# Help with AP statistics!!?

I need help with this problem in AP statistics.

A telephone survey of 1000 adults was taken shortly after the United States began bombing Iraq. If 832 voiced their support for this action, with what confidence can it be asserted that 83.2% +/- 3% of the adult U.S population supported the decision to go to war?

How do you do this on the ti 83/84 calculator?

Thank You soo much!!

Relevance

You can't, really. The problem relies on conceptual knowledge of confidence intervals, and asks you to determine the confidence level. While the TI-83/84/Plus/SE calculators can give you a confidence interval given proportions, a confidence level and sample size, it cannot give you individual components.

The margin of error for a z-confidence interval is ME = z*√(p(1 − p)/n), where z* is the critical z-score (which can be converted to a level of confidence), p is the probability, and n is the sample size. In this problem, ME is 3%. You should just be able to plug in the values and solve for z*, which you can then use to determine the confidence level.

ME = z*√(p(1 − p)/n)

.03 = z*√(.832(1 − .832)/n)

z* = .03/√(.832(1 − .832)/n)

z* ≈ 2.54

which corresponds to about a 99% confidence level.