Chapter 7, Survey Sampling
Solution 53
(a)
Proportional allocation:
| Farm Size | \( N_l \) | \( n_l = n \frac {N_l} N \) |
|---|---|---|
| 0-40 | 394 | \( 19.601 \approx 20 \) |
| 41-80 | 461 | \( 22.935 \approx 23 \) |
| 81-100 | 391 | \( 19.453 \approx 19 \) |
| 121-160 | 334 | \( 16.617 \approx 17 \) |
| 161-200 | 169 | \( 8.407 \approx 8 \) |
| 201-240 | 113 | \( 5.621 \approx 6 \) |
| 241 + | 148 | \( 7.363 \approx 7 \) |
Optimal allocation:
Since we need $\, \sum_{l=1}^{L} W_l \sigma_l \,$ in each of the computation, evalutating it gives $\, 17.017758 \,$
| Farm Size | \( N_l \) | \( \sigma_l \) | \( W_l = \frac {N_l} N \) | \( \sigma_l W_l \) | \( n_l = n \frac {W_l \sigma_l} {\sum_{l=1}^{L} W_l \sigma_l } \) |
|---|---|---|---|---|---|
| 0-40 | 394 | 8.3 | 0.19601 | 1.626883 | \( 09.5599 \approx 10 \) |
| 41-80 | 461 | 13.3 | 0.22935 | 3.050355 | \( 17.9245 \approx 18 \) |
| 81-100 | 391 | 15.1 | 0.19453 | 2.937403 | \( 17.2608 \approx 17 \) |
| 121-160 | 334 | 19.8 | 0.16617 | 3.290166 | \( 19.3337 \approx 19 \) |
| 161-200 | 169 | 24.5 | 0.08407 | 2.059715 | \( 12.1033 \approx 12 \) |
| 201-240 | 113 | 26.0 | 0.05621 | 1.461460 | \( 08.5878 \approx 9 \) |
| 241 + | 148 | 35.2 | 0.07363 | 2.591776 | \( 15.2298 \approx 15 \) |
(b)
Proportional allocation:
| Farm Size | \( N_l \) | \( \sigma_l \) | \( W_l = \frac {N_l} N \) | \( W_l \sigma_l^2 \) |
|---|---|---|---|---|
| 0-40 | 394 | 8.3 | 0.19601 | 13.503 |
| 41-80 | 461 | 13.3 | 0.22935 | 40.569 |
| 81-100 | 391 | 15.1 | 0.19453 | 44.355 |
| 121-160 | 334 | 19.8 | 0.16617 | 65.145 |
| 161-200 | 169 | 24.5 | 0.08407 | 50.463 |
| 201-240 | 113 | 26.0 | 0.05621 | 37.997 |
| 241 + | 148 | 35.2 | 0.07363 | 91.230 |
Thus $\, \Var(\bar X_{SP}) = \frac 1 n \sum_{l=1}^{L} W_l \sigma_l^2 = \frac {343.262} {100} \approx 3.433 \,$.
Optimal allocation:
| Farm Size | \( N_l \) | \( \sigma_l \) | \( W_l = \frac {N_l} N \) | \( \sigma_l W_l \) |
|---|---|---|---|---|
| 0-40 | 394 | 8.3 | 0.19601 | 1.626883 |
| 41-80 | 461 | 13.3 | 0.22935 | 3.050355 |
| 81-100 | 391 | 15.1 | 0.19453 | 2.937403 |
| 121-160 | 334 | 19.8 | 0.16617 | 3.290166 |
| 161-200 | 169 | 24.5 | 0.08407 | 2.059715 |
| 201-240 | 113 | 26.0 | 0.05621 | 1.461460 |
| 241 + | 148 | 35.2 | 0.07363 | 2.591776 |
Thus to compute $\, \Var(\bar X_{SO} = \frac 1 n {\Prn{\sum_{l=1}^{L} W_l \sigma_l} }^2 \,$, we square the sum of last column and divide it by $\, n \,$. Thus we get $\, 2.896 \,$.
