What does it mean that “point estimation resulted in $\hat{\mu} = 12.5 \pm 1.0$”?

I'm trying to figure out what the following statement means:

As a result of point estimation of the mean value of a certain quantity, the value $$\hat{\mu} = 12.5 \pm 1.0 \ [u]$$ was obtained. Assume $n=36$ measurements, each with the same precision.

I know that for a iid normal sample, the sample mean $\hat{\mu} \equiv \bar{X}$ (which is a random variable) has a $N(\mu, \sigma^2/n)$ distribution. Likewise, the sample variance $\hat{\sigma}^2 \equiv s^2$ is such that $(n-1) s^2/\sigma^2$ has a $\chi^2 _{n-1}$ distribution.

I'm confused because I thought "point estimation" referred to specifying one number, in this case $12.5$. So I'm not sure what the $1.0$ in this case - is it $s/ \sqrt{n}$, the standard error of the mean (which is the estimate for the standard deviation of the random variable $\bar{X}$ )? What is the convention used here?

As a result of point estimation of the variance of a certain quantity, the value $$\hat{\sigma} = 12.3 \pm 4.1 \ [u^2]$$ was obtained. Assume $n=36$ measurements, each with the same precision.

In this case, does the $\pm$ part refer to the estimate of the standard deviation of the random variable $s^2$, obtained from the $\chi^2$ distribution, which is $s^2 \sqrt{2/n-1}$ ?

• Can you provide the reference for the statements? – Thomas Jan 26 '18 at 13:00
• @Thomas Unfortunately there isn't any, it's just a problem from a problem set – Spine Feast Jan 26 '18 at 13:23
• You ought to consult the underlying authority, then: either the teacher or the textbook. – whuber Jan 26 '18 at 14:10
• @whuber So I take it this isn't a standard way of writing things? – Spine Feast Jan 26 '18 at 15:16
• It isn't nonstandard, either. It's a little idiosyncratic. It's also ambiguous, because the meaning of the numbers after the "$\pm$" isn't clear: would those be standard errors or confidence limits (and if they are the latter, with what confidence level)? The ambiguity might be resolved within the context; for instance, maybe your textbook has previously explained what these numbers mean. That's why you need to go back to it and why we can only guess. – whuber Jan 26 '18 at 16:37

From the prologue of Lehmann and Casella's Theory of Point Estimation:

Point estimation is one of the most common forms of statistical inference. One measures a physical quantity in order to estimate its value; surveys are conducted to estimate the proportion of voters favoring a candidate or viewers watching a television program; agricultural experiments are carried out to estimate the effect of a new fertilizer, and the clinical experiments to estimate the improved life expectancy or cure rate resulting from a medical treatment....

I take a "point estimate" to mean a very general sense of "things you estimate with statistics": unifying the concepts of means, proportions, rates, counts, hazards, differences, ratios, etc. as well as their respective methods: maximum likelihood, method of moments, regression, estimating equations, minimax, root finding algorithms, and so on.

This would seem to imply that that terminology is really only endemic to statistical theory and pedagogy. It would be totally out of place in the context of actually reporting statistics: you know the thing you estimated, why not tell us what it is?

The notation $\mu$ is highly suggestive of the population mean whose estimate $\hat{\mu}$ is usually the sample (arithmetic) mean and has a standard error estimated by the sample standard deviation divided by the root of the sample size. Confusingly, I have seen written $\hat{\mu} \pm \text{SE}$ as well as $\hat{\mu} \pm 1.96 \times \text{SE}$ in lieu of reporting an actual confidence interval, the former only really being appropriate if there is no set level of confidence or if it is in fact 66% and the latter only if the confidence is 95%. So alas, all I can definitively say is that the notation is ambiguous.