Non-standard choices for confidence intervals

[VantagePoint72]

Suppose I have a normally distributed random variable with unknown mean μ and known standard deviation σ. Based on n samples of the RV, I'd like to compute an estimate of μ and a confidence interval at confidence level P%.

The usual way of doing this would be to use the standard error in the mean: if my sample mean is $\bar{X}$ and P is the total Gaussian probability between standard scores -Z and Z, then a confidence interval $(\bar{X}- Z\sigma/\sqrt{n}, \bar{X}+ Z\sigma/\sqrt{n})$ does the job. That is, if I repeat this sampling process many times then I can expect about P% of these confidence intervals to cover μ.

Am I "allowed" to choose a different procedure for constructing confidence intervals? Suppose my chosen confidence level is 30% and $\sigma/\sqrt{n} = 1$. There is 30% (or close enough) probability that a given trial will yield a sample mean between $\mu$ and $\mu + 0.85$, which I computed with the Gaussian CDF. Thus, the interval $(\bar{X} - 0.85, \bar{X})$ will cover μ about 30% of the times I do this n-sample experiment. So, it satisfies the requirement to be a 30% confidence interval.

Is there anything I'm missing about the definition of a confidence interval that doesn't allow me to do this construction? Obviously, the symmetric version based on the the standard error in the mean has nicer properties and does a better job of capturing the reliability of the estimator. If I don't care about that, though, am I free to make this slightly perverse construction instead?

 

[Office_Shredder]

Yes, if someone just says "I want a confidence interval" then any interval will technically do. It is often assumed however when people say confidence interval and are talking about Gaussian random variables that they are talking about the interval which is symmetric about the sample mean.

 

[Hornbein]

Suppose I have a normally distributed random variable with unknown mean μ and known standard deviation σ. Based on n samples of the RV, I'd like to compute an estimate of μ and a confidence interval at confidence level P%.

The usual way of doing this would be to use the standard error in the mean: if my sample mean is $\bar{X}$ and P is the total Gaussian probability between standard scores -Z and Z, then a confidence interval $(\bar{X}- Z\sigma/\sqrt{n}, \bar{X}+ Z\sigma/\sqrt{n})$ does the job. That is, if I repeat this sampling process many times then I can expect about P% of these confidence intervals to cover μ.

Am I "allowed" to choose a different procedure for constructing confidence intervals? Suppose my chosen confidence level is 30% and $\sigma/\sqrt{n} = 1$. There is 30% (or close enough) probability that a given trial will yield a sample mean between $\mu$ and $\mu + 0.85$, which I computed with the Gaussian CDF. Thus, the interval $(\bar{X} - 0.85, \bar{X})$ will cover μ about 30% of the times I do this n-sample experiment. So, it satisfies the requirement to be a 30% confidence interval.

Is there anything I'm missing about the definition of a confidence interval that doesn't allow me to do this construction? Obviously, the symmetric version based on the the standard error in the mean has nicer properties and does a better job of capturing the reliability of the estimator. If I don't care about that, though, am I free to make this slightly perverse construction instead?

You may make a confidence interval any way you like. Usually the standard method is most useful, but if you have a good reason to due things otherwise then go ahead.

 

[Hornbein]

Yes, if someone just says "I want a confidence interval" then any interval will technically do. It is often assumed however when people say confidence interval and are talking about Gaussian random variables that they are talking about the interval which is symmetric about the sample mean.

I once taught statistics, and one of the main points was one-sided vs. two-sided confidence intervals. It depends on what hypothesis you are testing.