List of Practical Experiments (Official Syllabus)

1. Measurement of trend by method of moving averages (odd and even period).
2. Measurement of trend by method of Least squares (linear and parabola).
3. Determination of seasonal indices by method of simple averages.
4. Determination of seasonal indices by method of Ratio to Moving Averages.
5. Determination of seasonal indices by method of Ratio to Trend.
6. Determination of seasonal indices by method of Link Relatives.
7. Computation of simple index numbers.
8. Computation of all weighted index numbers.
9. Computation of reversal tests.
10. Computation of various Mortality rates.
11. Computation of various Fertility rates.
12. Computation of various Reproduction rates.
13. Construction of Life Table.

Experiment 1 — Trend by Moving Averages

1. Problem

(a) Production (lakh tonnes) over 8 years: 10, 12, 14, 16, 20, 22, 24, 26 — compute the 3-yearly moving average. (b) For the series 5, 7, 11, 17, 25, 31, 40, 50 — compute the 4-yearly centred moving average.

2. Aim

To smooth a time series and reveal its trend using odd-period (3-yearly) and even-period (4-yearly, centred) moving averages.

3. Formula

Applying it:

1. Odd period: the \(m\)-yearly moving total divided by \(m\) is placed against the middle year.
2. Even period: form the 4-yearly moving totals (they fall between years); average successive pairs to centre them against a year.

4. Calculation

Blank working table (4-yearly centred MA):

Year	Y	4-yr MA (between)	Centred MA
1	5
2	7
3	11
4	17
5	25
6	31
7	40
8	50

(a) 3-yearly MA (placed at years 2–7): 12.00, 14.00, 16.67, 19.33, 22.00, 24.00.

(b) 4-yearly centred MA:

Year	Y	4-yr MA (between)	Centred MA
1	5
2	7	10.00
3	11	15.00	12.500
4	17	21.00	18.000
5	25	28.25	24.625
6	31	36.50	32.375
7	40
8	50

The five 4-yearly moving averages are 10.00, 15.00, 21.00, 28.25, 36.50; centring successive pairs gives 12.50, 18.00, 24.625, 32.375 at years 3–6.

5. Result

Both series show a steadily rising trend; the smoothed (moving-average) values increase monotonically, confirming an upward long-term movement.

Experiment 2 — Trend by Least Squares

1. Problem

(a) Fit a straight-line trend to the sales 10, 12, 14, 16, 20, 22, 24, 26 (2018–2025). (b) Fit a second-degree (parabolic) trend to the production 5, 7, 11, 17, 25.

2. Aim

To fit linear and parabolic trends by the method of least squares using coded time \(X\) (so that \(\sum X = 0\)).

3. Formula

Applying it:

1. Code time so \(\sum X = 0\) (odd number of years: \(X = \dots,-1,0,1,\dots\); even number: \(X = \dots,-3,-1,1,3,\dots\) in half-year units).
2. Linear \(\hat Y = a + bX\): \(a = \sum Y/n,\; b = \sum XY/\sum X^2\).
3. Parabola \(\hat Y = a + bX + cX^2\): solve \(\sum Y = na + c\sum X^2\), \(\sum XY = b\sum X^2\), \(\sum X^2Y = a\sum X^2 + c\sum X^4\).

4. Calculation

Blank working table (linear):

\(X\)	\(Y\)	\(X^2\)	\(XY\)
−7	10
−5	12
−3	14
−1	16
1	20
3	22
5	24
7	26
Σ

(a) Linear:

\(X\)	\(Y\)	\(X^2\)	\(XY\)
−7	10	49	−70
−5	12	25	−60
−3	14	9	−42
−1	16	1	−16
1	20	1	20
3	22	9	66
5	24	25	120
7	26	49	182
Σ	144	168	200

\(a = 144/8 = 18\); \(b = 200/168 = 1.190\) per half-year. Trend: \(\hat Y = 18 + 1.190\,X\) (equivalently, a rise of about \(2.38\) per year).

(b) Parabola (\(X = -2,-1,0,1,2\)): \(\sum Y = 65,\; \sum X^2 = 10,\; \sum X^4 = 34,\; \sum XY = 50,\; \sum X^2Y = 144\).

