Statistics [13]: t Test - Comparing Two Samples

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Published: January 13, 2021

$t$ test is usually used in comparing two groups of samples, including two independent samples and paired sampels.

Comparing Two Independent Samples

Assume $X_1,X_2,...,X_n$ is drawn from a normal distribution $X\sim N(\mu_X,\sigma_X^2)$ and that an independent sample $Y_1,Y_2,...,Y_m$ is drawn from a normal distribution $Y\sim N(\mu_Y,\sigma_Y^2)$ , compare the two samples.

For the general case Assume $\mu_1\neq \mu_2$ and $\sigma_1\neq \sigma_2$ , the problem remains unsolved. Thus, we further assume that $\sigma_1= \sigma_2$ . In this case,

$\bar{X} - \bar{Y} \sim N\left[\mu_x-\mu_Y, \sigma^2\left(\dfrac{1}{n} + \dfrac{1}{m}\right)\right]$

If $\sigma^2$ is known, a confidence level for $\mu_X - \mu_Y$ can be based on

$Z = \dfrac{(\bar{X} - \bar{Y}) - (\mu_X-\mu_Y)}{\sigma\sqrt{\dfrac{1}{n} + \dfrac{1}{m}}}$

which follows a standard distribution. The caonfidence level would be

$\bar{X}-\bar{Y}\pm u_{\alpha\text{/}2}\sigma\sqrt{\dfrac{1}{n}+\dfrac{1}{m}}$

Generally, $\sigma^2$ is not known and must estimated from the data by

$\s_p^2 = \dfrac{(n-1)s_X^2 + (m-1)s_Y^2}{m + n - 2}$

and it can be seen that

$\dfrac{(n-1)s_X^2 + (m-1)s_Y^2}{\sigma^2} \sim \chi^2(m + n - 2)$

Let

$U = \dfrac{(\bar{X} - \bar{Y}) - (\mu_X-\mu_Y)}{\sigma\sqrt{\dfrac{1}{n} + \dfrac{1}{m}}}\sim N(0,1)$

$V = \dfrac{(n-1)s_X^2 + (m-1)s_Y^2}{\sigma^2} \sim \chi^2(m + n - 2)$

The statistic

$t = \dfrac{U}{\sqrt{ V /(m + n - 2)}} = \dfrac{(\bar{X} - \bar{Y}) - (\mu_X-\mu_Y)}{s_p\sqrt{\dfrac{1}{n} + \dfrac{1}{m}}}$

follows a $t$ distribution with $m + n - 2$ degrees of freedom.

Therefore, the confidence level would be

$(\bar{X}-\bar{Y})\pm t_{\alpha\text{/}2}s_p\sqrt{\dfrac{1}{n} + \dfrac{1}{m}}$

Comparing Paired Samples

Assume that the differences are a sample from a normal distribution with

$E(D_i) = \mu_X - \mu_Y = \mu_D$

$var(D_i) = \sigma^2_D$

Generally, $\sigma_D^2$ is unknown, and inferences will be based on

$t = \dfrac{\bar{D} - \mu_D}{s_{\bar{D}}}$

which follows a $t$ distribution with $n-1$ degrees of freedom. The confidence level would be

$\bar{D} \pm t_{\alpha\text{/}2}s_{\bar{D}}$

If the sample size n is large, the approximate validity of the confidence interval and hypothesis test follows from the central limit theorem. If the sample size is small and the true distribution of the differences is far from normal, the stated probability levels may be considerably in error.

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Chao Huang

Statistics [13]: t Test - Comparing Two Samples

Comparing Two Independent Samples

Comparing Paired Samples

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