Calculating level of confidence from an interval estimate I've recently been having trouble with this question on calculating confidence levels and am just wondering if anyone had any ideas on how to solve it.
"A market research company conducted a telephone survey of 255 Chicago households to determine the proportion of households seeking to purchase a new car in the next two years. An interval estimate of 0.151 to 0.289 was calculated. Determine the level of confidence."
 A: If it can be solved they are assuming something like (a) the interval is symmetric so you know the central estimate (b) they are using the normal approximation which involves $p$, $q = 1 - p$' and $n$ all of which you know. Of course if that is not what they are assuming it may be unsolvable.
A: This question appears to be asking for confidence limits (CL) from confidence interval (CI). The problem actually has no solution without making a lot of unstated assumptions. A confidence interval is sometimes the 95% range of standard error of the mean for a normal distribution. Standard error is $SE=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation, and $n$ is the sample size. Then 95%, and only 95% CI is $\mu +/- 1.96 SE$. Thus, $(\mu + 1.96 SE) +(\mu - 1.96 SE)=2\mu$ and $(\mu + 1.96 SE) -(\mu - 1.96 SE)=3.92SE$. From this find the mean $\mu$ and the standard error $SE$. Then $\sigma=\sqrt{n}SE$, and the 95% and only the 95% CL are $\mu+/-1.96\sigma$ for a normal distribution.
