My Statistics Text asks the following:
Consider a $t$-distribution $T$ with 8 degrees of freedom
$(a)$ Find $t_{0.025}$ so that $P(−t_{0.025} ≤ T ≤ t_{0.025}) = 0.95$. I found this to be $t=2.306004$
Now this is where I am confused
$(b)$ Solve the inequality $[−t_{0.025} ≤ T ≤ t_{0.025}]$ so that $μ$ is in the middle.
What does the author mean by "in the middle". Any insights much appreciated.
This is very vague as it stands, but from context it's relatively clear what they must be asking.
Presumably, $T$ is defined as $$ T = \frac{\sqrt{n}(\bar x-\mu)}{\hat{\sigma}}$$ where $\bar x$ and $\hat \sigma$ are the sample mean and standard deviation. So you can rewrite the inequality as $$ -t_{\alpha} < \frac{\sqrt{n}(\bar x-\mu)}{\hat{\sigma}} < t_\alpha$$ where $\alpha = .025.$
By 'put $\mu$ in the middle' they mean to rewrite this as and inequality of the form $a<\mu To do this, look at the inequalities separately. $$ \frac{\sqrt{n}(\bar x-\mu)}{\hat{\sigma}} < t_\alpha$$ can be solved to get $$\mu >\bar x-\frac{t_\alpha\hat\sigma}{\sqrt{n}}$$ and the other one gives $$\mu <\bar x+\frac{t_\alpha\hat\sigma}{\sqrt{n}}$$. So the inequality is $$\bar x-\frac{t_\alpha\hat\sigma}{\sqrt{n}}<\mu <\bar x+\frac{t_\alpha\hat\sigma}{\sqrt{n}}.$$ (This gives the $1-2\alpha$ confidence interval for $\mu.$)