List of Practical Experiments (Official Syllabus)

1. Show that the sample mean is an unbiased estimator of the population mean in SRSWOR and find its variance.
2. Show that the sample mean square is an unbiased estimator of the population mean square in SRSWOR.
3. Show that the sample mean is an unbiased estimator of the population mean in SRSWR and find its variance.
4. Compare means and variances between SRSWR and SRSWOR.
5. Allocate sample sizes among strata using proportional and optimum (Neyman) allocations.
6. Compare the precision of proportional and optimum allocations with SRSWOR and compute the gain in efficiency.
7. Systematic sampling with \(N = nk\) — compare its precision with Stratified and SRSWOR.

Experiment 1 — Unbiasedness of \(\bar y\) under SRSWOR

1. Problem

A finite population has four units \(Y_1 = 2,\; Y_2 = 4,\; Y_3 = 6,\; Y_4 = 8\) (\(N = 4\), population mean \(\bar Y = 5\)). Draw all possible simple random samples without replacement (SRSWOR) of size \(n = 2\) and verify that the sample mean is unbiased for \(\bar Y\); find its sampling variance.

2. Aim

To verify empirically that \(E(\bar y) = \bar Y\) under SRSWOR and that \(\text{Var}(\bar y) = \dfrac{N-n}{Nn}\,S^2\).

3. Formula

Applying it:

1. List all \(\binom{N}{n}\) possible samples of size \(n\).
2. Compute the mean \(\bar y\) of each sample.
3. Average the sample means to obtain \(E(\bar y)\) and compare with \(\bar Y\).
4. Compute the sampling variance directly as \(\dfrac{1}{\binom{N}{n}}\sum(\bar y - \bar Y)^2\) and check it against the formula.

4. Calculation

Blank working table (fill the mean of each of the \(\binom{4}{2}=6\) samples):

Sample	Units	\(\bar y\)	\((\bar y - \bar Y)^2\)
1
2
3
4
5
6
Total

Sample	Units	\(\bar y\)	\((\bar y - \bar Y)^2\)
1	{2, 4}	3	4
2	{2, 6}	4	1
3	{2, 8}	5	0
4	{4, 6}	5	0
5	{4, 8}	6	1
6	{6, 8}	7	4
Total		30	10

\(E(\bar y) = 30/6 = 5 = \bar Y\).

Direct variance \(= 10/6 = 1.667\).

Formula check: \(S^2 = \dfrac{(2-5)^2+(4-5)^2+(6-5)^2+(8-5)^2}{4-1} = \dfrac{20}{3} = 6.667\), so \(\text{Var}(\bar y) = \dfrac{4-2}{4\cdot 2}\cdot\dfrac{20}{3} = \dfrac{40}{24} = 1.667\).

5. Result

\(E(\bar y) = 5 = \bar Y\), so the sample mean is an unbiased estimator of the population mean under SRSWOR, and its variance is \(1.667\), matching the theoretical formula.

Experiment 2 — Unbiasedness of \(s^2\) under SRSWOR

1. Problem

For the same population (\(2, 4, 6, 8\)) and all six SRSWOR samples of size \(n = 2\), verify that the sample mean square \(s^2\) is an unbiased estimator of the population mean square \(S^2 = 6.667\).

2. Aim

To verify that \(E(s^2) = S^2\) under SRSWOR.

3. Formula

Applying it:

1. For each sample compute \(s^2 = \dfrac{1}{n-1}\sum (y_i - \bar y)^2\).
2. Average the six values of \(s^2\) and compare with \(S^2\).

4. Calculation

Blank working table:

Sample	\(\bar y\)	\(s^2\)
{2, 4}
{2, 6}
{2, 8}
{4, 6}
{4, 8}
{6, 8}
Total

Sample	\(\bar y\)	\(s^2\)
{2, 4}	3	2
{2, 6}	4	8
{2, 8}	5	18
{4, 6}	5	2
{4, 8}	6	8
{6, 8}	7	2
Total	—	40

\(E(s^2) = 40/6 = 6.667\), and \(S^2 = 20/3 = 6.667\).

