python compute sample size for one-sample Z-test

Question

I'm trying to see if there's an easy way to calculate minimum sample size required for a one-sample Z-test to reject the null hypothesis.

I know that we can reject the null hypothesis (i.e. the A/B test is successful) if

1 - scipy.stats.norm.cdf((x-mu)/(s/np.sqrt(n)) < alpha

where x is the sample mean, mu is the population mean, s is the population standard deviation and n the sample size.

Is there a way in python to solve the equation above for n?

To clarify, are you saying you want to vary n while keeping x fixed? — myrtlecat
– myrtlecat, Commented Jan 17, 2022 at 16:56
Ok thanks. One further clarification: you refer to a t-test but your equation implies your test statistic is normally distributed which would make this a Z-test not a t-test. I assume the equation is correct and you meant to say Z-test, but can you double check? — myrtlecat
– myrtlecat, Commented Jan 17, 2022 at 17:11

myrtlecat · Accepted Answer · 2022-01-17 17:49:30Z

1

Distributions in scipy.stats have an inverse of the cdf function, which is called ppf. ppf stands for "percentage point function" but this is a misnomer because it actually deals with quantiles, not percentiles. We can use this function and the fact that ppf(cdf(x)) = x to rearrange and solve your equation.

Assuming that alpha < 0.5, there are two cases to cases to consider. If x <= mu then there are no solutions, otherwise we can rearrange the equation to:

np.sqrt(n) > s * scipy.stats.norm.ppf(1 - alpha) / (x - mu)

An aside:

If alpha >= 0.5 then the solutions are strange: either any n will do (if x >= mu) or you get an upper bound on n (if x < mu). But that's what you get for treating your null hypothesis with such contempt!

edited Jan 17, 2022 at 17:49

answered Jan 17, 2022 at 17:27

myrtlecat

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