"Consider an estimator $T(X), X = \{X_1, X_2, ..., X_n\}$, for a parameter $μ$. For $T(X)$, it is given that, $P(3T(X) + μ > 8) = 0.025$ and $P(−3T(X) − μ < −2) = 0.975$. Calculate a $95\%$ confidence interval for $μ$, given that $T(X) = 2$."

I understand the basics behind CIs but I am very confused how to attempt this question because I am unsure how I can use $T(X)=2$ and the probabilities to attain a CI when there is no sample size, mean, or standard deviation to work with.

I am tearing my hair out over this so any help would be greatly appreciated. I would post my attempts but I haven't got very far... is $T(X) = 2$ the sample mean in this case? I am not sure if it is the notation that is confusing me or the question itself. Any help would be really great, thanks guys.

I understand that is customary to show at least an attempt at the question - but in this case I really do have no idea where to start. The issue lies with interpreting the probabilities, I have no idea what they actually represent - are they the upper and lower bound of the interval?

  • $\begingroup$ You don't need a sample size (in any case, what could you do with it?). $T(X)$ is not necessarily the mean, it's just some statistic, and it doesn't matter which statistic it is. The required information that it might lead you to is already in the question. You might start with asking yourself "What is a confidence interval?" If you modify your question to include information as discussed at the link I gave (what you've done, and where exactly your problems lie), more detailed help will be more likely to occur. $\endgroup$ – Glen_b Feb 23 '14 at 21:37
  • $\begingroup$ Thanks again for the response Glen_b, added some more information - wish I could give a better attempt! $\endgroup$ – Casev Feb 23 '14 at 21:57
  • $\begingroup$ Those probability statements are not the endpoints of the intervals, they're probability statements. You manipulate them to work out the endpoints of the intervals. Again, "What is a confidence interval?" $\endgroup$ – Glen_b Feb 23 '14 at 22:30
  • $\begingroup$ Okay so if T(X) = 2 then 3T(X) = 6, so we have a probability of 0.025 that 6+µ > 8 and a probability of 0.975 that -6-µ <-2, is this correct? and if so how does this help? Surely I need a probability so that the true mean lies within the interval 95% of the time, how can I manipulate these probabilities/inequalities to achieve this? $\endgroup$ – Casev Feb 23 '14 at 22:40
  • $\begingroup$ yes, I have managed to get that P(µ>8) = 0.025 and P(µ>-4) = 0.975, so now is there some way that I can manipulate the probabilities? This new information is helpful but how do I get some interval with probability 0.95? and also I will surely need an upper bound. i.e. p(µ<x) $\endgroup$ – Casev Feb 23 '14 at 23:05

(Since the OP has solved the question above, here's a brief answer.)

This is actually a very simple question because it gives you everything you need. You don't need a sample size, and $T(X)$ isn't necessarily the mean. It's really just about what a confidence interval is.

For $T(X)$, it is given that $P(3T(X)+μ>8)=0.025$ and $P(−3T(X)−μ<−2)=0.975$.

So $P(μ>2−3T(X))=0.975$ and $P(μ>8-3T(X))=0.025$

As a result, $P( 2−3T(X)<\mu<8-3T(X)) = 0.975-0.025 = 0.95$.

Calculate a 95% confidence interval for μ, given that T(X)=2."

A 95% CI for $\mu$ is given by $(2−3T(X),8-3T(X))$, so for the particular data at hand,

the 95% CI for $\mu$ is $(2−3\times 2,8-3\times 2)=(-4,2)$

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