5. Result

\(E(s^2) = 6.667 = S^2\), so the sample mean square is an unbiased estimator of the population mean square under SRSWOR.

Experiment 3 — Unbiasedness of \(\bar y\) under SRSWR

1. Problem

For the population \(2, 4, 6, 8\), draw all ordered samples of size \(n = 2\) with replacement (SRSWR) and verify that \(\bar y\) is unbiased for \(\bar Y = 5\); find its variance.

2. Aim

To verify that \(E(\bar y) = \bar Y\) under SRSWR and that \(\text{Var}(\bar y) = \sigma^2/n\).

3. Formula

Applying it:

1. There are \(N^n = 4^2 = 16\) equally likely ordered samples; each unit is equally represented, so the mean of all sample means equals \(\bar Y\).
2. Compute the population variance \(\sigma^2 = \dfrac{1}{N}\sum(Y_i - \bar Y)^2\).
3. Obtain the sampling variance from \(\sigma^2/n\).

4. Calculation

Blank working table:

Quantity	Value
Number of ordered samples \(N^n\)
\(E(\bar y)\)
\(\sigma^2\)
\(\text{Var}_{WR}(\bar y) = \sigma^2/n\)

Quantity	Value
Number of ordered samples \(N^n\)	16
\(E(\bar y)\)	5
\(\sigma^2 = 20/4\)	5
\(\text{Var}_{WR}(\bar y) = 5/2\)	2.5

By the symmetry of replacement each unit appears with equal frequency across the 16 samples, so \(E(\bar y) = \bar Y = 5\).

5. Result

\(E(\bar y) = 5 = \bar Y\) (unbiased) and \(\text{Var}_{WR}(\bar y) = 2.5\). Since \(2.5 > 1.667\), SRSWOR is more precise than SRSWR for a fixed \(n\).

Experiment 4 — Comparison of SRSWR and SRSWOR

1. Problem

Using the results of Experiments 1 and 3, compare the mean and variance of \(\bar y\) under SRSWR and SRSWOR and compute the relative efficiency of SRSWOR.

2. Aim

To show that both designs give unbiased means but SRSWOR has the smaller variance, by the factor \((N-n)/(N-1)\).

3. Formula

Applying it:

1. Tabulate \(E(\bar y)\), \(\text{Var}(\bar y)\) and the finite-population-correction (FPC) factor for both designs.
2. Compute the relative efficiency \(\text{RE} = \text{Var}_{WR}/\text{Var}_{WOR}\).

4. Calculation

Blank working table:

	SRSWR	SRSWOR
\(E(\bar y)\)
\(\text{Var}(\bar y)\)
FPC factor
Relative efficiency

	SRSWR	SRSWOR
\(E(\bar y)\)	5	5
\(\text{Var}(\bar y)\)	2.500	1.667
FPC factor	1	\((N-n)/N = 0.5\)
Relative efficiency	1.00	1.50

\(\text{RE} = 2.5/1.667 = 1.50\).

5. Result

Both designs are unbiased. SRSWOR has the smaller variance and is 50 % more efficient than SRSWR for this population.

Experiment 5 — Allocation of Sample Sizes (Proportional & Optimum)

1. Problem

A population is divided into three strata with \(N_1 = 200,\; N_2 = 300,\; N_3 = 500\) and within-stratum standard deviations \(S_1 = 4,\; S_2 = 6,\; S_3 = 8\). A total sample of \(n = 50\) is to be drawn. Allocate the sample among the strata by proportional and by optimum (Neyman) allocation.

2. Aim

To compute stratum sample sizes under proportional allocation \((n_h = nN_h/N)\) and Neyman allocation \((n_h = nN_hS_h/\sum N_kS_k)\).

