Calculating Confidence Interval How would one go solving for this?
You are told that a 95% CI for expected lead content when traffic flow is 15, based on a sample of n = 12 observations, is (468.1, 591.7). Calculate a CI with confidence level 99% for expected lead content when traffic flow is 15. (Round your answers to one decimal place.)
Thanks
 A: A $c$% confidence interval for data $x_1,\ldots,x_n$ is defined as
$$ (\bar{x} - q_c\frac{s}{\sqrt{n}}, \bar{x} + q_c\frac{s}{\sqrt{n}})$$
where


*

*$n$ is the number of samples,

*$s$ is the sample standard deviation, and

*$q_c$ is the appropriate quantile, the $(100-(100-c)/2)$th percentile of an appropriate distribution (usually the normal or t distribution, see below). In the case of $c=95\%$, $q_c$ is the $100-(5/2)=97.5$th percentile.
Assuming your data is normally distributed, you can use the t distribution with $n-1$ degrees of freedom for the quantiles: you can look the quantiles up in tables or use a software package.
Defining the standard error as $\sigma_M=s/\sqrt{n}$, the confidence interval can be written
$$ \;(\bar{x} - q_c\sigma_M, \bar{x} + q_c\sigma_M)$$
For further background search for "t confidence intervals". Khan academy is good and wikipedia is worth a look too.
That's the theory. To solve your problem what you need to do is:


*

*Figure out what $q_c$ is for the given confidence interval with $c=95\%$ and $n=12$, looking up using the appropriate t distribution.

*Given the width of the confidence interval, and the distance either side of the mean to the boundaries, solve for $\sigma_M$ and $\bar{x}$.

*Figure out what $q_c$ is for the required confidence interval with $c=99\%$ in a similar way to what you did in step 1.

*Use your new $q_c$ and $\sigma_M$, $\bar{x}$ to calculate the new confidence interval.
