### Chapter 7, Survey Sampling

#### Solution 48

Note: Initially I planned to solve problem-49 instead of 48 as both are similar. In my copy of the book, it seems that the answer mentioned in the book for problem-49 are actually for problem-48 or perhaps while computing the answers data of problem 48 is used.

#### (a)

The estimate for the ratio $\, r \,$ is:

$\, R = \frac {\bar Y} {\bar X} = \frac {\sum_{i=1}^{n} Y_i} {\sum_{i=1}^{n} X_i} \,$. Thus we only need to plugin the given values to get $\, \frac {10000} {320} = 31.25 \,$.

#### (b)

To compute $\, 95\% \,$ confidence interval, first we need to find the standard error, $\, s_R \,$ of our esitmate $\, R \,$. Now we just need to plugin the values in the expression derived in the book for $\, s_R^2 = \frac 1 n \Prn{1-\frac {n-1} {N-1} } \frac 1 {\bar X} \Prn{ R^2 s_x^2 + s_y^2 - 2 R s_{xy} }\,$.

To compute $\, s_x^2 \,$, we use $\, s_x^2 = \frac 1 {n-1} \sum_{i=1}^{n} (X_i - \bar X)^2 \,$ which simplifies to $\, \frac 1 {n-1} \sum_{i=1}^{n} X_i^2 \; - \; n {\bar X}^2 \,$.

Similarly, $\, s_y^2 = \frac 1 {n-1} \sum_{i=1}^{n} Y_i^2 \; - \; n {\bar Y}^2 \,$

and, $\, s_{xy}^2 = \frac 1 {n-1} \sum_{i=1}^{n} X_i Y_i \; - \; n {\bar X}{\bar Y} \,$

Pluging the values we get $\, s_R^2 = 0.69743455 \,$, which gives $\, s_R = 0.835125 \,$.

Thus the $\, 95\% \,$ confidence interval is $\, R \pm 1.96 s_R = 31.25 \pm 1.636845 \,$.

#### (c)

We can estimate the total $\, \tau \,$ using $\, T = N \bar Y \,$.

Thus $\, T = 100,000 \times \frac {10,000} {100} = {10}^7 \,$

To compute confidence interval,

$\, \Var(T) = N^2 \Var(\bar Y) \,$.

To compute variance of the sample mean $\, \bar Y \,$, we use $\, \Var(\bar Y) = \frac {s_y^2} n \Prn{1 - \frac n N} \,$, where $\, s_y^2 \,$, variance of the sample, is already computed in part-b. This gives $\, \Var(\bar Y) = 10.1010 \,$(ignoring the finite population correction).

Thus we get $\, \Var(T) = N^2 \Var(\bar Y) = 101, 010, 000 \,$. Thus $\, s_T = \sqrt{101,010, 000} = 317,820.704 \,$. Thus the $\, 90\% \,$ condidence interval is $\, 10^7 \pm 1.645 \times 317,820.704 = {10}^7 \pm 522,815.05808 \,$.

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