# Sample size (CI 95%)

I have two groups (treatment and placebo) and for each group I have the 95% CI mean calculated as per here. I want to do the same for calculating the 95% sample size as per table 11 of this paper: they clearly write Sample size (95% CI), but I cannot find anywhere how to do it and which SD to use. Other mention the SD difference while others the pooled SD. This formula gives the sample size for a given effect size which is fairly straightforward to calculate in Python as it solves the formula for n.

def get_n(mean_diff, sd_diff):
std_effect_size = mean_diff / sd_diff

n = tt_ind_solve_power(effect_size=std_effect_size, alpha=0.05, power=0.8, ratio=1, alternative='two-sided')
print('Number in *each* group: {:.5f}'.format(n))


P.S: I really do net get the negative votes. It's a niche question that is very Pythonic and statistic at the same time.

• I didn't down-vote the question, but I confess that amongst the two multi-click links and the Python code, I am not exactly sure what you're asking. Perhaps my Answer is useful. – BruceET Mar 26 '19 at 14:23

If you have a sample of size $$n$$ from a normal distribution with unknown mean $$\mu$$ and standard deviation $$\sigma,$$ then a 95% confidence interval for $$\mu$$ is of the form $$\bar X \pm t^*S/\sqrt{n},$$ where $$\bar X$$ is the sample mean, $$S$$ is the sample standard deviation, and $$t*$$ cuts probability $$0.25$$ from the upper tail of the symmetrical Student t distribution with $$n-1$$ degrees of freedom.
If $$n > 30,$$ then $$t^* \approx 2.0.$$ (This 'rule of 30' works only for 95% CIs.)
In this situation, the margin of error (half-width) of the CI is $$M = 2S/\sqrt{n}.$$ If you are able to make a reasonable guess for the value of $$\sigma$$ and want the sample size $$n$$ necessary for a particular margin of error $$M,$$ then
$$n \approx 4(\sigma/M)^2.$$
• If the sample sizes and variances are equal for the two groups, then the approximate margin of error is $M = 2\sigma\sqrt{2/n}$ and the CI is $D \pm M,$ where $D = \bar X_1 - \bar X_2.$ From that you can solve for $n$ in terms of $M,$ etc. // Perhaps you would do better to frame this as a test of hypothesis and do a computation for $n$ required to give reasonable power against a particular difference in means. – BruceET Mar 26 '19 at 14:37