# How to calculate the 95% CI of the difference in population means?

Suppose I want to determine if treatment T affects parameter X, and by how much. I have a sample of subjects that received no treatment, and can calculate mean X(control) with associated variability. This allows me to estimate the true population mean of untreated population. The same goes for the treated population.

Now, if I want to know the magnitude of the effect I can calculate the difference between the means of two groups. However, this ignores the sampling variability - the true population means are represented by 95% CI around the sample mean. My question is, how to properly calculate the difference between population means? I would expect to obtain not a single number but rather again, a mean of the estimated difference and some measure of variability, like 95% CI.

Analogously, how do I determine the ratio (instead of the difference) of the population means?

• Is T dichotomous, polytomous, or continuous? Ratios of means (geometric means) can be calculated using log transformations. Jan 13, 2014 at 22:52

My question is, how to properly calculate the difference between population means?

Gosset asked himself the same question and came up with the t-test (or in this case the two sample t-test), which means that you are thinking about this right!

I would expect to obtain not a single number but rather again, a mean of the estimated difference and some measure of variability, like 95% CI.

And you obtain exactly that.

As the sample mean is a function of random variables, and has its own distribution, so is the difference of means. In case your two samples are large enough, the central limit theorem states that the sample mean has approximately normal distribution. Additionally, sums of independent Gaussians are also Gaussian distributed.

$\hat{x} \sim N(\mu_X,\frac{\sigma^2_X}{n_1})$

$\hat{x}_c \sim N(\mu_{X_c},\frac{\sigma^2_{X_c}}{n_2})$

$\hat{x} - \hat{x}_c \sim N(\mu_{X} - \mu_{X_c},\frac{\sigma^2_{X}}{n_1}+\frac{\sigma^2_{X_c}}{n_2})$

Observe that the difference of the sample means $\hat{x}$ for treatment group and $\hat{x}_c$ for control i. does have a Gaussian distribution, ii. its mean lies between the two population means $\mu_X$ and $\mu_{X_c}$, iii. it has larger variance than any of the sample means.

The statistical test mentioned above is based on calculating the probability of outcomes using exactly the above probability distribution.

As far as the ratio $\frac{\hat{x}}{\hat{x}_c}$ is concerned, it is a much more complicated story. Basically, except for very specific cases, the distribution is not Gaussian and the estimator is biased.