ₹23,000, ₹26,000, ₹30,000 and the consumer price index (2018 = 100) is 100, 110, 125, 140. Compute the real wages.

2. Aim

To convert nominal values to real values by deflating with a price index.

3. Formula

Applying it:

1. Divide each nominal value by the price index of that year and multiply by 100.
2. Compare the real series to judge purchasing-power change.

4. Calculation

Blank working table:

Year	Nominal wage (₹)	CPI	Real wage (₹)
2018	20,000	100
2019	23,000	110
2020	26,000	125
2021	30,000	140

Year	Nominal wage (₹)	CPI	Real wage (₹)
2018	20,000	100	20,000
2019	23,000	110	20,909
2020	26,000	125	20,800
2021	30,000	140	21,429

E.g. 2021: \(30000/140\times100 = 21{,}429\).

5. Result

Real wages rose only from ₹20,000 to about ₹21,429 over the period, so most of the nominal wage growth was offset by inflation.

Phase 3 — Demand Analysis and Educational Test Analytics

Experiment 10 — Income and Price Elasticities of Demand

1. Problem

When price rises from ₹10 to ₹12, quantity demanded falls from 100 to 80 units. When income rises from ₹5,000 to ₹6,000, quantity rises from 100 to 115. Compute the arc price and income elasticities.

2. Aim

To measure the responsiveness of demand to price and to income using the arc-elasticity formula.

3. Formula

Applying it:

1. Price elasticity: \(E_p = \dfrac{\Delta Q}{\Delta P}\cdot\dfrac{P_1 + P_2}{Q_1 + Q_2}\).
2. Income elasticity: \(E_m = \dfrac{\Delta Q}{\Delta M}\cdot\dfrac{M_1 + M_2}{Q_1 + Q_2}\).
3. Interpret: \(|E| > 1\) elastic, \(< 1\) inelastic.

4. Calculation

Blank working table:

Elasticity	Value	Interpretation
Price \(E_p\)
Income \(E_m\)

\(E_p = \dfrac{-20}{2}\cdot\dfrac{10+12}{100+80} = -10\cdot\dfrac{22}{180} = -1.222\).

\(E_m = \dfrac{15}{1000}\cdot\dfrac{5000+6000}{100+115} = 0.015\cdot\dfrac{11000}{215} = 0.767\).

Elasticity	Value	Interpretation
Price \(E_p\)	−1.222	elastic (|E| > 1)
Income \(E_m\)	0.767	normal good, income-inelastic

5. Result

Demand is price-elastic (\(E_p = -1.22\)): a 1 % price rise cuts quantity by about 1.22 %. It is a normal but income-inelastic good (\(E_m = 0.77\)).

Experiment 11 — Fitting Pareto's Curve to Income Distribution

1. Problem

The number \(N\) of persons with income at least \(X\) is:

Income \(X\) (₹)	1000	2000	4000	8000
Persons \(N\)	10000	2500	625	156.25

Fit Pareto's law \(N = A\,X^{-\alpha}\).

2. Aim

To estimate the Pareto parameters \(A\) and \(\alpha\) by least-squares regression on a double-log scale.

3. Formula

Applying it:

1. Take logs: \(\log N = \log A - \alpha\,\log X\) — a straight line.
2. Regress \(\log N\) on \(\log X\); the slope is \(-\alpha\) and the intercept is \(\log A\).

4. Calculation

Blank working table (\(\log_{10}\)):

\(\log X\)	\(\log N\)
3.000
3.301
3.602
3.903

\(\log X\)	\(\log N\)
3.000	4.000
3.301	3.398
3.602	2.796
3.903	2.194

The points are exactly collinear with slope \((2.194 - 4.000)/(3.903 - 3.000) = -1.806/0.903 = -2.0\), so \(\alpha = 2\) and \(\log A = 10 \Rightarrow A = 10^{10}\).

5. Result

The fitted Pareto law is \(N = 10^{10}\,X^{-2}\), i.e. \(\alpha = 2\). A higher \(\alpha\) means less income concentration in the upper tail.

Experiment 12 — Test Reliability Computations

1. Problem

A 5-item test has item difficulty (proportion correct) \(p = 0.8, 0.6, 0.5, 0.7, 0.4\) and total test-score variance \(\sigma_X^{2} = 2.5\). Compute the KR-20 reliability. Also, if the two half-tests correlate \(r_{hh} = 0.6\), find the Spearman–Brown full-test reliability.

2. Aim

To evaluate the internal-consistency reliability of a test using KR-20 and the split-half (Spearman–Brown) method.

3. Formula

Applying it:

1. For each item compute \(p_i q_i\) (with \(q_i = 1 - p_i\)) and sum.
2. Apply KR-20 with \(k\) items and total-score variance \(\sigma_X^2\).
3. Split-half: correct the half-test correlation with the Spearman–Brown formula.

4. Calculation

Blank working table:

Item	\(p_i\)	\(q_i\)	\(p_i q_i\)
1	0.8
2	0.6
3	0.5
4	0.7
5	0.4
Σ	—	—

Item	\(p_i\)	\(q_i\)	\(p_i q_i\)
1	0.8	0.2	0.16
2	0.6	0.4	0.24
3	0.5	0.5	0.25
4	0.7	0.3	0.21
5	0.4	0.6	0.24
Σ	—	—	1.10

\(r_{KR\text{-}20} = \dfrac{5}{4}\!\left(1 - \dfrac{1.10}{2.5}\right) = 1.25\,(1 - 0.44) = 1.25\times0.56 = 0.70\).

Split-half: \(r_{SB} = \dfrac{2(0.6)}{1 + 0.6} = \dfrac{1.2}{1.6} = 0.75\).

5. Result

KR-20 reliability = 0.70 and Spearman–Brown split-half reliability = 0.75 — both indicate acceptable (moderate) internal consistency for the test.

Lab Record Format (to be followed for every experiment)

1. 1. Problem — the data given and what is to be fitted/computed.
2. 2. Aim — the parameter, index or coefficient the experiment produces.
3. 3. Formula — the formula, then the numbered steps that apply it.
4. 4. Calculation — the filled table with the arithmetic worked through.
5. 5. Result — the final value(s) with interpretation.