Where n is the number of trials, and p is the probability that nofollow

[zihadhasan019]

Wins a trial, the normal approximation to the binomial distribution is: Where W is the number of wins, and z is the number of standard deviations above the mean, the formula for the number of wins is thus: The null hypothesis is that nofollow wins at p = 0.5 (even odds). To reject the null hypothesis in favor of the hypothesis that nofollow is an effective PageRank sculpting method with 95% confidence, we would need to see a minimum of W(n,0.


5, 1.645) wins. How many wins will we see? We australia business email lists are 95% sure to see at least W(n, p, -1.645), where p of the actual chance that nofollow win a given trial. If we are setting a lower bound of p = 5/8 = 0.625 for what we are trying to detect, then we have a lower bound of seeing W(n, 0.625, -1.645) wins (with 95% likelihood) if in fact nofollow is at least that much better.





We can set this lower bound on the number of wins we expect equal to the number of wins we need to see to have 95% confidence nofollow is better. After that we can then solve for the number of trials. So we conclude we need 168 trials. If this test fails to show nofollow is better, then we are 95% sure that nofollow wins trials less than 62.5% of the time. We wouldn't be able to say nofollow sculpting doesn't matter, but this does say it doesn't seem large in comparison to the other factors we were unable to control for in our experiment.