If the value of the null hypothesis is outside the 95% confidence interval can it be rejected at the 95% level? To illustrate: I want to test that the relationship between two variables is directly proportional. So I want to test if the coefficient, β, of log(X) on log(Y) is 1. 
I know that my 95% confidence interval for the coefficient for β is 0.81, 0.85.
I can therefore confidently reject the null. But is this the same as failing a 95% t-test?
To clarify, I know that my calculated value of β does fail a t-test at 95% confidence, I'm interested in whether or not saying that my null falls outside the 95% interval is mathematically/statistically equivalent to failing the t-test and why.
Many thanks in advance.
 A: For a coefficient in a regression model, yes, if the $1-\alpha$ interval doesn't contain the value of the parameter under the null hypothesis, that same null value would be rejected by a test at significance level $\alpha$.
Indeed, both procedures are based on the same pivotal quantity
In the case of the hypothesis test, the statistic is $T=\frac{\hat{\beta}-\beta_0}{s_\hat{\beta}}$, which is rejected when $|T|\geq t_{1-\alpha/2}$ (at the relevant degrees of freedom; n-2 for simple linear regression).
The $1-\alpha$ confidence interval has bounds of $\hat\beta\pm s_\hat{\beta}\cdot t_{1-\alpha/2}$. 
If $\beta_0$ is outside the interval $\hat\beta\pm s_\hat{\beta}\cdot t_{1-\alpha/2}$, that's equivalent to saying $|\hat{\beta}-\beta_0|\geq {s_\hat{\beta}}\cdot t_{1-\alpha/2}$.
Now let's show that rejecting in the hypothesis test is the same:
$|T|\geq t_{1-\alpha/2}\implies |\frac{\hat{\beta}-\beta_0}{s_\hat{\beta}}|\geq t_{1-\alpha/2}$
$\hspace{2.55cm}\implies|\hat{\beta}-\beta_0|\geq s_\hat{\beta}\cdot t_{1-\alpha/2}$

(Not all tests have this property, but many do. There's an explicit example of one that doesn't in comments here)
Note that both inferential procedures rely on the same set of assumptions; if the assumptions don't hold (or reasonably nearly so), the actual type I error rate and the actual coverage probability might not be very close to their nominal values.