\(b = 50/10 = 5\). From \(65 = 5a + 10c\) and \(144 = 10a + 34c\): \(c = 1,\; a = 11\). Trend: \(\hat Y = 11 + 5X + X^2\).

5. Result

The linear trend is \(\hat Y = 18 + 1.190\,X\) (upward). The parabola \(\hat Y = 11 + 5X + X^2\) reproduces the data exactly (5, 7, 11, 17, 25), so growth is accelerating.

Experiment 3 — Seasonal Indices by Simple Averages

1. Problem

Determine the quarterly seasonal indices from three years of sales (₹ lakhs) by the method of simple averages.

Year	Q1	Q2	Q3	Q4
2023	60	80	72	68
2024	65	88	76	71
2025	70	92	80	78

2. Aim

To compute seasonal indices (summing to 400 for quarterly data) by averaging each quarter across years and expressing it relative to the grand mean.

3. Formula

Applying it:

1. Average each quarter over the years.
2. Compute the grand mean of the four quarterly averages.
3. Seasonal index of a quarter \(= (\text{quarter average}/\text{grand mean})\times 100\).

4. Calculation

Blank working table:

	Q1	Q2	Q3	Q4
Quarter average
Seasonal index

	Q1	Q2	Q3	Q4
Quarter average	65.00	86.67	76.00	72.33
Seasonal index	86.67	115.56	101.33	96.44

Grand mean \(= (65 + 86.67 + 76 + 72.33)/4 = 75\); indices sum to 400.

5. Result

Seasonal indices are Q1 = 86.67, Q2 = 115.56, Q3 = 101.33, Q4 = 96.44. Sales peak in Q2 and are weakest in Q1.

Experiment 4 — Seasonal Indices by Ratio to Moving Averages

1. Problem

Using the same quarterly data as Experiment 3, obtain the seasonal indices by the ratio-to-moving- average method.

2. Aim

To remove trend by a 4-quarter centred moving average and isolate the seasonal component as the ratio of actual value to moving average.

3. Formula

Applying it:

1. Compute 4-quarter moving totals, then the centred moving average (CMA).
2. Form the percentage ratio \((Y/\text{CMA})\times 100\) for each period.
3. Average the ratios by quarter; adjust so the four indices sum to 400.

4. Calculation

Blank working table (averaged ratios by quarter):

	Q1	Q2	Q3	Q4
Mean ratio
Adjusted index

Centred moving averages (Q3 2023 → Q2 2025): 70.625, 72.25, 73.75, 74.625, 75.625, 76.75, 77.75, 79.125. The corresponding \((Y/\text{CMA})\times 100\) ratios average by quarter as below (raw sum 400.72, adjustment factor \(400/400.72 = 0.9982\)):

	Q1	Q2	Q3	Q4
Mean ratio	89.08	117.10	101.22	93.31
Adjusted index	88.93	116.89	101.04	93.15

5. Result

Ratio-to-moving-average seasonal indices are Q1 = 88.93, Q2 = 116.89, Q3 = 101.04, Q4 = 93.15 (sum 400) — the same seasonal shape as the simple-average method, but trend-adjusted.

Experiment 5 — Seasonal Indices by Ratio to Trend

1. Problem

For the same data, obtain the seasonal indices by the ratio-to-trend method.

2. Aim

To fit a linear trend to the yearly averages, read off the quarterly trend values, and express each actual value as a percentage of its trend.

3. Formula

Applying it:

1. Compute the yearly average and fit a linear trend to it (coded \(X\)).
2. Distribute the annual trend to quarters (per-quarter increment \(= b/4\); the annual value sits mid-year, between Q2 and Q3).
3. Ratio \(= (\text{actual}/\text{trend})\times 100\); average by quarter; adjust to sum 400.

4. Calculation

Blank working table:

	Q1	Q2	Q3	Q4
Mean ratio
Seasonal index

Yearly averages 70, 75, 80 give the trend \(\hat Y = 75 + 5X\) (per-quarter increment 1.25). The quarterly ratios average by quarter as:

	Q1	Q2	Q3	Q4
Mean ratio	88.85	116.51	100.56	94.08
Seasonal index	88.85	116.51	100.56	94.08

The ratios already sum to 400, so no further adjustment is needed.

