Finding a t test statistic from a gamma distribution Suppose that $T_1,…,T_{10}$ are iid $Exp(λ)$ and the goal is to test if $H_0:λ≤1$ versus $H_a:λ≥2$. Suppose that the test statistic is $S=\sum_{i=1}^{10}T_i$, and rejection of the null occurs when $S≤7$.

I realize that $T_1...T_{10}$ is a gamma distribution  of $S =Gamma(10, \lambda$),
but I am not sure how to use this information to find the p-value for the type 1 error.
 A: You can't compute a $p$-value because you have no data.  You can compute the significance level $\alpha$ corresponding to the rejection criterion $S \le 7$.  These two things are different.
To be specific, the $\alpha$ level is the maximum Type I error for the test, meaning that the probability of (incorrectly) rejecting the null hypothesis when it is true is at most $\alpha$.  So if your criterion for rejection is $S \le 7$, and the null is $H_0 : \lambda \le 1$, then in the worst-case* scenario, the true value of $\lambda$ under the null hypothesis is $\lambda = 1$, and you want to compute $$\alpha = \sup_{\lambda_0 \le 1} \Pr[\text{reject } H_0 \mid \lambda = \lambda_0] = \Pr[S \le 7 \mid \lambda = 1].$$  Since you know $S \mid \lambda = 1$ is gamma distributed with shape $10$ and rate $\lambda$, this probability is an integral of the gamma density function; i.e. it is simply a cumulative distribution function evaluated at $7$: $$\alpha = \Pr[S \le 7 \mid \lambda = 1] = F_S(7) = \int_{s=0}^7 \frac{\lambda^{10} s^9 e^{-\lambda s}}{\Gamma(10)} \, ds = \frac{1}{9!} \int_{s=0}^7 s^9 e^{-s} \, ds.$$

Note.  Why do I say "worst-case" scenario above and why is it $\lambda = 1$?  The reason is because as $\lambda$ is a rate parameter, larger values will result in higher probabilities for the exponentially distributed $X_i$s to be smaller.  This in turn means $S$ will have a larger probability of being small, and so the rejection criterion is easier to observe.  This is consistent with the way the hypothesis test is structured:  we reject $H_0$ in favor of $H_a : \lambda \ge 2$ if $S$ is "sufficiently small," where in this case, we have quantified "sufficiently small" as meaning $S \le 7$.  So the Type I error of the test is greatest when the true value of the parameter $\lambda$ is as large as possible under the null, namely $\lambda = 1$.  This is the "worst-case" scenario, and again makes intuitive sense because it is also the value that is "closest" to the alternative hypothesis $\lambda \ge 2$.
To summarize:  large $\lambda$ = small $S$ = easier rejection of $H_0$ = higher chance of Type I error.
A: You reject when $S \le 7$ where, under $H_0; \lambda \le 1,$
$S \sim \mathsf{Gamma}(10, 1),$ so the probability of Type I Error
(rejecting $H_0$ when $H_0$ is true) is
$\alpha = P(S \le 7\, |\, H_0) = 0.1695.$ [Computation in R, where
pgamma is a gamma CDF.]
pgamma(7, 10, 1)
[1] 0.1695041

Then $\beta = P(x \ge 7\, |\, H_a) = 1 - P(S \le 7\, |\, H_a) =  0.109,$
whee $H_a: \lambda \ge 2.$
1 - pgamma(7, 10, 2)
[1] 0.1093994

In the figure below $\alpha$ is the area under the blue density curve to the
left of $s = 7$ and $\beta$ is the area under the brown density curve to the right
of $s=7.$

R code for figure:
hdr = "Density functions of GAMMA(10,1) [null, blue] and GAMMA(10,2)"
curve(dgamma(x, 10, 1), 0, 20, ylim=c(0,.3), col="blue", lwd=2,
      ylab="Density", xlab="s", main=hdr)
  curve(dgamma(x, 10, 2), add=T, col="brown", lwd=2)
  abline(v=0, col="green2");  abline(h=0, col="green2")
  abline(v=7, col="red", lwd=2, lty="dashed")

Note: See Note in @Heropup's (+1) Answer.
