Answer:
about 50 percent of its former width.
Step-by-step explanation:
Let's assume that our parameter of interest is given by 
and in order to construct a confidence interval we can use the following formula:
Where 
is an estimator for the parameter of interest and the margin of error is defined usually if the distribution for the parameter is normal as:
Where 
is a quantile from the normal standard distribution that accumulates 
of the area on each tail of the distribution. And 
represent the standard error for the parameter.
If our parameter of interest is the population proportion the standard of error is given by:
And if our parameter of interest is the sample mean the standard error is given by:
As we can see the standard error for both cases assuming that the other things remain the same are function of n the sample size and we can write this as:
And since the margin of error is a multiple of the standard error we have that 
Now if we find the width for a confidence interval we got this:
![Width = \hat \theta + ME(\hat \theta) -[\hat \theta -ME(\hat \theta)]](https://tex.z-dn.net/?f=%20Width%20%3D%20%5Chat%20%5Ctheta%20%2B%20ME%28%5Chat%20%5Ctheta%29%20-%5B%5Chat%20%5Ctheta%20-ME%28%5Chat%20%5Ctheta%29%5D)
And we can express this as:
And we can define the function 
since as we can see the margin of error and the standard error are function of the inverse square root of n. So then we have this:
The subscript i is in order to say that is with the sample size n
If we increase the sample size from n to 4n now our width is:
The subscript f is in order to say that is the width for the sample size 4n.
So then as we can see the width for the sample size of 4n is the half of the wisth for the width obtained with the sample size of n. So then the best option for this case is:
about 50 percent of its former width.