5. Result

Ratio-to-trend seasonal indices are Q1 = 88.85, Q2 = 116.51, Q3 = 100.56, Q4 = 94.08, in close agreement with the ratio-to-moving-average result.

Experiment 6 — Seasonal Indices by Link Relatives

1. Problem

For the same data, obtain the seasonal indices by Pearson's method of link relatives.

2. Aim

To compute link relatives, convert them to chain relatives, correct for the trend carried over one cycle, and express the corrected values as seasonal indices.

3. Formula

Applying it:

1. Link relative \(= (Y_t/Y_{t-1})\times 100\); average by quarter.
2. Chain relative: CR of the first quarter \(= 100\); each subsequent \(\text{CR} = (\text{avg LR}\times \text{previous CR})/100\).
3. Correction: compute the "return" \(\text{CR}\) of Q1; the excess over 100, divided by 4, is the per-quarter correction \(d\); subtract \((q-1)d\) from each CR.
4. Seasonal index \(= (\text{corrected CR}/\text{mean corrected CR})\times 100\).

4. Calculation

Blank working table:

	Q1	Q2	Q3	Q4
Avg link relative
Chain relative
Corrected CR
Seasonal index

Return CR of Q1 \(= (97.09\times 111.36)/100 = 108.12\), so \(d = (108.12-100)/4 = 2.03\); mean corrected CR \(= 112.41\).

	Q1	Q2	Q3	Q4
Avg link relative	97.09	133.38	87.77	95.12
Chain relative	100.00	133.38	117.07	111.36
Corrected CR	100.00	131.35	113.01	105.27
Seasonal index	88.96	116.85	100.54	93.65

5. Result

Link-relative seasonal indices are Q1 = 88.96, Q2 = 116.85, Q3 = 100.54, Q4 = 93.65 (sum 400). All four methods (Experiments 3–6) agree: Q2 is the peak season and Q1 the trough.

Experiment 7 — Simple Index Numbers

1. Problem

Prices in the base year \(p_0\): 5, 8, 10, 12 and in the current year \(p_1\): 6, 10, 14, 15. Compute the simple aggregate and simple average-of-relatives price index.

2. Aim

To construct unweighted (simple) index numbers by the aggregate and the average-of-relatives methods.

3. Formula

Applying it:

1. Simple aggregate: \((\sum p_1/\sum p_0)\times 100\).
2. Average of price relatives: for each item \((p_1/p_0)\times 100\), then take the arithmetic mean.

4. Calculation

Blank working table:

Item	\(p_0\)	\(p_1\)	\((p_1/p_0)\times100\)
1	5	6
2	8	10
3	10	14
4	12	15
Σ

Item	\(p_0\)	\(p_1\)	\((p_1/p_0)\times100\)
1	5	6	120
2	8	10	125
3	10	14	140
4	12	15	125
Σ	35	45	510

Simple aggregate \(= (45/35)\times 100 = 128.57\); average of relatives \(= 510/4 = 127.5\).

5. Result

The price level rose about 28.6 % (aggregate) or 27.5 % (average of relatives) over the base year.

Experiment 8 — Weighted Index Numbers (All Formulae)

1. Problem

Compute Laspeyres, Paasche, Marshall–Edgeworth, Fisher and Bowley price index numbers from:

Item	\(p_0\)	\(q_0\)	\(p_1\)	\(q_1\)
A	4	10	6	12
B	5	15	8	14
C	10	5	12	4

2. Aim

To construct all the standard weighted price index numbers and compare them.

3. Formula

Applying it:

1. Form the four aggregates \(\sum p_0q_0,\; \sum p_1q_0,\; \sum p_0q_1,\; \sum p_1q_1\).
2. Apply each formula.

4. Calculation

Blank working table:

Item	\(p_0q_0\)	\(p_1q_0\)	\(p_0q_1\)	\(p_1q_1\)
A
B
C
Σ

Item	\(p_0q_0\)	\(p_1q_0\)	\(p_0q_1\)	\(p_1q_1\)
A	40	60	48	72
B	75	120	70	112
C	50	60	40	48
Σ	165	240	158	232

• Laspeyres: \(240/165\times 100 = \mathbf{145.45}\)
• Paasche: \(232/158\times 100 = \mathbf{146.84}\)
• Marshall–Edgeworth: \(472/323\times 100 = \mathbf{146.13}\)
• Fisher: \(\sqrt{145.45\times 146.84} = \mathbf{146.14}\)
• Bowley: \((145.45+146.84)/2 = \mathbf{146.15}\)

5. Result

All weighted indices cluster near 146, indicating a price rise of about 46 %. Fisher's index (146.14) is the geometric "ideal" figure.

