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 in each of the computation, evalutating it gives

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 .

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 , we square the sum of last column and divide it by . Thus we get .

Simple Random Sampling:

It was shown in the book that . Thus .

We can compute , which gives .

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 is the sum of and times sum of last column, which gives .

(c)

Population mean we already computed in part-c, . Population Variance, .

(d)

We have .

We have for all .

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 .

To get same variance of from the simple random sampling, we need . Putting value of and (computed in last part-c). We get . Thus using stratified sampling we used elements in the sample while to get same variance we needed elements in simple random sampling.

$$\tag*{$\blacksquare$} $$