confidence intervalprobabilitystatistics
Given a normally distributed population with mean $\mu$ and say I constructed a confidence interval of $\mu \pm a$ with a confidence level of 95%. Is the following statement correct?
There is a 0.95 probability/chance that the population mean is captured within this confidence interval?
This is a common error in the interpretation of a confidence interval. Consider the explanation from the following site: http://onlinestatbook.com/2/estimation/confidence.html
"Confidence intervals for means are intervals constructed using a procedure that will contain the population mean a specified proportion of the time, typically either 95% or 99% of the time.
It is natural to interpret a 95% confidence interval as an interval with a 0.95 probability of containing the population mean. However, the proper interpretation is not that simple. One problem is that the computation of a confidence interval does not take into account any other information you might have about the value of the population mean. For example, if numerous prior studies had all found sample means above 110, it would not make sense to conclude that there is a 0.95 probability that the population mean is between 72.85 and 107.15. What about situations in which there is no prior information about the value of the population mean? Even here the interpretation is complex. The problem is that there can be more than one procedure that produces intervals that contain the population parameter 95% of the time. Which procedure produces the "true" 95% confidence interval? Although the various methods are equal from a purely mathematical point of view, the standard method of computing confidence intervals has two desirable properties: each interval is symmetric about the point estimate and each interval is contiguous."
While the article you refer to correctly defines the concept of confidence interval (your highlighted text) it does not correctly treat the case of a normal distribution with unknown standard deviation. You may want to search "Neyman confidence interval" to see an approach that produces confidence intervals with the property you highlighted.
The Neyman procedure selects a region containing 95% of outcomes, for each true value of the parameter of interest. The confidence interval is the union of all parameter values for which the observation is within the selected region. The probability for the observation to be within the selected region for the true parameter value is 95%, and only for those observations, will the confidence interval contain the true value. Therefore the procedure guarantees the property you highlight.
If the standard deviation is known and not a function of the mean, the Neyman central confidence intervals turn out to be identical to those described in the article.
Thank you for the link to Neyman's book - interesting to read from the original source! You ask for a simple description, but that is what my second paragraph was meant to be. Perhaps a few examples will help illustrate: Example 1 and 1b could be considered trivial, whereas 2 would not be handled correctly by the article you refer to.
Example 1. Uniform random variable. Let X follow a uniform distribution, $$f(x)=1/2 {\mathrm{\ \ for\ \ }}\theta-1\le x\le \theta+1 $$ and zero otherwise. We can make a 100% confidence interval for $\theta$ by considering all possible outcomes $x$, given $\theta$, ie. $x \in [\theta-1,\theta+1]$. Now consider an observed value, $x_0$. The union of all possible values of $\theta$ for which $x_0$ is a possible outcome is $[x_0-1,x_0+1]$. That is the 100% confidence interval for $\theta$ for this problem.
Example 1b. Uniform random variable. Let X follow the same uniform distribution. We can make a 95% central confidence interval for $\theta$ by selecting the 95% central outcomes $x$, given $\theta$, ie. $x \in [\theta-0.95,\theta+0.95]$. Now consider an observed value, $x_0$. The union of all possible values of $\theta$ for which $x_0$ is within the selected range is $[x_0-0.95,x_0+0.95]$. That is the 95% confidence interval for $\theta$ for this problem.
Example 2. Uniform random variable. Let X follow a uniform distribution, $$f(x)=1/\theta {\mathrm{\ \ for\ \ }}{1\over2}\theta \le x \le {3\over2}\theta $$ and zero otherwise. We can make a 100% confidence interval for $\theta$ by considering all possible outcomes $x$, given $\theta$, ie. $x \in [{1\over2}\theta,{3\over2}\theta]$. Now consider an observed value, $x_0$. The union of all possible values of $\theta$ for which $x_0$ is a possible outcome is $[{2\over3}x_0,2x_0]$. That is the 100% confidence interval for $\theta$ for this problem. (You can confirm this by inserting the endpoints of the confidence interval into the pdf and see they are at the boundaries of the pdf). Note that the central confidence interval is not centered on the point estimate for $\theta$, $\hat\theta = x_0$.
Example 3. Normal distribution with mean $\theta$ and standard deviation $1$. The 68% central confidence interval would be constructed identically to example 1, that is the selected region for $X$ would be $[\theta-1,\theta+1]$. The 68% central confidence interval is therefore the same as in Example 1, $[x_0-1,x_0+1]$. You can extend this to 95% and arbitrary KNOWN standard deviation $\sigma$ to be $[x_0-1.96\sigma,x_0+1.96\sigma]$.
Example 4. Normal distribution with mean $\theta$ and standard deviation $\theta/2$. The 68% central confidence interval would be constructed identically to example 2. The 68% central confidence interval for $\theta$ is therefore the same as in Example 2, $[{2\over3}x_0,2x_0]$.
The authors of the article you refer to and the other commenters to your question would not get Example 2 or 4 right. Only following a procedure like Neyman's will the confidence interval have the property that you highlighted in your post. The other methods are approximations for the general problem of building confidence intervals.
The exact solution to the problem with a normal distribution and UNKNOWN standard deviation is more difficult to work out than the examples above.
Best Answer
In the sense of the Bayesian interpretation of probability, that is actually correct. But it is easy to be misled by that if you don't also understand how frequentists look at it.
In the sense of the frequentist interpretation of probability, it is not correct. Because from the frequentist point of view, you construct one confidence interval, and now the mean is either in that interval or it isn't. Where is the repeatable experiment? The repeatable experiment is the construction of the confidence interval itself. That is, from the frequentist point of view, we "have 95% confidence that the population mean is in the confidence interval" because if we were to take many samples and construct many confidence intervals, 95% of the samples would produce 95% confidence intervals which overlap with $\mu$.