Experiment 9 — Reversal Tests

1. Problem

Using the aggregates of Experiment 8, verify that Fisher's ideal index satisfies the time-reversal and factor-reversal tests.

2. Aim

To check the two consistency tests: TRT \((P_{01}\times P_{10} = 1)\) and FRT \((P_{01}\times Q_{01} = V_{01})\).

3. Formula

Applying it:

1. Time reversal: show \(P_{01}^F\times P_{10}^F = 1\) (the current-to-base index is the reciprocal).
2. Factor reversal: show \(P_{01}^F\times Q_{01}^F = \sum p_1q_1/\sum p_0q_0\), the value index.

4. Calculation

Blank working table:

Quantity	Value
Fisher price index \(P_F\)
Fisher quantity index \(Q_F\)
\(P_F\times Q_F\)
Value index \(V = \sum p_1q_1/\sum p_0q_0\)

TRT: \(P_{01}^F\times P_{10}^F = \sqrt{\tfrac{240}{165}\cdot\tfrac{232}{158}}\cdot\sqrt{\tfrac{158}{232}\cdot\tfrac{165}{240}} = 1\) ✓.

Quantity	Value
Fisher price index \(P_F\)	1.4614
Fisher quantity index \(Q_F = \sqrt{(158/165)(232/240)}\)	0.9621
\(P_F\times Q_F\)	1.406
Value index \(V = 232/165\)	1.406

5. Result

Fisher's index satisfies both the time-reversal test (\(P_{01}\times P_{10} = 1\)) and the factor-reversal test (\(P_F\times Q_F = V = 1.406\)); hence it is the "ideal" index.

Experiment 10 — Mortality Rates

1. Problem

Compute the crude death rate, age-specific death rates and standardised death rates (direct and indirect) from:

Age	Population \(P_x\)	Deaths \(D_x\)
0–14	20 000	100
15–59	50 000	250
60+	30 000	900
Total	100 000	1 250

2. Aim

To compute CDR, ASDRs and the standardised death rates using a standard population (30 000, 55 000, 15 000) and standard ASDRs (4, 6, 25 per 1 000, standard CDR 8).

3. Formula

Applying it:

1. CDR \(= (\sum D/\sum P)\times 1000\); ASDR of a group \(= (D_x/P_x)\times 1000\).
2. Direct SDR \(= \sum(P_x^{std}\, m_x)/\sum P_x^{std}\).
3. Indirect: expected deaths \(= \sum P_x\, m_x^{std}\); SMR \(=\) observed/expected; indirect SDR \(=\) SMR × standard CDR.

4. Calculation

Blank working table:

Age	ASDR (per 1000)	\(P_x^{std}\)	\(P_x^{std}\times m_x\)
0–14		30 000
15–59		55 000
60+		15 000
Total	—	100 000

CDR \(= (1250/100000)\times 1000 = 12.5\) per 1 000. ASDRs \(= 5,\; 5,\; 30\) per 1 000.

Age	ASDR (per 1000)	\(P_x^{std}\)	\(P_x^{std}\times m_x\)
0–14	5	30 000	150
15–59	5	55 000	275
60+	30	15 000	450
Total	—	100 000	875

Direct SDR \(= 875/100000\times 1000 = 8.75\) per 1 000.

Indirect: expected deaths \(= 20000(0.004)+50000(0.006)+30000(0.025) = 80+300+750 = 1130\); SMR \(= 1250/1130 = 1.106\); indirect SDR \(= 1.106\times 8 = 8.85\) per 1 000.

5. Result

CDR = 12.5 per 1 000; after standardisation the death rate is about 8.75 (direct) or 8.85 (indirect) per 1 000 — the crude rate overstates mortality because this population is older than the standard.

