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\[ T = \frac{\Delta}{\sqrt{Var(\Delta)}} \]
Let’s expand on the previous relationship:
\[ T = \frac{ \bar{X}_t - \bar{X_c}}{\sqrt{Var(\bar{X}_t) + Var(\bar{X}_c) + 2COV(\bar{X}_t, \bar{X}_c)}} \] Since the two samples are independent, the covariance of the means is zero. The variance of the mean is:
\[ Var(X_t) = Var(\frac{X_1+X_2+X_{n_t}}{n_t}) = \frac{1}{n_t^2} \bigg[ Var(X_1) + Var(X_2) + ... \bigg] = \frac{n_t}{n_t^2} s_t^2 = \frac{s_t^2}{n_t} \]
where we assumed that all samples are independent and come from the same distribution, with empirical variance \(s_t^2\). Similarly, we can obtain the same result for the control group. Hence, our t-test can be written as:
\[ T = \frac{ \bar{X}_t - \bar{X_c}}{\sqrt{\frac{s_t^2}{n_t}+ \frac{s_c^2}{n_c}}} \]
This is an AI-enhanced solution that took as input the original solution.
A two-sample t-test is a statistical procedure used to determine whether there is a significant difference between the means of two independent groups. Consider two samples, \(X_t, X_c\); for the sake of clarity, let's refer to them as treatment and control samples. The difference in their means is expressed as: \(\Delta = \bar{X}_t - \bar{X_c}\). The formula for the two-sample t-test is presented as follows (referencing Kohavi, Tang, and Xu, 2020):
\[ T = \frac{\Delta}{\sqrt{Var(\Delta)}} \]
We can further expand this relationship to:
\[ T = \frac{ \bar{X}_t - \bar{X_c}}{\sqrt{Var(\bar{X}_t) + Var(\bar{X}_c) + 2COV(\bar{X}_t, \bar{X}_c)}} \]
Given the two samples are independent, the covariance of the means is zero. The variance of the mean is computed as:
\[ Var(X_t) = Var\left(\frac{X_1+X_2+X_{n_t}}{n_t}\right) = \frac{1}{n_t^2} \left[Var(X_1) + Var(X_2) + \ldots \right] = \frac{n_t}{n_t^2} s_t^2 = \frac{s_t^2}{n_t} \]
We assume here that all sample values are independent and identically distributed, each with an empirical variance \(s_t^2\). A similar analysis applies to the control group, yielding the following expression for the t-test:
\[ T = \frac{ \bar{X}_t - \bar{X_c}}{\sqrt{\frac{s_t^2}{n_t}+ \frac{s_c^2}{n_c}}} \]
Hypothesis testing