3. Formula

Applying it:

1. Find the total population size \(N = \sum N_h\).
2. Proportional allocation: \(n_h = n\,N_h/N\).
3. Neyman allocation: form \(N_hS_h\), sum them, then \(n_h = n\,N_hS_h/\sum N_kS_k\).
4. Round each \(n_h\) so that the totals equal \(n = 50\).

4. Calculation

Blank working table:

Stratum	\(N_h\)	\(S_h\)	\(N_h S_h\)	\(n_h\) (prop.)	\(n_h\) (Neyman)
1
2
3
Total

Stratum	\(N_h\)	\(S_h\)	\(N_h S_h\)	\(n_h\) (prop.)	\(n_h\) (Neyman)
1	200	4	800	10	6
2	300	6	1800	15	14
3	500	8	4000	25	30
Total	1000	—	6600	50	50

Proportional: \(n_1 = 50\cdot 200/1000 = 10,\; n_2 = 15,\; n_3 = 25\).

Neyman: \(n_1 = 50\cdot 800/6600 = 6.06 \approx 6,\; n_2 = 50\cdot 1800/6600 = 13.64 \approx 14,\; n_3 = 50\cdot 4000/6600 = 30.30 \approx 30\).

5. Result

Under proportional allocation the sample is \((10, 15, 25)\); under Neyman allocation it is \((6, 14, 30)\). The most variable stratum (stratum 3, largest \(S_h\)) receives a larger share under Neyman allocation.

Experiment 6 — Precision Comparison & Gain in Efficiency

1. Problem

For the strata of Experiment 5 with weights \(W_1 = 0.2,\; W_2 = 0.3,\; W_3 = 0.5\) and variances \(S_1^2 = 16,\; S_2^2 = 36,\; S_3^2 = 64\), compare the variance of the estimated mean under proportional and optimum allocation with SRSWOR (\(n = 50\)) and find the gain in efficiency.

2. Aim

To compute \(\text{Var}_{prop}\), \(\text{Var}_{opt}\) and \(\text{Var}_{SRS}\), and the percentage gain of stratified over SRSWOR.

3. Formula

Applying it:

1. Form \(\sum W_h S_h^2\) and \(\sum W_h S_h\).
2. Ignoring the FPC (large \(N\)), compute \(\text{Var}_{prop} = \frac{1}{n}\sum W_h S_h^2\) and \(\text{Var}_{opt} = \frac{1}{n}\left(\sum W_h S_h\right)^2\).
3. Take \(\text{Var}_{SRS} = S^2/n\) with \(S^2 \approx 50\) for this population.
4. Compute the gain \(= \text{Var}_{SRS}/\text{Var}_{strat} - 1\).

4. Calculation

Blank working table:

Stratum	\(W_h\)	\(S_h\)	\(S_h^2\)	\(W_h S_h\)	\(W_h S_h^2\)
1
2
3
Total		—	—

Stratum	\(W_h\)	\(S_h\)	\(S_h^2\)	\(W_h S_h\)	\(W_h S_h^2\)
1	0.2	4	16	0.8	3.2
2	0.3	6	36	1.8	10.8
3	0.5	8	64	4.0	32.0
Total	1.0	—	—	6.6	46.0

\(\text{Var}_{prop} = 46/50 = 0.920\).

\(\text{Var}_{opt} = 6.6^2/50 = 43.56/50 = 0.871\).

\(\text{Var}_{SRS} = 50/50 = 1.000\) (taking \(S^2 \approx 50\)).

Gain: proportional \(= 1.000/0.920 - 1 = 8.7\,\%\); optimum \(= 1.000/0.871 - 1 = 14.8\,\%\).

5. Result

The precision ordering is \(\text{Var}_{opt}(0.871) < \text{Var}_{prop}(0.920) < \text{Var}_{SRS}(1.000)\). Stratification gives a gain of about 8.7 % (proportional) and 14.8 % (optimum) over SRSWOR.

Experiment 7 — Systematic Sampling Comparison

1. Problem

A population of \(N = 16\) units (arranged in order) has the values 2, 3, 5, 7, 9, 10, 12, 13, 15, 17, 18, 20, 22, 24, 25, 27. Draw a linear systematic sample of size \(n = 4\) (so \(k = N/n = 4\)) and compare its precision with stratified sampling (one unit per group of \(k\)) and SRSWOR.

2. Aim

To obtain all \(k\) possible systematic samples, verify unbiasedness, and compare \(\text{Var}_{sys}\), \(\text{Var}_{stratified}\) and \(\text{Var}_{SRSWOR}\).

3. Formula

Applying it:

1. With \(k = 4\), the \(r\)-th systematic sample \((r = 1,\dots,4)\) contains units \(r, r+k, r+2k, r+3k\); compute its mean \(\bar y_r\).
2. Average the four means to get \(E(\bar y_{sys})\) and compare with \(\bar Y\).
3. Systematic variance: \(\text{Var}_{sys} = \frac{1}{k}\sum_r(\bar y_r - \bar Y)^2\).
4. SRSWOR variance: \(S^2 = \frac{1}{N-1}\sum(Y_i-\bar Y)^2\), \(\text{Var}_{SRSWOR} = \frac{N-n}{Nn}S^2\).
5. Stratified variance: treat each consecutive group of \(k\) units as a stratum and draw one unit per stratum; \(\text{Var}_{st} = \sum W_h^2\frac{N_h-1}{N_h}S_h^2\) with \(N_h = k,\; W_h = 1/n\).

4. Calculation

Blank working table (systematic samples):

\(r\)	Units \((r, r+k, r+2k, r+3k)\)	\(\bar y_r\)	\((\bar y_r - \bar Y)^2\)
1
2
3
4
Total

Population mean \(\bar Y = 229/16 = 14.3125\).

\(r\)	Units	\(\bar y_r\)	\((\bar y_r - \bar Y)^2\)
1	{2, 9, 15, 22}	12.00	5.348
2	{3, 10, 17, 24}	13.50	0.660
3	{5, 12, 18, 25}	15.00	0.473
4	{7, 13, 20, 27}	16.75	5.941
Total		57.25	12.422

\(E(\bar y_{sys}) = 57.25/4 = 14.3125 = \bar Y\) (unbiased).

\(\text{Var}_{sys} = 12.422/4 = 3.11\).

SRSWOR: \(\sum(Y_i-\bar Y)^2 = 955.44\), so \(S^2 = 955.44/15 = 63.70\) and \(\text{Var}_{SRSWOR} = \dfrac{16-4}{16\cdot 4}\cdot 63.70 = \dfrac{12}{64}\cdot 63.70 = 11.94\).

Stratified (strata {2,3,5,7}, {9,10,12,13}, {15,17,18,20}, {22,24,25,27}; \(S_h^2 = 4.917, 3.333, 4.333, 4.333\)): \(\text{Var}_{st} = (0.25)^2\cdot\tfrac{3}{4}\,(4.917+3.333+4.333+4.333) = 0.0625\cdot 0.75\cdot 16.917 = 0.79\).

5. Result

All three designs are unbiased. The precision ordering is \(\text{Var}_{stratified}(0.79) < \text{Var}_{sys}(3.11) < \text{Var}_{SRSWOR}(11.94)\). For this ordered, near-linear population, stratified sampling is the most precise and systematic sampling is far better than SRSWOR.

Lab Record Format (to be followed for every experiment)

1. 1. Problem — the population/data given and what is to be verified or computed.
2. 2. Aim — the property or estimate the experiment demonstrates.
3. 3. Formula — the estimating formula, then the numbered steps that apply it.
4. 4. Calculation — the filled table with the arithmetic worked through.
5. 5. Result — the final estimate/variance and the conclusion drawn.