# Creating a CI for the mean from an approximation of x-bar

The "classical" $100(1-\alpha)\%$-confidence interval starts from the Student's t statistic $$t=\frac{\bar{x}-\mu}{s/\sqrt{n}}.$$ Then, one obtains the desired result, e.g. $\bar{x}\pm t_{1-\alpha/2;n-1}\frac{s}{\sqrt{n}}$ by simple algebraic manipulation.

Here's my problem. I am using an approximation for $\bar{x}$, which leads me to the following modified $t$ statistic: $$t'=\frac{a(\bar{x}-\mu)^2+(\bar{x}-\mu)+b}{s/\sqrt{n}}$$ where $a$ and $b$ are parameters depending on sample size, and some central moments. I would like to derive a CI for $\mu$ using $t'$, but the squared term gives me a little trouble.

Is it as simple as treating the enumerator as a quadratic equation in $\bar{x}-\mu$, solving it and moving terms around? That is, let $$a(\bar{x}-\mu)^2+(\bar{x}-\mu)+b=t' \frac{s}{\sqrt{n}},$$ find the roots $r_1, r_2$ and then write, e.g., $\bar{x}-\mu=r_1$, yielding $\mu = \bar{x}-r_1$ and thus the CI is $(\bar{x}-r_1) \pm t' (s n^{-1/2})$ ? If both roots are real, how do I decide on which to pick?

If your statistic actually has a t-distribution, then you can identify the parts of the $\mu$-line corresponding to insignificant t-statistics (ones that correspond to the usual interval) by evaluating $t'$ as a function over $\mu$ and seeing where $|t'|<t_{crit}$. This may be a single $\mu$-interval or may consist of two separate ones, or sometimes it might give no interval at all.

This is no different from generating the interval in the ordinary $t$ - the region of $\mu$-values that would result in acceptance of the null at significance level $\alpha$ are those regions which would be inside a $100(1-\alpha)\%$ confidence region.

I mention it because it's much easier to see how to answer your question about two real roots.

Strictly speaking what we're doing is taking a pivotal quantity Q, and finding an interval for the pivotal quantity and from that backing out an interval for $\mu$. It just happens that in this case a suitable pivotal quantity happens to look exactly like the test statistic in a hypothesis test ... (though the fact that the same quantity based on ML estimation and arranged to give a pivot turns up in both isn't really coincidence). As a result, we can back the interval out the the test statistic, just as you did in your first paragraph.

The values of Q that result in it being inside an interval for it then correspond to values of $\mu$ inside our confidence region, even when that's complicated.

[While - despite initial appearances - I didn't actually suggest it as a principle here, more generally you can use many test statistics to produce acceptance regions for parameter values (including multiple parameters); e.g you can produce acceptance regions for a pair of location shifts in the Kruskal-Wallis test. These regions are sometimes called consonance intervals (or regions) rather than confidence intervals, though they are in many cases the same.]

• 1) How can $\text{abs}(|t'|-t_{crit})$ be negative? 2) If I understand correctly, it means that the procedure only makes sense in a hypothesis test setting, where I have some null hypothesis about $\mu$? – baudolino Mar 21 '13 at 3:37
• You're right, serves me right for doing it tired. I munged together two different things I was thinking about. Fixing it now. – Glen_b Mar 21 '13 at 4:40
• Makes sense now. Basically the discriminant of the quadratic tells me how many intervals I should expect to find where $|t'| < t_{crit}$. In regards to your comment about ensuring that $t'$ follows a Student distribution, one could always partition the real axis along several critical values of interest, simulate from the parent distribution, locate in which interval the computed $t'$ falls, and then do a simple $\chi^2$ with the observed numbers in each bin vs. the expected numbers assuming Student (e.g. $5\%$ between $(-\infty, t_{.05; n-1}]$). – baudolino Mar 22 '13 at 1:30