# Mathematical Statistics and Data Analysis - Solutions

### Chapter 7, Survey Sampling

#### (a)

$$\, N = \sum_{l=1}^{L} N_l = 2010 \,$$

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.

$$\tag*{\blacksquare}$$