In this paper, we establish the following two congruences: \begin{gather*} \sum_{k=0}^{(p+1)/2}(3k-1)\frac{\left(-\frac{1}{2}\right)_k^2\left(\frac{1}{2}\right)_k4^k}{k!^3}\equiv p-6p^3\left(\frac{-1}{p}\right)+2p^3\left(\frac{-1}{p}\right)E_{p-3}\pmod{p^4},\\ \sum_{k=0}^{p-1}(3k-1)\frac{\left(-\frac{1}{2}\right)_k^2\left(\frac{1}{2}\right)_k4^k}{k!^3}\equiv p-2p^3\pmod{p^4}, \end{gather*} where $p>3$ is a prime, $E_{p-3}$ is the $(p-3)$-th Euler number and $\left(-\right)$ is the Legendre symbol. The first congruence modulo $p^3$ was conjectured by Guo and Schlosser recently.