Experiment 11 — Fertility Rates

1. Problem

Total population 1 000 000; live births 18 000; female population aged 15–49 is 250 000. The age-specific fertility rates (per 1 000 women) for the seven 5-year groups 15–19, …, 45–49 are 40, 120, 140, 80, 40, 10, 2. Compute CBR, GFR and TFR.

2. Aim

To compute the crude birth rate, general fertility rate and total fertility rate.

3. Formula

Applying it:

1. CBR \(= (B/P)\times 1000\).
2. GFR \(= (B/W_{15\text{–}49})\times 1000\).
3. TFR \(= (\sum \text{ASFR})\times \text{class width}/1000\).

4. Calculation

Blank working table:

Rate	Value
CBR (per 1000)
GFR (per 1000 women)
\(\sum\)ASFR
TFR (children/woman)

Rate	Value
CBR \(= (18000/1000000)\times 1000\)	18
GFR \(= (18000/250000)\times 1000\)	72
\(\sum\)ASFR \(= 40+120+140+80+40+10+2\)	432
TFR \(= 432\times 5/1000\)	2.16

5. Result

CBR = 18 per 1 000, GFR = 72 per 1 000 women, TFR = 2.16 children per woman — close to the replacement level of about 2.1.

Experiment 12 — Reproduction Rates

1. Problem

From Experiment 11 (TFR = 2.16), with a female-birth proportion of 0.487 and a mean survival probability of 0.92 to the mean age of childbearing, compute the gross and net reproduction rates and Pearl's vital index (deaths 8 000).

2. Aim

To compute GRR, NRR and Pearl's vital index and interpret the replacement level.

3. Formula

Applying it:

1. GRR \(=\) TFR × (proportion of female births).
2. NRR \(=\) GRR × (mean survival probability).
3. Pearl's vital index \(= (B/D)\times 100\).

4. Calculation

Blank working table:

Measure	Value
GRR (daughters/woman)
NRR
Pearl's vital index

Measure	Value
GRR \(= 2.16\times 0.487\)	1.052
NRR \(= 1.052\times 0.92\)	0.968
Pearl's vital index \(= (18000/8000)\times 100\)	225

5. Result

GRR = 1.052 daughters per woman, but NRR = 0.968 < 1, so the population fails to fully replace itself and will decline in the long run despite a vital index of 225.

Experiment 13 — Construction of a Life Table

1. Problem

Given the survivorship column \(l_x\), construct the remaining life-table columns (\(d_x,\; q_x,\; L_x\)) and outline the computation of \(T_x\) and \(e_x^0\).

Age \(x\)	\(l_x\)
0	100 000
1	98 800
5	98 300
10	98 000
15	97 500
20	97 000

2. Aim

To build a life table from the survivorship column and compute the expectation of life \(e_x^0 = T_x/l_x\).

3. Formula

Applying it:

1. \(d_x = l_x - l_{x+n}\) (deaths in the interval).
2. \(q_x = d_x/l_x\) (probability of dying).
3. \(L_x \approx (l_x + l_{x+n})/2\) (central survivors of the interval).
4. \(T_x = \sum_{a\ge x} L_a\) (cumulate from the oldest age up); \(e_x^0 = T_x/l_x\).

4. Calculation

Blank working table:

\(x\)	\(l_x\)	\(d_x\)	\(q_x\)	\(L_x\ (\approx)\)
0	100 000
1	98 800
5	98 300
10	98 000
15	97 500

\(x\)	\(l_x\)	\(d_x\)	\(q_x\)	\(L_x\ (\approx)\)
0	100 000	1 200	0.0120	99 400
1	98 800	500	0.0051	98 550
5	98 300	300	0.0031	98 150
10	98 000	500	0.0051	97 750
15	97 500	500	0.0051	97 250

(\(L_x\) is shown as the central survivorship \((l_x+l_{x+n})/2\); for the wider age intervals it would be multiplied by the interval width \(n\) to obtain person-years.) Cumulating \(L_x\) from the oldest age gives \(T_x\), and \(e_x^0 = T_x/l_x\).

5. Result

The completed columns give the probability of dying \(q_x\) at each age and, through \(T_x\), the expectation of life \(e_x^0 = T_x/l_x\) — the key output of the life table.

Lab Record Format (to be followed for every experiment)

1. 1. Problem — the data given and what is to be computed.
2. 2. Aim — the measure(s) 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 figures, with units and interpretation.