We are given a 95% confidence interval with some range - in this particular case (4.0; 11.0) - and asked to find the 90% and 99% confidence intervals for the same parameter, assuming normal distribution, without any aditional information.
I am not sure how to solve this: interval = $\bar x \pm (t)(s_\bar x)$ So even if I am to get the new t-value based on the new percentage, I wouldn't know how to calculate the new intervals, since it's only multiplied by the standard error.. Maybe I'm missing something?
Based on your formulae $\bar{x} \pm t \times s_{\bar{x}}$, it seems like you are dealing with a $t$ confidence interval. If so, you will need the sample size $n$ to accurately get the 90% or 99% confidence interval.
But, just to clarify the thought process, suppose the 95% confidence interval given $(4, 11)$ is a $z$ confidence interval. If so, you can get the other confidence intervals based on the following logic (some of which also apply in the $t$ confidence interval scenario as well).
First, the mid-point of that interval is the sample mean $\bar{x}$ by design. So $\bar{x} = (4+11)/2 = 7.5$.
Second, the half-length of the interval is $1.96 \times \frac{\sigma}{\sqrt{n}}$.
So $\frac{\sigma}{\sqrt{n}} = \frac{(11-7.5)}{1.96}= \frac{3.5}{1.96}$.
Then, 90% confidence interval:
$$\bar{x} \pm 1.645 \times \frac{\sigma}{\sqrt{n}} \quad \Rightarrow \quad 7.5 \pm 1.645 \times \frac{3.5}{1.96} $$