Quiz #3

Question:
Calculate the margin of error for a 95% confidence interval where:
the sample size is 64
the sample mean is 350
the population standard deviation is 64

Answer:
The formula for margin of error = x-bar +/- (z-alpha/2)(stddev/sqrt(n))

We find z-alpha/2 for a 95% confidence interval = 1.96 from our class notes. Manually, you work it out as:
alpha = 1-0.95 = 0.05
alpha/2 = 0.05/2 = 0.025
Since the z-scored table gives us the area from 0 to z, we need to use the "complement" of 0.025 which is 0.5-0.025 = 0.475
Do a reverse lookup in a z-score table for 0.475 and we find that it corresponds to a z-score of 1.96.

So,
margin of error = 350 +/- 1.96(64/(sqrt(64)) = 350 +/- (1.96)(8)
= 350 +/- 15.68*

* technically, our textbook refers to this as a "95% confidence interval for the mean". Although our text doesn't use the term "margin of error" explicitly, it seems to me that the margin of error is just the 15.68 term. I've looked at some web sites that define the terms, but there's still some ambiguity - is the margin the range of values or just the half-width of the range? Perhaps that's why our text avoids the issue by not using the term.