### Chapter 7, Survey Sampling

#### Solution 23

#### (a)

The standard error of an estimated proportion is given by $\, \sigma_{\bar p} = \sqrt{ \frac {p(1-p)} n \left( \frac {N-n} {N-1} \right) }\,$. To maximize it by changing $\, p \,$, we need to maximize the term containing $\, p \,$ and can ignore the other terms as constant.

Thus we shall maximize $\, p(1-p) \,$ by differentiating it:

Equating the first derivative to zero, we get $\, p=\frac 1 2 \,$. Since the first derivative $\, 1-2p \,$ is a decreasing function, it follows that at $\, p=\frac 1 2 \,$, the function $\, p(1-p) \,$ will be maximum.

#### (b)

Corollary B says that $\, Var(\hat p) \,$ is:

Similar to part (a), we can find the value of $\, \hat p \,$ where $\, s^2_{\hat p} \,$ is maximum. Since the steps are similar - skipping the steps, we have $\, \hat p = \frac 1 2 \,$ maximizes $\, s_{\hat p} \,$.

Now, putting this value in $\, s_{\hat p} \,$, gives:

Denoting the above quantity, maximum value of $\, s_{\hat p} \,$ by $\, M_{s} \,$.

Also, putting $\, p = \frac 1 2 \,$ in $\, \sigma_{\bar p} = \sqrt{ \frac {p(1-p)} n \left( \frac {N-n} {N-1} \right) } \,$, as shown in part-(a) will give the maximum value of $\, \sigma_{\hat p} \,$:

Denoting this quantity, maximum value of $\, \sigma_{\hat p} \,$ by $\, M_{\sigma} \,$.

Comparing the quantities, $\, M_{\sigma} \,$ and $\, M_s \,$ and since $\, n < N \,$, it follows that $\, M_{\sigma} \le M_s \,$. Since $\, \sigma_{\hat p} \le M_{\sigma} \,$(from part-(a)), it follows that, $\, \sigma_{\hat p} \le M_s \,$. Since $\, \sigma_{\hat p} \,$ is not associated with a particular value of $\, p \,$, it follows that the result holds for all $\, p \,$.

Note: By conservative estimate, it means we need to prove that the estimate will never over-estimate the quantity in consideration(here it is standard error).

#### (c)

We know that the $\, 100(1-\alpha) \,$% confidence interval for $\, \hat p \,$ is $\, \hat p \pm z(\alpha/2) s_{\hat p} \,$. Now if we use the maximum possible value of $\, s_{\hat p} \,$, it will give the interval that contains $\, p \,$ with atleast $\, 100(1-\alpha) \,$. This is because the maximum value of $\, s_{\hat p} \,$will give the widest interval for the given width $\, 100(1-\alpha) \,$.

Comparing this with the given interval in the problem, $\, \hat p \pm \sqrt{ \frac {N-n} {N(n-1)} } \,$, it follows that $\, z(\alpha/2) = 2 \,$, since the maximum possible value of $\, s_{\hat p} \,$ is as found in part (b) equals $\, \frac 1 2 \sqrt{ \frac {N-n} {N(n-1)} } \,$.

Now we shall compute $\, \alpha/2 \,$ using $\, z(\alpha/2) = 2 \,$. By definition of $\, z \,$, we get $\, \Phi^{-1}(1-\alpha/2) = 2\,$, which gives $\alpha = 0.0456$. Thus the confidence interval is $\, 100(1-\alpha) = 95.44 \,$. Thus it is even slightly better than, $\, 95 \,$, mentioned in the problem.

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