Chapter 7, Survey Sampling

Solution 23


The standard error of an estimated proportion is given by . To maximize it by changing , we need to maximize the term containing and can ignore the other terms as constant.

Thus we shall maximize by differentiating it:

Equating the first derivative to zero, we get . Since the first derivative is a decreasing function, it follows that at , the function will be maximum.


Corollary B says that is:

Similar to part (a), we can find the value of where is maximum. Since the steps are similar - skipping the steps, we have maximizes .

Now, putting this value in , gives:

Denoting the above quantity, maximum value of by .

Also, putting in , as shown in part-(a) will give the maximum value of :

Denoting this quantity, maximum value of by .

Comparing the quantities, and and since , it follows that . Since (from part-(a)), it follows that, . Since is not associated with a particular value of , it follows that the result holds for all .

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).


We know that the % confidence interval for is . Now if we use the maximum possible value of , it will give the interval that contains with atleast . This is because the maximum value of will give the widest interval for the given width .

Comparing this with the given interval in the problem, , it follows that , since the maximum possible value of is as found in part (b) equals .

Now we shall compute using . By definition of , we get , which gives . Thus the confidence interval is . Thus it is even slightly better than, , mentioned in the problem.

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