To reduce the margin of error to 1/3 of its original size for a fixed confidence level and population standard deviation, what should the new sample size be?

To understand how to reduce the margin of error for a given confidence level and population standard deviation, it's essential to look at the relationship between sample size and margin of error in the context of confidence intervals.

The margin of error in estimating a population mean is typically represented by the formula:

[ E = z \cdot \frac{\sigma}{\sqrt{n}} ]

where ( E ) is the margin of error, ( z ) is the z-value corresponding to the desired confidence level, ( \sigma ) is the population standard deviation, and ( n ) is the sample size.

If you want to reduce the margin of error to ( \frac{1}{3} ) of its original size, you will need to adjust the sample size accordingly. When the margin of error shrinks, it does so inversely to the square root of the sample size. Specifically, if we denote the original margin of error as ( E ) and the new desired margin of error as ( E' = \frac{E}{3} ), we can set up the following relationship:

[ E' = z \cdot \frac{\sigma}{\sqrt{n'}} ]

Setting these two equations for (