I want to investigate the power of t test in detecting shift so I have a sample $X_1, \dots,X_n \sim N(\mu_x,\sigma)$ and other sample $Y_1, \dots,Y_n \sim N(\mu_y,\sigma)$.

My null hypothesis is $H_0:\mu_Y-\mu_Y=0$ the alternative is $H_1:\mu_Y-\mu_Y=\delta$. Now I want to simulate these samples. I know it's a non-central t-distribution but I do not know how the non-centrality parameter relates to the difference in the means.

What's the distribution of the t-statistic for a two-sample test under the alternative hypothesis where the two samples are from Normal distributions differing only in their means?


1 Answer 1


Below is a geometric view of the t-test (a similar view is also expressed here).

The t-statistic, which is a ratio of the sample mean and the sample standard deviation, follows a ratio distribution, which is when properly scaled the t-distribution.

$$T \sim \frac{Z+\mu}{\sqrt{V_\nu/\nu}} \sim \text{t-distribution with $\nu$ degrees of freedom} $$

where $Z \sim N(0,1)$ is standard normal distributed, $V_\nu \sim \chi^2_\nu$ is $\chi^2$-distributed with $\nu$ degrees of freedom, and $\mu$ is a non-centrality parameter.


In the image above we plotted simulations sample $X$ and $Y$ with sample size $n=5$.

  • On the x-axis is the difference in the sample means. The distribution is $$\bar{y}-\bar{x} \sim N(\mu_Y-\mu_X, \frac{2}{n} \sigma^2)$$
  • On the y-axis is the pooled variance. The distribution is $$\sqrt{\sum_{i=1}^n (x_i-\bar{x})^2+\sum_{i=1}^n (y_i-\bar{y})^2} \sim \sigma \chi_{2n-2}$$ where $\sigma \chi_{2n-2}$ is a scaled $\chi_{2n-2}$ distribution.

Both the 'difference in the sample means' and the 'pooled variance' relate to the unknown deviation parameter $\sigma$ which is a nuisance parameter. But, their ratio does not depend on $\sigma$, and that is how we can perform a t-test. The ratio, the t-statistic (note we apply scaling to the numerator and denominator), follows a t-distribution.

$$\begin{array}{rcl}T &= & \frac{1/(\sqrt{2/n})}{1/\sqrt{2n-2}} \frac{\bar{y}-\bar{x}}{\sqrt{\sum_{i=1}^n (x_i-\bar{x})^2+\sum_{i=1}^n (y_i-\bar{y})^2}}\\ & \sim& \frac{Z}{\sqrt{V_\nu/\nu}} \\&\sim &\text{t-distribution with $\nu$ degrees of freedom} \hphantom{\text{and noncentrality parameter $\frac{\sqrt{2n-2}}{\sqrt{2/n}}$ }} \end{array}$$

In the lower image we showed how the distribution will look like when the difference $\mu_y-\mu_x \neq 0$ and would be shifted by 2 standard deviations. Then the distribution of the t-statistic would become

$$\begin{array}{rcl}T &= & \frac{1/(\sqrt{2/n})}{1/\sqrt{2n-2}} \frac{\bar{y}-\bar{x}+ c\sigma}{\sqrt{\sum_{i=1}^n (x_i-\bar{x})^2+\sum_{i=1}^n (y_i-\bar{y})^2}}\\ & \sim& \frac{Z+c \frac{1}{\sqrt{2/n}} }{\sqrt{V_\nu/\nu}} \\&\sim &\text{t-distribution with $\nu$ degrees of freedom and noncentrality parameter $\frac{1}{\sqrt{2/n}}$ } \end{array}$$

So the power for a 95% confidence interval can be expressed in terms of the shift in standard deviations:


You do not know the standard deviation, so you can not express the power in terms of the effect size in absolute terms, but you can express the power in terms of the relative effect size (relative to the standard deviation).

You can generalize this result for different sample sizes, but not for different populations variances (which is the Behrens Fisher problem)


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