### Chapter 7, Survey Sampling

#### Solution 33

We have $\, \hat d = \hat p_1 - \hat p_2 \,$, and similar to Problem-32, we get $\, \Var(\hat d) = \Var(p_1) + \Var(\hat p_2) = \frac 1 n \frac {N-n} {N-1} (\hat p_1(1-\hat p_1) + \hat p_2(1-\hat p_2))\,$. Since $\, \frac {N-n} {N-1} \approx 1 \,$ and $\, \hat p_1 \,$ and $\, \hat p_2 \,$ are close to $\, \frac 1 2 \,$, it follows that $\, \Var(\hat d) = \frac 1 n(\frac 1 2 \times \frac 1 2 + \frac 1 2 \times \frac 1 2) = \frac 1 {2n} \,$.

Thus $\, \Var(\hat d) = \frac 1 {2n} \,$. Since we need standard error to be less than $\, 0.02 \,$, it follows that we need variance,$\, \Var(\hat d) \,$, to be less than $\, {0.02}^2 = 0.0004 \,$. Thus the minimum amount of sample size $\, n \,$ that we need is $\, \frac 1 {2n} \le 0.0004 \,$, which results in $\, n \ge 1250 \,$.

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