Areport states that 46% of home owners have a vegetable garden. how large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to within 4 percentage points with 98% confidence?

Our sample size needs to be 843.

Our margin of error in this case is 4% or 0.04.  We use the formula:



To find the z-score:
Convert 98% to a decimal:  0.98
Subtract from 1:  1-0.98 = 0.02
Divide by 2:  0.02/2 = 0.01
Subtract from 1:  1-0.01 = 0.99

Using a z-table (http://www.z-table.com) we see that the z-score for this value is approximately 2.33.

Using this as well as our proportion and margin of error we have:


Divide both sides by 2.33:


Square both sides:


Multiply both sides by n:


Divide both sides to isolate n: