Interval Estimation Question

Hi! This is my first topic here.
I have the following doubt:
When you construct a confidence interval for estimate the mean of a population, for example with a 95% of confidence, you can say that 95% of the samples with the same size, will contain the population mean within the interval.
So if you take 100 samples of the same size, 95 of them will contain the mean within the estimation interval. Let’s select one of that 100 samples, then I have 0.95 probability of having selected a sample whose estimation interval will contain the mean of the population.
So, if i’m right, why can’t we say that there is 95% of probability that the mean is in the interval?
Thanks!!
Regards,
Alter

Offhand, I’d guess the answer depends on what the distribution is. Your proposed answer may only apply to a uniform distribution.

If this relates to a question in the course, please identify it specifically.

Thanks for yor response!
I copy the transcript of the class 7 of lesson 1, week 4: ‘’ Saying that you’re 95% confident has to do with repeating the sampling experiment many times and calculating the intervals for each sample estimate. 95% of the time those confidence intervals will contain the mean. This confidence level has to do with the success rate of contracting the confidence interval.
It is not the probability that a specific interval contains the population mean because as we’ve seen, the population mean is either on the interval or not.‘’

My doubt is, if 95 percent of the times the interval contains the population mean, why can’t I say that there is 95 % of probabilty that the mean is within the interval that I calculate with a random sample. It has to be with the population distribution?

I understand the concept, I just want to be sure why I can’t say, when I calculate a confidence interval, that there is a 95% probability that the mean lies in this interval.

Thanks again!

Sorry, I can’t give a more definite answer. Probability is not my best subject.

Maybe someone else will answer.

Hello @Alter_Caimi

This is wrong. It should be “95 percent of THOSE INTERVALS contain the population mean”.

I am splitting the transcript into steps:

Step 1: repeating the sampling experiment many times
Step 2: calculating the intervals for each sample estimate
Step 3: 95% of the time those confidence intervals will contain the mean.

The problem with your if statement is that the words were twisted from “those confidence intervals” to “the confidence interval” where the latter deviates from the original meaning and is thus wrong.

When I learnt probability, I had also wanted to rephrase the statement to make it more intuitive and easy to digest, but anything deviates from the 3-step is not going to work.

Cheers,
Raymond

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You may try to rephrase it again, but please inspect carefully whether your new statement is consistent with the 3-step. Of course you may also show me it if you are not 100% sure. :wink:

This is all you are going to get within the frequentist framework.

Hi @Alter_Caimi

A possibility just crossed my mind and I was wondering if you might be actually looking for something like this?

“If we draw a future sample, there is a 95% probability that its 95% confident interval covers the population mean.”

Note that, if I have a sample in my hand, everything about it is certain. It has a certain 95% confident interval, and it is either covering the population mean or not. Your two statements I quoted above seem to be such case to me.

It’s like after you roll a dice :game_die: and get a “six”, then it is 100% a “six”. However, if you do many experiments, and find that 16.67% the dice gave a “six”, then you may say, if I am to roll the dice again, there is a 16.67% that I will get a “six”.

It might sound too picky with words, but wording is in question, isn’t it?

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Hi Raymond!
In fact, wording is in question, especially considering my English :rofl:
Thank you very much for your explaining!
I now understand this subtlety in the frequentist interpretation.
Regards!

You are welcome, @Alter_Caimi! :wink:

Cheers,
Raymond