The general formula for the margin of error is e = zσ/√n. We know that e = 0.8, and that σ = 4.6.
We still need the z-statistic for a 93% confidence level. This can be obtained by a standard normal table or by technology. We need the area to be symmetrical around the mean, so it needs to have an area of half of .93 on each side. So to convert to a cumulative area, it needs to end at 0.465 (half of 0.93), which makes a total cumulative area of 0.965 (0.5 + 0.465).
From here, we can use either invNorm on the TI-84 or by looking up 0.965 on the normal table. Both ways we get about 1.81.
Now we have enough to plug in what we know and solve for what we don't know.
e = zσ/√n
0.8 = (1.81 · 4.6)/√n
0.8√n = 8.326
√n = 10.4075
n ≈ 108.3
This figure will always be rounded up, because it is a minimum sample size to provide the desired result. So if we rounded down, the sample size would be too small. So our final result would be 109.