Simple Random Sampling:
It was shown in the book that $\, \Var(\bar X) - \Var(\bar X_{SP} = \frac 1 n \sum_{l=1}^{L} W_l (\mu_l - \mu)^2 \,$. Thus $\, \Var(\bar X) = \Var(\bar X_{SP}) + \sum_{l=1}^{L} W_l (\mu_l - \mu)^2 \,$.
We can compute $\, \mu = \frac 1 N \sum_{l=1}^{L} N_l \mu_l \,$, which gives $\, \mu = 26.3108 \,$.
| Farm Size | \( N_l \) | \( \mu_l \) | \( W_l = \frac {N_l} N \) | \( W_l (\mu_l - \mu)^2 \) |
|---|---|---|---|---|
| 0-40 | 394 | 5.4 | 0.19601 | 85.707 |
| 41-80 | 461 | 16.3 | 0.22935 | 22.984 |
| 81-100 | 391 | 24.3 | 0.19453 | 0.786 |
| 121-160 | 334 | 34.5 | 0.16617 | 11.144 |
| 161-200 | 169 | 42.1 | 0.08407 | 20.958 |
| 201-240 | 113 | 50.1 | 0.05621 | 31.811 |
| 241 + | 148 | 63.8 | 0.07363 | 103.482 |
Thus $\, \Var(\bar X) \,$ is the sum of $\, \Var(\bar X_{SP}) \,$ and $\, \frac 1 n \,$ times sum of last column, which gives $\, \Var(\bar X) = 6.202 \,$.
(c)
Population mean we already computed in part-c, $\, \mu = 26.3108 \,$. Population Variance, $\, \sigma^2 = n \Var(\bar X) = 100 \times 6.202 = 620.2 \,$.
(d)
We have $\, \Var(\bar X_s) = \sum^{L}_{l=1} W_l^2 \frac 1 {n_l} \Prn{ 1 - \frac {n_l-1} {N_l-1} } \sigma_l^2 \,$.
We have $\, n_l = 10 \,$ for all $\, l \,$.
| Farm Size | \( N_l \) | \( \sigma_l \) | \( \sigma^2_l \) | \( W_l = \frac {N_l} N \) | \( \frac {W_l^2} {n_l} \) | \( \Prn{1 - \frac {n_l-1} {N_l-1} } \) | \( W_l^2 \frac 1 {n_l} \Prn{ 1 - \frac {n_l-1} {N_l-1} } \sigma_l^2 \) |
|---|---|---|---|---|---|---|---|
| 0-40 | 394 | 8.3 | 68.89 | 0.19601 | 0.00384 | 0.977 | 0.2584 |
| 41-80 | 461 | 13.3 | 176.89 | 0.22935 | 0.00526 | 0.980 | 0.9118 |
| 81-100 | 391 | 15.1 | 228.01 | 0.19453 | 0.00378 | 0.977 | 0.8420 |
| 121-160 | 334 | 19.8 | 392.04 | 0.16617 | 0.00276 | 0.972 | 1.0517 |
| 161-200 | 169 | 24.5 | 600.5 | 0.08407 | 0.00070 | 0.946 | 3.9765 |
| 201-240 | 113 | 26.0 | 676.0 | 0.05621 | 0.00031 | 0.920 | 0.1927 |
| 241 + | 148 | 35.2 | 1239.04 | 0.07363 | 0.00054 | 0.939 | 0.6282 |
Thus $\, \Var(\bar X_s) = 7.8614 \,$.
To get same variance of $\, 7.8614 \,$ from the simple random sampling, we need $\, \Var(\bar X) = \frac {\sigma^2} n \,$. Putting value of $\, \Var(\bar X) \,$ and $\, \sigma^2 = 620.2\,$(computed in last part-c). We get $\, n \approx 79 \,$. Thus using stratified sampling we used $\, 10*7 = 70 \,$ elements in the sample while to get same variance we needed $\, 79 \,$ elements in simple random sampling.
$$\tag*{$\blacksquare$} $$